A Sabbath-rest for the people of God

我们既蒙留下，有进入他安息的应许，就当畏惧，免得我们（原文是你们） 中间或有人似乎是赶不上了。 (希伯来书 4:1 和合本)
Therefore, since the promise of entering his rest still stands, let us be careful that none of you be found to have fallen short of it. (Hebrews 4:1 NIV)
因为有福音传给我们，像传给他们一样；只是所听见的道与他们无益，因为他们没有信心与所听见的道调和。 (希伯来书 4:2 和合本)
For we also have had the good news proclaimed to us, just as they did; but the message they heard was of no value to them, because they did not share the faith of those who obeyed. (Hebrews 4:2 NIV)
但我们已经相信的人得以进入那安息，正如　神所说："我在怒中起誓说：'他们断不可进入我的安息！'"其实造物之工，从创世以来已经成全了。 (希伯来书 4:3 和合本)
Now we who have believed enter that rest, just as God has said, "So I declared on oath in my anger, 'They shall never enter my rest.' " And yet his works have been finished since the creation of the world. (Hebrews 4:3 NIV)
论到第七日，有一处说："到第七日，　神就歇了他一切的工。" (希伯来书 4:4 和合本)
For somewhere he has spoken about the seventh day in these words: "On the seventh day God rested from all his works." (Hebrews 4:4 NIV)
又有一处说："他们断不可进入我的安息！" (希伯来书 4:5 和合本)
And again in the passage above he says, "They shall never enter my rest." (Hebrews 4:5 NIV)
既有必进安息的人，那先前听见福音的，因为不信从，不得进去。 (希伯来书 4:6 和合本)
Therefore since it still remains for some to enter that rest, and since those who formerly had the good news proclaimed to them did not go in because of their disobedience, (Hebrews 4:6 NIV)
所以过了多年，就在大卫的书上，又限定一日，如以上所引的说："你们今日若听他的话，就不可硬着心。" (希伯来书 4:7 和合本)
God again set a certain day, calling it "Today." This he did when a long time later he spoke through David, as in the passage already quoted: "Today, if you hear his voice, do not harden your hearts." (Hebrews 4:7 NIV)
若是约书亚已叫他们享了安息，后来　神就不再提别的日子了。 (希伯来书 4:8 和合本)
For if Joshua had given them rest, God would not have spoken later about another day. (Hebrews 4:8 NIV)
这样看来，必另有一安息日的安息为　神的子民存留。 (希伯来书 4:9 和合本)
There remains, then, a Sabbath-rest for the people of God; (Hebrews 4:9 NIV)

Photo by Natalie Toombs / Unsplash

bring good news

那报佳音，传平安，报好信，传救恩的，对锡安说：你的　神作王了！这人的脚登山何等佳美！ (以赛亚书 52:7 和合本)
How beautiful on the mountains are the feet of those who bring good news, who proclaim peace, who bring good tidings, who proclaim salvation, who say to Zion, "Your God reigns!" (Isaiah 52:7 NIV)

Photo by Flo Dahm / Unsplash

Correct usage of tzset(), TZ env

https://knowledgebase.progress.com/articles/Article/000044163
TZ variable can be set on on for a session, user profile or use it in the Database start-up script to start the database for the specific timezone or country or city.

Following are some examples can be tested in a terminal session of UNIX.

For China is UTC+8:00 or GMT+8:00

TZ needs to set it with opposite sign as follows:

TZ=UTC-8:00; export TZ
or,
TZ=GMT-8:00; export TZ
or,
TZ=Asia/Shanghai;export TZ

setenv("TZ", "GMT-8:00", 1) tzset(); 

For US EST (GMT-5:00, New York):

TZ=UTC+5:00; export TZ
Or,
TZ=GMT+5:00; export TZ
Or,
TZ=America/New_York;export TZ

setenv("TZ", "UTC+5:00", 1) tzset(); 

Trapping Rain Water

Question:

http://www.lintcode.com/en/problem/trapping-rain-water/
Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

Example

Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

Answer

RunTime : 1272ms

class Solution {
public:
 /*
 * @param heights: a list of integers
 * @return: a integer
 */
 int trapRainWater(vector<int> &heights) {
  // write your code here
  if (heights.size() < 3)
   return 0;
  int result = 0;
  size_t l, r;
  l = 0;
  r = heights.size() - 1;
  while (l < r) {
   while (l < r && heights[l] == 0) l++;
   while (l < r && heights[r] == 0) r--;
   int min = heights[l] < heights[r] ? 

                                  heights[l] : heights[r];
   for (size_t i = l; i <= r; i++) {
    if (heights[i] >= min)
     heights[i] -= min;
    else {
     result += min - heights[i];
     heights[i] = 0;
    }
   }
  }
  return result;
 }
};

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RIGHT ACTIONS FROM WRONG MOTIVES

"你们祷告的时候，不可像那假冒为善的人，爱站在会堂里和十字路口上祷告，故意叫人看见。我实在告诉你们，他们已经得了他们的赏赐。 (马太福音 6:5 和合本)
"And when you pray, do not be like the hypocrites, for they love to pray standing in the synagogues and on the street corners to be seen by others. Truly I tell you, they have received their reward in full. (Matthew 6:5 NIV)
你祷告的时候，要进你的内屋，关上门，祷告你在暗中的父；你父在暗中察看，必然报答你。 (马太福音 6:6 和合本)
But when you pray, go into your room, close the door and pray to your Father, who is unseen. Then your Father, who sees what is done in secret, will reward you. (Matthew 6:6 NIV)
你们祷告，不可像外邦人，用许多重复话，他们以为话多了必蒙垂听。 (马太福音 6:7 和合本)
And when you pray, do not keep on babbling like pagans, for they think they will be heard because of their many words. (Matthew 6:7 NIV)
你们不可效法他们；因为你们没有祈求以先，你们所需用的，你们的父早已知道了。 (马太福音 6:8 和合本)
Do not be like them, for your Father knows what you need before you ask him. (Matthew 6:8 NIV)
所以，你们祷告要这样说：我们在天上的父：愿人都尊你的名为圣。 (马太福音 6:9 和合本)
"This, then, is how you should pray: " 'Our Father in heaven, hallowed be your name, (Matthew 6:9 NIV)
愿你的国降临；愿你的旨意行在地上，如同行在天上。 (马太福音 6:10 和合本)
your kingdom come, your will be done, on earth as it is in heaven. (Matthew 6:10 NIV)
我们日用的饮食，今日赐给我们。 (马太福音 6:11 和合本)
Give us today our daily bread. (Matthew 6:11 NIV)
免我们的债，如同我们免了人的债。 (马太福音 6:12 和合本)
And forgive us our debts, as we also have forgiven our debtors. (Matthew 6:12 NIV)
不叫我们遇见试探；救我们脱离凶恶（或译：脱离恶者） 。因为国度、权柄、荣耀，全是你的，直到永远。阿们（有古卷没有因为…阿们等字） ！ (马太福音 6:13 和合本)
And lead us not into temptation, but deliver us from the evil one. ' (Matthew 6:13 NIV)

Photo by Tom Barrett / Unsplash

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AT THE BETHESDA POOL IN JERUSALEM

这事以后，到了犹太人的一个节期，耶稣就上耶路撒冷去。 (约翰福音 5:1 和合本)
Some time later, Jesus went up to Jerusalem for one of the Jewish festivals. (John 5:1 NIV)
在耶路撒冷，靠近羊门有一个池子，希伯来话叫作毕士大，旁边有五个廊子； (约翰福音 5:2 和合本)
Now there is in Jerusalem near the Sheep Gate a pool, which in Aramaic is called Bethesda and which is surrounded by five covered colonnades. (John 5:2 NIV)
里面躺着瞎眼的、瘸腿的、血气枯干的许多病人。（有古卷加：等候水动；4因为有天使按时下池子搅动那水，水动之后，谁先下去，无论害什么病就痊愈了。） (约翰福音 5:3 和合本)
Here a great number of disabled people used to lie—the blind, the lame, the paralyzed.
(John 5:3 NIV)
(John 5:4 NIV)
在那里有一个人，病了三十八年。 (约翰福音 5:5 和合本)
One who was there had been an invalid for thirty-eight years. (John 5:5 NIV)
耶稣看见他躺着，知道他病了许久，就问他说："你要痊愈吗？" (约翰福音 5:6 和合本)
When Jesus saw him lying there and learned that he had been in this condition for a long time, he asked him, "Do you want to get well?" (John 5:6 NIV)
病人回答说："先生，水动的时候，没有人把我放在池子里；我正去的时候，就有别人比我先下去。" (约翰福音 5:7 和合本)
"Sir," the invalid replied, "I have no one to help me into the pool when the water is stirred. While I am trying to get in, someone else goes down ahead of me." (John 5:7 NIV)
耶稣对他说："起来，拿你的褥子走吧！" (约翰福音 5:8 和合本)
Then Jesus said to him, "Get up! Pick up your mat and walk." (John 5:8 NIV)
那人立刻痊愈，就拿起褥子来走了。 (约翰福音 5:9 和合本)
At once the man was cured; he picked up his mat and walked. The day on which this took place was a Sabbath, (John 5:9 NIV)
那天是安息日，所以犹太人对那医好的人说："今天是安息日，你拿褥子是不可的。" (约翰福音 5:10 和合本)
and so the Jewish leaders said to the man who had been healed, "It is the Sabbath; the law forbids you to carry your mat." (John 5:10 NIV)
他却回答说："那使我痊愈的，对我说：'拿你的褥子走吧。'" (约翰福音 5:11 和合本)
But he replied, "The man who made me well said to me, 'Pick up your mat and walk.'" (John 5:11 NIV)
他们问他说："对你说'拿褥子走'的是什么人？" (约翰福音 5:12 和合本)
So they asked him, "Who is this fellow who told you to pick it up and walk?" (John 5:12 NIV)
那医好的人不知道是谁；因为那里的人多，耶稣已经躲开了。 (约翰福音 5:13 和合本)
The man who was healed had no idea who it was, for Jesus had slipped away into the crowd that was there. (John 5:13 NIV)
后来耶稣在殿里遇见他，对他说："你已经痊愈了，不要再犯罪，恐怕你遭遇的更加利害。" (约翰福音 5:14 和合本)
Later Jesus found him at the temple and said to him, "See, you are well again. Stop sinning or something worse may happen to you." (John 5:14 NIV)

Photo by James Donovan / Unsplash

Word Break III

Question

http://www.lintcode.com/en/problem/word-break-iii/
Give a dictionary of words and a sentence with all whitespace removed, return the number of sentences you can form by inserting whitespaces to the sentence so that each word can be found in the dictionary.

Example

Given a String CatMat
Given a dictionary ["Cat", "Mat", "Ca", "tM", "at", "C", "Dog", "og", "Do"]
return 3
we can form 3 sentences, as follows:
CatMat = Cat Mat
CatMat = Ca tM at
CatMat = C at Mat

Answer

RunTime : 288ms

class Solution {
public:
 /*
 * @param s: A string
 * @param dict: A dictionary of words dict
 * @return: A boolean
 */
 int wordBreak3(string &s, unordered_set<string> &dict) {
  // write your code here
  if (s.length() <= 0)
   return 0;
  bool found = true;
  for (auto c : s) {
   bool in = false;
   for (auto d : dict) {
    if (in)
     break;
    for (auto e : d) {
     if (e == c) {
      in = true;
      break;
     }
    }
   }
   if (!in) {
    found = false;
    break;
   }
  }
  if (!found)
   return 0;
  Trie trie;
  for (auto d : dict)
   trie.insert(d);
  result = 0;
  wordBreakHelp(s, trie, 0);
  return result;
 }
private:
 int result;
 struct TrieNode {
  TrieNode() {
   for (int i = 0; i < 256; i++) {
    next[i] = 0;
   }
   s = 0;
  }
  int s;
  TrieNode* next[256];
 };
 class Trie {
 public:
  Trie() {
   root = new TrieNode();
  }
  inline void deleteHelper(TrieNode* root) {
   if (root) {
    for (int i = 0; i < 256; i++) {
     deleteHelper(root->next[i]);
    }
    delete root;
   }
  }
  ~Trie() {
   deleteHelper(root);
   root = 0;
  }
  void insert(string& s) {
   TrieNode* iter = root;
   for (auto c : s) {
    if (iter->next[c] == 0)
     iter->next[c] = new TrieNode();
    iter = iter->next[c];
   }
   iter->s = s.length();
  }
  unordered_set<int> search(string& s, int pos) {
   TrieNode* iter = root;
   unordered_set<int> words;
   for (size_t i = pos; i < s.length(); i++) {
    char c = s[i];
    if (iter->next[c] == 0)
     break;
    if (iter->s > 0) {
     words.insert(iter->s);
    }
    iter = iter->next[c];
   }
   if (iter->s > 0) {
    words.insert(iter->s);
   }
   return words;
  }
 private:
  TrieNode* root;
 };
 inline void wordBreakHelp(string& s, Trie& trie, int pos) {
  unordered_set<int> words = trie.search(s, pos);
  if (words.size() > 0) {
   for (auto word : words) {
    if ((pos + word) == s.length()) {
     result++;
    }
    else {
     wordBreakHelp(s, trie, pos + word);
    }
   }
  }
 }
};

Word Break II

Question

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.
Return all such possible sentences.

Example

Gieve s = lintcode,
dict = ["de", "ding", "co", "code", "lint"].
A solution is ["lint code", "lint co de"].

Answer

RunTime : 902ms

class Solution {
public:
 /*
 * @param s: A string
 * @param dict: A dictionary of words dict
 * @return: A boolean
 */
 vector<string> wordBreak(string &s, unordered_set<string> &dict) {
  // write your code here
  result.clear();
  if (s.length() <= 0)
   return result;
  bool found = true;
  for (auto c : s) {
   bool in = false;
   for (auto d : dict) {
    if (in)
     break;
    for (auto e : d) {
     if (e == c) {
      in = true;
      break;
     }
    }
   }
   if (!in) {
    found = false;
    break;
   }
  }
  if (!found)
   return result;
  Trie trie;
  for (auto d : dict)
   trie.insert(d);
  string bw;
  wordBreakHelp(s, trie, 0, bw);
  return result;
 }
private:
 vector<string> result;
 struct TrieNode {
  TrieNode() {
   for (int i = 0; i < 26; i++) {
    next[i] = 0;
   }
   s = 0;
  }
  int s;
  TrieNode* next[26];
 };
 class Trie {
 public:
  Trie() {
   root = new TrieNode();
  }
  inline void deleteHelper(TrieNode* root) {
   if (root) {
    for (int i = 0; i < 26; i++) {
     deleteHelper(root->next[i]);
    }
    delete root;
   }
  }
  ~Trie() {
   deleteHelper(root);
   root = 0;
  }
  void insert(string& s) {
   TrieNode* iter = root;
   for (auto c : s) {
    if (iter->next[c - 'a'] == 0)
     iter->next[c - 'a'] = new TrieNode();
    iter = iter->next[c - 'a'];
   }
   iter->s = s.length();
  }
  unordered_set<int> search(string& s, int pos) {
   TrieNode* iter = root;
   unordered_set<int> words;
   for (size_t i = pos; i < s.length(); i++) {
    char c = s[i];
    if (iter->next[c - 'a'] == 0)
     break;
    if (iter->s > 0) {
     words.insert(iter->s);
    }
    iter = iter->next[c - 'a'];
   }
   if (iter->s > 0) {
    words.insert(iter->s);
   }
   return words;
  }
 private:
  TrieNode* root;
 };
 inline void wordBreakHelp(string& s, Trie& trie, int pos, string& bw) {
  unordered_set<int> words = trie.search(s, pos);
  if (words.size() > 0) {
   for (auto word : words) {
    string bwd = bw;
    if (bwd.empty())
     bwd = s.substr(pos, word);
    else {
     bwd.append(" ");
     bwd.append(s.substr(pos, word));
    }
    if ((pos + word) == s.length()) {
     result.push_back(bwd);
    }
    else {
     wordBreakHelp(s, trie, pos + word, bwd);
    }
   }
  }
 }
};

Word Break

Question

http://www.lintcode.com/en/problem/word-break/
Given a string s and a dictionary of words dict, determine if s can be break into a space-separated sequence of one or more dictionary words.

Example

Given s = "lintcode", dict = ["lint", "code"].
Return true because "lintcode" can be break as "lint code".

Answer

Run Time : 582ms 

class Solution {
public:
 /*
 * @param s: A string
 * @param dict: A dictionary of words dict
 * @return: A boolean
 */

 bool wordBreak(string &s, unordered_set<string> &dict) {
  // write your code here
  if (dict.size() <= 0) {
   if (s.length() <= 0)
    return true;
   return false;
  }
  Trie trie;
  for (auto d : dict)
   trie.insert(d);
  return wordBreakHelp(s, trie, 0);
 }
private:
 struct TrieNode {
  TrieNode() {
   for (int i = 0; i < 26; i++) {
    next[i] = 0;    
   }
   s = 0;
  }
  int s;
  TrieNode* next[26];
 };
 class Trie {
 public:
  Trie() {
   root = new TrieNode();
  }
  inline void deleteHelper(TrieNode* root) {
   if (root) {
    for (int i = 0; i < 26; i++) {
     deleteHelper(root->next[i]);
    }
    delete root;
   }
  }
  ~Trie() {
   deleteHelper(root);
   root = 0;
  }
  void insert(string& s) {
   TrieNode* iter = root;
   for (auto c : s) {
    if (iter->next[c - 'a'] == 0)
     iter->next[c - 'a'] = new TrieNode();
    iter = iter->next[c - 'a'];
   }
   iter->s = s.length();
  }
  unordered_set<int> search(string& s, int pos) {
   TrieNode* iter = root;
   unordered_set<int> words;
   for (int i = pos; i < s.length(); i++) {
    char c = s[i];
    if (iter->next[c - 'a'] == 0)
     break;
    if (iter->s > 0) {
     words.insert(iter->s);
    }
    iter = iter->next[c - 'a'];
   }
   if (iter->s > 0) {
    words.insert(iter->s);
   }
   return words;
  }
 private:
  TrieNode* root;
 };
 inline bool wordBreakHelp(string& s, Trie& trie, int pos) {
  unordered_set<int> words = trie.search(s, pos);
  if (s.length() > 0 && words.size() > 0) {
   for (auto word : words) {
    if ((pos + word) == s.length())
     return true;    
    if (wordBreakHelp(s, trie, pos+word))
     return true;
   }
  }
  return false;
 }
}; 

Lord's will

其实明天如何，你们还不知道。你们的生命是什么呢？你们原来是一片云雾，出现少时就不见了。 (雅各书 4:14 和合本)
Why, you do not even know what will happen tomorrow. What is your life? You are a mist that appears for a little while and then vanishes. (James 4:14 NIV)
你们只当说："主若愿意，我们就可以活着，也可以做这事，或做那事。" (雅各书 4:15 和合本)
Instead, you ought to say, "If it is the Lord's will, we will live and do this or that." (James 4:15 NIV)

Photo by James Chou / Unsplash

Unique Paths III

Question

http://www.lintcode.com/en/problem/unique-paths-iii/
Follow up for "Unique Paths II": https://acm.errong.win/uniquepathsii/
Now each grid contains a value, so each path also has a value. Find the sum of all the unique values paths.

Example

For example,
[
[1,1,2],
[1,2,3],
[3,2,4]
]
There are 2 unique value path:
[1,1,2,3,4] = 11
[1,1,2,2,4] = 10
return 21

Answer

class Solution {
public:
 /*
 * @param : an array of arrays
 * @return: the sum of all unique weighted paths
 */
 int uniqueWeightedPaths(vector<vector<int>> &obstacleGrid) {
  int m = obstacleGrid.size();
  if (m <= 0)
   return 0;
  int n = obstacleGrid[0].size();
  if (n <= 0)
   return 0;
  struct Up {
   int step;
   int sum;
   Up(int st, int su) : step(st), sum(su) {}
  };
  map<int, vector<Up>> sum_paths;
  sum_paths[0] = vector<Up>();
  sum_paths[0].push_back(Up(0, obstacleGrid[0][0]));
  for (int i = 1; i < m + n - 1; i++) {
   map<int, vector<Up>> new_sum_paths;
   for (auto sp : sum_paths) {
    int pos = sp.first;
    int x = pos / n;
    int y = pos % n;
    int nx = x + 1;
    int ny = y;    
    if (nx < m) {
     int npos = nx * n + ny;
     for (auto up : sp.second) {
      if (up.step < i) {
       Up nup = up;
       nup.sum += obstacleGrid[nx][ny];
       nup.step += 1;
       bool found = false;
       if (!new_sum_paths.count(npos)) {
        new_sum_paths[npos] = vector<Up>();
       }
       for (auto eup : new_sum_paths[npos]) {
        if (eup.sum == nup.sum) {
         found = true;
         break;
        }
       }
       if (!found)
        new_sum_paths[npos].push_back(nup);
      }
     }
    }
    nx = x;
    ny = y + 1;
    if (ny < n) {
     int npos = nx * n + ny;
     for (auto up : sp.second) {
      if (up.step < i) {
       Up nup = up;
       nup.sum += obstacleGrid[nx][ny];
       nup.step += 1;
       bool found = false;
       if (!new_sum_paths.count(npos)) {
        new_sum_paths[npos] = vector<Up>();
       }
       for (auto eup : new_sum_paths[npos]) {
        if (eup.sum == nup.sum) {
         found = true;
         break;
        }
       }
       if (!found)
        new_sum_paths[npos].push_back(nup);
      }
     }
    }
   }
   sum_paths = new_sum_paths;
  }
   int sum = 0;
  int pos = m*n - 1;
  for (auto sp : sum_paths[pos]) {
   sum += sp.sum;
  }
  return sum;
 }
};

JESUS THE SOUL WINNER

在那里有雅各井；耶稣因走路困乏，就坐在井旁。那时约有午正。 (约翰福音 4:6 和合本)
Jacob's well was there, and Jesus, tired as he was from the journey, sat down by the well. It was about noon. (John 4:6 NIV)
有一个撒玛利亚的妇人来打水。耶稣对她说："请你给我水喝。"（ (约翰福音 4:7 和合本)
When a Samaritan woman came to draw water, Jesus said to her, "Will you give me a drink?" (John 4:7 NIV)
那时门徒进城买食物去了。） (约翰福音 4:8 和合本)
(His disciples had gone into the town to buy food.) (John 4:8 NIV)
撒玛利亚的妇人对他说："你既是犹太人，怎么向我一个撒玛利亚妇人要水喝呢？"原来犹太人和撒玛利亚人没有来往。 (约翰福音 4:9 和合本)
The Samaritan woman said to him, "You are a Jew and I am a Samaritan woman. How can you ask me for a drink?" (For Jews do not associate with Samaritans. ) (John 4:9 NIV)
耶稣回答说："你若知道　神的恩赐，和对你说'给我水喝'的是谁，你必早求他，他也必早给了你活水。" (约翰福音 4:10 和合本)
Jesus answered her, "If you knew the gift of God and who it is that asks you for a drink, you would have asked him and he would have given you living water." (John 4:10 NIV)

Photo by Gerrie van der Walt / Unsplash

Unique Paths II

Question

http://www.lintcode.com/en/problem/unique-paths-ii/
Follow up for "Unique Paths":
Now consider if some obstacles are added to the grids. How many unique paths would there be?
An obstacle and empty space is marked as 1 and 0 respectively in the grid.

Example

For example,
There is one obstacle in the middle of a 3x3 grid as illustrated below.
[
[0,0,0],
[0,1,0],
[0,0,0]
]
The total number of unique paths is 2.

Answer

class Solution {
public:
    /*
     * @param obstacleGrid: A list of lists of integers
     * @return: An integer
     */
    int uniquePathsWithObstacles(vector<vector<int>> &obstacleGrid) {
        int m = obstacleGrid.size();
        if (m <= 0)
            return 0;
        int n = obstacleGrid[0].size();
        if (obstacleGrid[0][0] == 1 || obstacleGrid[m-1][n-1] == 1)
            return 0;
        vector<vector<int>> res(m, vector<int>(n, 0));
        res[0][0] = 1;
        for (int j = 1; j < n; j++) {
             if (obstacleGrid[0][j] == 1) {
                 for (int i = j; i < n; i++)
                      res[0][i] = 0;
                 break;
             } else
                 res[0][j] = 1;
        }
        for (int j = 1; j < m; j++) {
             if (obstacleGrid[j][0] == 1) {
                 for (int i = j; i < m; i++)
                      res[i][0] = 0;
                 break;
             } else {
                 res[j][0] = 1;
             }
        }
        for (int i = 1; i < m; i++) {
             for (int j = 1; j < n; j++) {
                 if (obstacleGrid[i][j] == 1)
                     res[i][j] = 0;
                 else
                     res[i][j] = res[i-1][j] + res[i][j-1];
             }
        }
        return res[m-1][n-1];
    }
};

Unique Paths

Question

http://www.lintcode.com/en/problem/unique-paths/
A robot is located at the top-left corner of a m x n grid.
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid.
How many possible unique paths are there?

Example

Given m = 3 and n = 3, return 6.
Given m = 4 and n = 5, return 35.
Given m = 6 end = 63, return 9657648

Answer

// brute force solution, Time Limit Exceeded
// help us get dp formula pattern
// res[i][j] = res[i-1][j] + res[i][j-1];
public int uniquePaths(int m, int n) {
    return helper(1, 1, m, n);
}
private int helper(int row, int col, int m, int n)
{
    if(row == m && col == n)
        return 1;  
    if( row > m || col > n)
        return 0;
    return helper(row+1, col, m, n) + helper(row, col+1, m, n);
}

// dp solution
class Solution {
public:
    /*
     * @param m: positive integer (1 <= m <= 100)
     * @param n: positive integer (1 <= n <= 100)
     * @return: An integer
     */
    int uniquePaths(int m, int n) {
        // write your code here
        if (m <= 0 || n <= 0)
            return 0;
        vector<int> res(n);
        res[0] = 1;
        for (int i = 0; i < m; i++) {
             for (int j = 1; j < n; j++) {
                  res[j] += res[j-1];
             }
        }
        return res[n-1];
    }
};

Next Sparse Number

Question

http://www.lintcode.com/en/problem/next-sparse-number/
A number is Sparse if there are no two adjacent 1s in its binary representation. Given a number n, find the smallest Sparse number which greater than or equal to n.
eg. 5 (binary representation: 101) is sparse, but 6 (binary representation: 110) is not sparse.

Example

Given n = 6, return 8
Next Sparse Number is 8

Given n = 4, return 4
Next Sparse Number is 4

Given n = 38, return 40
Next Sparse Number is 40

Given n = 44, return 64
Next Sparse Number is 64

Given n = 341381939, return 343932928
Next Sparse Number is 343932928

Answer

class Solution {
public:
    /*
     * @param : a number
     * @return: return the next sparse number behind x
     */
    int nextSparseNum(int x) {
        // write your code here
        if (x == 0 || x == 1) return x;
        string binary;
        int n = x;
        while (n) {
            if (n&1)
                binary.append("1");
            else
                binary.append("0");
            n >>= 1;
        }
        for (int i = 1; i < binary.length(); i++) {
            if (binary[i] == '1' && binary[i-1] == '1') {
                // plus 1 from i-1
                int c = 1;
                int j = i-1;
                while (c == 1 && j < binary.length()) {
                    int t = binary[j] - '0' + c;
                    if (t < 2) {
                        c = 0;
                        binary[j] = t + '0';
                    } else {
                        c = 1;
                        binary[j] = '0';
                    }
                    j++;
                }
                if (c == 1)
                    binary.append("1");
            }
        }
        int result = 0;
        for (int i = 0; i < binary.length() - 1; i++) {
            if (binary[i] == '1') {
                string nb = binary;
                nb[i] = '0';
                int nx = bintoint(nb);
                if (nx >= x) {
                    binary[i] = '0';
                    if (result == 0)
                        result = nx;
                    else if (result > nx)
                        result = nx;
                }
            }
        }
        if (result == 0)
            result = bintoint(binary);
        return result;
    }
    inline int bintoint(string& binary) {
       int result = 0;
       for (int i = binary.length(); i > 0; i--)
            result += (binary[i-1] - '0') << (i-1);
       return result;
    }
};

How to verify your ownership of the web server powered by Ghost via Search Console

Since I want to do Verify my ownership of http://www.errong.win/ via Search Console.

I tried to upload the file google9aedadfa5be8a815.html under my site's working directory where I installed via ghost instal command.
but only get 404 error while visiting http://www.errong.win/google9aedadfa5be8a815.html

Solution

Thanks to Nistor Cristian, kevin, vikas.potluri, they replied my thread quickly in ghost.slack.org and I got the issue fixed.

Nistor Cristian
simply include it in your theme and should work
kevin
@Errong add the file to the root of your theme
vikas.potluri
If you're not sure which theme to use, go to the ghost admin panel, download the current theme, drop in your google verification and upload it back to the admin interface

For my case:
I use the default casper theme, its path is :
blog/content/themes/casper/
I moved google9aedadfa5be8a815.html there and then it works.
You do not even need to restart your ghost(ghost restart).

You can use below command to check it.
curl https://www.errong.win/google9aedadfa5be8a815.html
google-site-verification: google9aedadfa5be8a815.html

CHRIST’S DEATH MEANS OUR LIFE

有一个法利赛人，名叫尼哥底母，是犹太人的官。 (约翰福音 3:1 和合本)
Now there was a Pharisee, a man named Nicodemus who was a member of the Jewish ruling council. (John 3:1 NIV)
这人夜里来见耶稣，说："拉比，我们知道你是由　神那里来作师傅的；因为你所行的神迹，若没有　神同在，无人能行。" (约翰福音 3:2 和合本)
He came to Jesus at night and said, "Rabbi, we know that you are a teacher who has come from God. For no one could perform the signs you are doing if God were not with him." (John 3:2 NIV)
耶稣回答说："我实实在在地告诉你，人若不重生，就不能见　神的国。" (约翰福音 3:3 和合本)
Jesus replied, "Very truly I tell you, no one can see the kingdom of God unless they are born again. " (John 3:3 NIV)
尼哥底母说："人已经老了，如何能重生呢？岂能再进母腹生出来吗？" (约翰福音 3:4 和合本)
"How can someone be born when they are old?" Nicodemus asked. "Surely they cannot enter a second time into their mother's womb to be born!" (John 3:4 NIV)
耶稣说："我实实在在地告诉你，人若不是从水和圣灵生的，就不能进　神的国。 (约翰福音 3:5 和合本)
Jesus answered, "Very truly I tell you, no one can enter the kingdom of God unless they are born of water and the Spirit. (John 3:5 NIV)
从肉身生的就是肉身；从灵生的就是灵。 (约翰福音 3:6 和合本)
Flesh gives birth to flesh, but the Spirit gives birth to spirit. (John 3:6 NIV)
我说：'你们必须重生'，你不要以为希奇。 (约翰福音 3:7 和合本)
You should not be surprised at my saying, 'You must be born again.' (John 3:7 NIV)
风随着意思吹，你听见风的响声，却不晓得从哪里来，往哪里去；凡从圣灵生的，也是如此。" (约翰福音 3:8 和合本)
The wind blows wherever it pleases. You hear its sound, but you cannot tell where it comes from or where it is going. So it is with everyone born of the Spirit." (John 3:8 NIV)
尼哥底母问他说："怎能有这事呢？" (约翰福音 3:9 和合本)
"How can this be?" Nicodemus asked. (John 3:9 NIV)
耶稣回答说："你是以色列人的先生，还不明白这事吗？ (约翰福音 3:10 和合本)
"You are Israel's teacher,"said Jesus, "and do you not understand these things? (John 3:10 NIV)
我实实在在地告诉你，我们所说的是我们知道的；我们所见证的是我们见过的；你们却不领受我们的见证。 (约翰福音 3:11 和合本)
Very truly I tell you, we speak of what we know, and we testify to what we have seen, but still you people do not accept our testimony. (John 3:11 NIV)
我对你们说地上的事，你们尚且不信，若说天上的事，如何能信呢？ (约翰福音 3:12 和合本)
I have spoken to you of earthly things and you do not believe; how then will you believe if I speak of heavenly things? (John 3:12 NIV)
除了从天降下、仍旧在天的人子，没有人升过天。 (约翰福音 3:13 和合本)
No one has ever gone into heaven except the one who came from heaven—the Son of Man. (John 3:13 NIV)
摩西在旷野怎样举蛇，人子也必照样被举起来， (约翰福音 3:14 和合本)
Just as Moses lifted up the snake in the wilderness, so the Son of Man must be lifted up, (John 3:14 NIV)
叫一切信他的都得永生（或译：叫一切信的人在他里面得永生） 。 (约翰福音 3:15 和合本)
that everyone who believes may have eternal life in him." (John 3:15 NIV)
"　神爱世人，甚至将他的独生子赐给他们，叫一切信他的，不至灭亡，反得永生。 (约翰福音 3:16 和合本)
For God so loved the world that he gave his one and only Son, that whoever believes in him shall not perish but have eternal life. (John 3:16 NIV)

Photo by diego / Unsplash

Maximum Subarray VI

Question

Given an array of integers. find the maximum XOR subarray value in given array.
What's the XOR: https://en.wikipedia.org/wiki/Exclusive_or

Notice

Expected time complexity O(n).

Answer

https://threads-iiith.quora.com/Tutorial-on-Trie-and-example-problems
Let's say F(L,R) is XOR of subarray from L to R.
Here we use the property that F(L,R)=F(1,R) XOR F(1,L-1). How?
Let's say our subarray with maximum XOR ended at position i. Now, we need to maximise F(L,i) ie. F(1,i) XOR F(1,L-1) where L<=i. Suppose, we inserted F(1,L-1) in our trie for all L<=i, then it's just problem1.

class Solution {
public:
    /*
     * @param : the array
     * @return: the max xor sum of the subarray in a given array
     */    
    int maxXorSubarray(vector<int> &nums) {
        // write code here
        int ans = INT_MIN;
        int pre = 0;
        Trie trie;
        for (auto n : nums) {
            pre = pre ^ n;
            trie.insert(pre);
            ans = max(ans, trie.query(pre));
        }
        return ans;
    }
private:    
    class Trie {
    public:
        Trie() {
            root = new TrieNode(0);
            INTBITS = sizeof(int) * 8;
        }
        void insert(int x) {
            TrieNode* iter = root;
            for (int i = INTBITS - 1; i >= 0; i--) {
                bool v = x & (1 << i);
                if (iter->arr[v] == 0)
                    iter->arr[v] = new TrieNode(0);
                iter = iter->arr[v];
            }
            iter->val = x;
        }
        int query(int x) {
            TrieNode* iter = root;
            for (int i = INTBITS - 1; i >= 0; i--) {
                bool v = x & (1 << i);
                if (iter->arr[1-v] != 0)
                    iter = iter->arr[1-v];
                else if (iter->arr[v] != 0)
                    iter = iter->arr[v];
            }
            return x ^ iter->val;
        }
    private:
        class TrieNode {
        public:
            int val;
            TrieNode *arr[2];
            TrieNode(int v)
                : val(v) {
                arr[0] = arr[1] = 0;
            }
        };
        TrieNode* root;
        int INTBITS;
    };    
};

Maximum Subarray III

Question:

http://www.lintcode.com/en/problem/maximum-subarray-iii/
Given an array of integers and a number k, find k non-overlapping subarrays which have the largest sum.
The number in each subarray should be contiguous.
Return the largest sum.

Answer

localMax[i][j], means the maximum sum we can get while select j subarrays 

from first i elements, including element i 

globalMax[i][j], means the maximum sum we can get while select j subarrays 

from first i elements, may not including element i 

// state transfer 

localMax[i][j] = max(localMax[i - 1][j] + nums[i - 1], globalMax[i - 1][j - 1] + nums[i - 1]) 

globalMax[i][j] = max(globalMax[i - 1][j], localMax[i][j]) 

class Solution {
public:
    /*
     * @param nums: A list of integers
     * @param k: An integer denote to find k non-overlapping subarrays
     * @return: An integer denote the sum of max k non-overlapping subarrays
     */
    int maxSubArray(vector<int> &nums, int k) {
        // write your code here
        int n = nums.size();
        if (n == 0 || k > n)
            return 0;
        // Use INT_MIN might cause numeric overflow 
        vector<vector<int>> localMax(n+1, vector<int>(k+1, INT_MIN/2));
        vector<vector<int>> globalMax(n+1, vector<int>(k+1, INT_MIN/2));

        localMax[0][0] = globalMax[0][0] = 0;

        for (int i = 1; i <= n; i++) {
             localMax[i][0] = 0;
             globalMax[i][0] = 0;
             for (int j = 1; j <= k; j++) {
                  localMax[i][j] = max(localMax[i-1][j] + nums[i-1], 
                                       globalMax[i-1][j-1] + nums[i-1]);
                  globalMax[i][j] = max(globalMax[i-1][j], localMax[i][j]);
             }
        }
        return globalMax[n][k];
    }
};

Maximum Subarray II

Question

Given an array of integers, find two non-overlapping subarrays which have the largest sum.
The number in each subarray should be contiguous.
Return the largest sum.

Notice

The subarray should contain at least one number

Answer

class Solution {
public:
    /*
     * @param nums: A list of integers
     * @return: An integer denotes the sum of max two non-overlapping subarrays
     */
    int maxTwoSubArrays(vector<int> &nums) {
        // write your code here
        int len = nums.size();
        vector<int> suml(len);
        int cur_max = nums[0];
        int max = cur_max;
        suml[0] = max;
        for (int i = 1; i < len; i++) {
            if (cur_max <= 0) {
                cur_max = nums[i];
            } else {
                cur_max += nums[i];
            }
            if (max < cur_max)
                max = cur_max;
            suml[i] = max;
        }
        vector<int> sumr(len);
        cur_max = nums[len-1];
        max = cur_max;
        sumr[len-1] = max;
        for (int i = len - 2; i >= 0; i--) {
             if (cur_max <= 0) {
                 cur_max = nums[i];
             } else {
                 cur_max += nums[i];
             }
             if (max < cur_max)
                 max = cur_max;
             sumr[i] = max;
        }
        int result = suml[0] + sumr[1];
        for (int i = 1; i < len - 1; i++) {
             int temp = suml[i] + sumr[i+1];
             if (result < temp)
                 result = temp;
        }
        return result;
    }
};

Look at the new growth in the valley

那向外观看、如晨光发现、美丽如月亮、皎洁如日头、威武如展开旌旗军队的是谁呢？ (雅歌 6:10 和合本)
Who is this that appears like the dawn, fair as the moon, bright as the sun, majestic as the stars in procession? He (Song of Songs 6:10 NIV)
我下入核桃园，要看谷中青绿的植物，要看葡萄发芽没有，石榴开花没有。 (雅歌 6:11 和合本)
I went down to the grove of nut trees to look at the new growth in the valley, to see if the vines had budded or the pomegranates were in bloom. (Song of Songs 6:11 NIV)
不知不觉，我的心将我安置在我尊长的车中。〔耶路撒冷的众女子〕 (雅歌 6:12 和合本)
Before I realized it, my desire set me among the royal chariots of my people. Friends (Song of Songs 6:12 NIV)

Photo by 贝莉儿 NG / Unsplash

Candy

Question:

http://www.lintcode.com/en/problem/candy/
There are N children standing in a line. Each child is assigned a rating value.
You are giving candies to these children subjected to the following requirements:

Each child must have at least one candy. 

Children with a higher rating get more candies than their neighbors. 

What is the minimum candies you must give?

Answer:

class Solution {
public:
    /*
     * @param ratings: Children's ratings
     * @return: the minimum candies you must give
     */
    int candy(vector<int> &ratings) {
        // write your code here
        vector<int> candys(ratings.size(), 1);
        for (int i = 0; i < ratings.size() - 1; i++) {
            if (ratings[i+1] > ratings[i])
                candys[i+1] = candys[i] + 1;
        }
        for (int i = ratings.size() - 1; i > 0; i--) {
            if (ratings[i-1] > ratings[i])
                candys[i-1] = max(candys[i] + 1, candys[i-1]);
        }
        int count = 0;
        for (auto i : candys)
            count += i;
        return count;
    }
};

Course Schedule III

Question:

http://www.lintcode.com/en/problem/course-schedule-iii/
There are ·n· different online courses numbered from 1 to n. Each course has some duration(course length) t and closed on dth day. A course should be taken continuously for t days and must be finished before or on the dth day. You will start at the 1st day.
Given n online courses represented by pairs (t,d), your task is to find the maximal number of courses that can be taken.

Notice

The integer 1 <= d, t, n <= 10,000. You can't take two courses simultaneously. 

Example

Given [[100, 200], [200, 1300], [1000, 1250], [2000, 3200]]
return 3
There're totally 4 courses, but you can take 3 courses at most:
First, take the 1st course, it costs 100 days so you will finish it on the 100th day, and ready to take the next course on the 101st day.
Second, take the 3rd course, it costs 1000 days so you will finish it on the 1100th day, and ready to take the next course on the 1101st day.
Third, take the 2nd course, it costs 200 days so you will finish it on the 1300th day.
The 4th course cannot be taken now, since you will finish it on the 3300th day, which exceeds the closed date.

Answer:

class Solution {
public:
    /*
     * @param : duration and close day of each course
     * @return: the maximal number of courses that can be taken
     */
    int scheduleCourse(vector<vector<int>> &courses) {
        // write your code here
        int curTime = 0;
        priority_queue<int> q;
        sort(courses.begin(), courses.end(), [](vector<int>& a, vector<int>& b) {return a[1] < b[1];});
        for (auto course : courses) {
            curTime += course[0];
            q.push(course[0]);
            if (curTime > course[1]) {
                curTime -= q.top(); q.pop();
            }
        }
        return q.size();
    }
};

Course Schedule II

Question:

http://www.lintcode.com/en/problem/course-schedule-ii/
There are a total of n courses you have to take, labeled from 0 to n - 1.
Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]
Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses.
There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

Example

Given n = 2, prerequisites = [[1,0]]
Return [0,1]
Given n = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
Return [0,1,2,3] or [0,2,1,3]

Answer:

class Solution {
public:
    /*
     * @param numCourses: a total of n courses
     * @param prerequisites: a list of prerequisite pairs
     * @return: the course order
     */
    vector<int> findOrder(int numCourses, vector<pair<int, int>> &prerequisites) {
        // write your code here
        vector<vector<int>> adj(numCourses);
        vector<int> indegree(numCourses);
        for (int i = 0; i < numCourses; i++)
             indegree[i] = 0;
        for (auto pq : prerequisites) {
            adj[pq.second].push_back(pq.first);
            indegree[pq.first]++;
        }
        stack<int> st;
        for (int i = 0; i < numCourses; i++) {
            if (indegree[i] == 0)
                st.push(i);
        }
        int count = 0;
        vector<int> result;
        while (!st.empty()) {
            int v1 = st.top();
            st.pop();
            count++;
            result.push_back(v1);
            for (auto v2 : adj[v1]) {
                indegree[v2]--;
                if (indegree[v2] == 0)
                    st.push(v2);
            }
        }
        if (count != numCourses) {
            return vector<int>();
        }
        return result;
    }
};

Topological Sorting

Question:

http://www.lintcode.com/en/problem/topological-sorting/
Given an directed graph, a topological order of the graph nodes is defined as follow:
For each directed edge A -> B in graph, A must before B in the order list.
The first node in the order can be any node in the graph with no nodes direct to it.
Find any topological order for the given graph.

Answer:

class Solution {
public:
    /*
     * @param ratings: Children's ratings
     * @return: the minimum candies you must give
     */
    int candy(vector<int> &ratings) {
        // write your code here
        vector<int> candys(ratings.size(), 1);
        for (int i = 0; i < ratings.size() - 1; i++) {
            if (ratings[i+1] > ratings[i])
                candys[i+1] = candys[i] + 1;
        }
        for (int i = ratings.size() - 1; i > 0; i--) {
            if (ratings[i-1] > ratings[i])
                candys[i-1] = max(candys[i] + 1, candys[i-1]);
        }
        int count = 0;
        for (auto i : candys)
            count += i;
        return count;
    }
};

Course Schedule

Question:

http://www.lintcode.com/en/problem/course-schedule/
There are a total of n courses you have to take, labeled from 0 to n - 1.
Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]
Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses?

Answer:

class Solution {
public:
    /*
     * @param numCourses: a total of n courses
     * @param prerequisites: a list of prerequisite pairs
     * @return: true if can finish all courses or false
     */
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        // Write your code here
        vector<vector<int>> adj(numCourses);
        vector<int> indegree(numCourses);
        for (int i = 0; i < numCourses; i++)
             indegree[i] = 0;
        for (auto pq : prerequisites) {
            adj[pq.first].push_back(pq.second);
            indegree[pq.second]++;
        }
        stack<int> st;
        for (int i = 0; i < numCourses; i++) {
            if (indegree[i] == 0)
                st.push(i);
        }
        int count = 0;
        while (!st.empty()) {
            int v1 = st.top();
            st.pop();
            count++;
            for (auto v2 : adj[v1]) {
                indegree[v2]--;
                if (indegree[v2] == 0)
                    st.push(v2);
            }
        }
        return count == numCourses;
    }
}

How to be Jesus's disciples

我是葡萄树，你们是枝子。常在我里面的，我也常在他里面，这人就多结果子；因为离了我，你们就不能作什么。 (约翰福音 15:5 和合本)
"I am the vine; you are the branches. If you remain in me and I in you, you will bear much fruit; apart from me you can do nothing. (John 15:5 NIV)
人若不常在我里面，就像枝子丢在外面枯干，人拾起来，扔在火里烧了。 (约翰福音 15:6 和合本)
If you do not remain in me, you are like a branch that is thrown away and withers; such branches are picked up, thrown into the fire and burned. (John 15:6 NIV)
你们若常在我里面，我的话也常在你们里面，凡你们所愿意的，祈求，就给你们成就。 (约翰福音 15:7 和合本)
If you remain in me and my words remain in you, ask whatever you wish, and it will be done for you. (John 15:7 NIV)
你们多结果子，我父就因此得荣耀，你们也就是我的门徒了。 (约翰福音 15:8 和合本)
This is to my Father's glory, that you bear much fruit, showing yourselves to be my disciples. (John 15:8 NIV)

Photo by Bill Williams / Unsplash

Wildcard Matching

Question:
http://www.lintcode.com/en/problem/wildcard-matching/
Implement wildcard pattern matching with support for '?' and '*'.

    '?' Matches any single character. 

    '*' Matches any sequence of characters (including the empty sequence). 

The matching should cover the entire input string (not partial).


Answer:

class Solution {
public:
    /*
     * @param s: A string 
     * @param p: A string includes "?" and "*"
     * @return: is Match?
     */
    bool isMatch(string &s, string &p) {
        if (p.empty())
            return false;
        int n = s.length();
        int m = p.length();
        bool f[n + 1][m + 1];
        memset(f, false, sizeof(f));
        f[0][0] = true;
        for (int i = 1; i <= n; i++)
            f[i][0] = false;
        for (int i = 1; i <= m; i++)
            f[0][i] = f[0][i - 1] && p[i - 1] == '*';
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                if (p[j - 1] == '*') {
                    f[i][j] = f[i - 1][j] || f[i][j - 1];
                } else if (p[j - 1] == '?') {
                    f[i][j] = f[i - 1][j - 1];
                } else {
                    f[i][j] = f[i - 1][j - 1] && (s[i - 1] == p[j - 1]);
                }
            }
        } 
        return f[n][m];
    }
}; 

Maximum Gap

Question:
http://www.lintcode.com/en/problem/maximum-gap/
Given an unsorted array, find the maximum difference between the successive elements in its sorted form.
Return 0 if the array contains less than 2 elements.
Sort is easy but will cost O(nlogn) time. Try to solve it in linear time and space.
Answer:

class Solution {  
public:  
    int maximumGap(vector<int> &num) {  
       if (num.size() < 2) return 0;  
        int maxNum = num[0];  
        int minNum = num[0];  
        for (int x : num) {  
            maxNum = max(maxNum, x);  
            minNum = min(minNum, x);  
        }  
        int len = (maxNum - minNum) / num.size() + 1;  
        vector<vector<int>> buckets((maxNum - minNum) / len + 1);  
        for (int x : num) {  
            int i = (x - minNum) / len;  
            if (buckets[i].empty()) {  
                buckets[i].reserve(2);  
                buckets[i].push_back(x);  
                buckets[i].push_back(x);  
            } else {  
                if (x < buckets[i][0]) buckets[i][0] = x;  
                if (x > buckets[i][1]) buckets[i][1] = x;  
            }  
        }  
        int gap = 0;  
        int prev = 0;  
        for (int i = 1; i < buckets.size(); i++) {  
            if (buckets[i].empty()) continue;  
            gap = max(gap, buckets[i][0] - buckets[prev][1]);  
            prev = i;  
        }  
        return gap;  
    }  
};

HOW CHRIST MET A REAL NEED

第三日，在加利利的迦拿有娶亲的筵席，耶稣的母亲在那里。 (约翰福音 2:1 和合本)
On the third day a wedding took place at Cana in Galilee. Jesus' mother was there, (John 2:1 NIV)
耶稣和他的门徒也被请去赴席。 (约翰福音 2:2 和合本)
and Jesus and his disciples had also been invited to the wedding. (John 2:2 NIV)
酒用尽了，耶稣的母亲对他说："他们没有酒了。" (约翰福音 2:3 和合本)
When the wine was gone, Jesus' mother said to him, "They have no more wine." (John 2:3 NIV)
耶稣说："母亲（原文是妇人） ，我与你有什么相干？我的时候还没有到。" (约翰福音 2:4 和合本)
"Woman, why do you involve me?"Jesus replied. "My hour has not yet come." (John 2:4 NIV)
他母亲对用人说："他告诉你们什么，你们就做什么。" (约翰福音 2:5 和合本)
His mother said to the servants, "Do whatever he tells you." (John 2:5 NIV)
照犹太人洁净的规矩，有六口石缸摆在那里，每口可以盛两三桶水。 (约翰福音 2:6 和合本)
Nearby stood six stone water jars, the kind used by the Jews for ceremonial washing, each holding from twenty to thirty gallons. (John 2:6 NIV)
耶稣对用人说："把缸倒满了水。"他们就倒满了，直到缸口。 (约翰福音 2:7 和合本)
Jesus said to the servants, "Fill the jars with water"; so they filled them to the brim. (John 2:7 NIV)
耶稣又说："现在可以舀出来，送给管筵席的。"他们就送了去。 (约翰福音 2:8 和合本)
Then he told them, "Now draw some out and take it to the master of the banquet."They did so, (John 2:8 NIV)
管筵席的尝了那水变的酒，并不知道是哪里来的，只有舀水的用人知道。管筵席的便叫新郎来， (约翰福音 2:9 和合本)
and the master of the banquet tasted the water that had been turned into wine. He did not realize where it had come from, though the servants who had drawn the water knew. Then he called the bridegroom aside (John 2:9 NIV)
对他说："人都是先摆上好酒，等客喝足了，才摆上次的，你倒把好酒留到如今！" (约翰福音 2:10 和合本)
and said, "Everyone brings out the choice wine first and then the cheaper wine after the guests have had too much to drink; but you have saved the best till now." (John 2:10 NIV)
这是耶稣所行的头一件神迹，是在加利利的迦拿行的，显出他的荣耀来；他的门徒就信他了。 (约翰福音 2:11 和合本)
What Jesus did here in Cana of Galilee was the first of the signs through which he revealed his glory; and his disciples believed in him. (John 2:11 NIV)

Photo by Roberta Sorge / Unsplash

The LORD is great

愿一切寻求你的，因你高兴欢喜；愿那些喜爱你救恩的，常说：当尊　神为大！ (诗篇 70:4 和合本)
But may all who seek you rejoice and be glad in you; may those who long for your saving help always say, "The Lord is great!" (Psalms 70:4 NIV)
但我是困苦穷乏的；　神啊，求你速速到我这里来！你是帮助我的，搭救我的。耶和华啊，求你不要耽延！ (诗篇 70:5 和合本)
But as for me, I am poor and needy; come quickly to me, O God. You are my help and my deliverer; Lord , do not delay. (Psalms 70:5 NIV)

Photo by Justin Kauffman / Unsplash

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