A New Collection of Thoughtful Learning Apps — Now Available on iOS & Android

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I’m excited to share a set of mobile apps I’ve recently completed and published on both the Google Play Store and the Apple App Store. These apps are designed with a simple goal in mind: to make meaningful, structured content more accessible, whether you’re studying theology or improving your English vocabulary. 📱 Now Available on Both Platforms All apps are live and available for download: Google Play Developer Page: https://play.google.com/store/apps/dev?id=5835943159853189043 Apple App Store Developer Page: https://apps.apple.com/ca/developer/q-z-l-corp/id1888794100 📖 Theology & Confession Study Apps For those interested in Reformed theology and classical Christian teachings, I’ve developed a series of apps that present foundational texts in a clean, focused reading format: The Belgic Confession Canons of Dort Heidelberg Catechism Westminster Shorter Catechism Each app is designed to provide a distraction-free experience, making it easier to read, reflect, and revisit these im...

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];
    }
};

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