7 lines C++ DP


  • 0

    There are only 2 directions to reach a given cell[i][j], either from above it or from left,

    • dp[i][j - 1] (paths from its left)

    • dp[i - 1][j] (paths from its above)

    Total path = Sum of path from its left + Sum of path from its above

    dp[i][j] = dp[i][j - 1] + dp[i - 1][j];

    class Solution {
    public:
        int uniquePaths(int m, int n) {
            vector<vector<int>>dp(m, vector<int>(n));
            for(int i = 0; i < m; i++) dp[i][0] = 1;
            for(int i = 0; i < n; i++) dp[0][i] = 1;
            for(int i = 1; i < m; i++)
                for(int j = 1; j < n; j++)
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            return dp[m - 1][n - 1];
        }
    };
    

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