64. Minimum Path Sum

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

Example:

Input:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
         int m = grid.size();
        if(m == 0)
            return 0;
        int n = grid[0].size();
        int dp[m][n];
        for(int i = 0; i < m; ++i) {
            for(int j = 0; j < n; ++j) {
                if(i == 0 && j == 0)
                    dp[i][j] = grid[0][0];
                else if(i == 0)
                    dp[i][j] = grid[i][j] + dp[i][j - 1];
                else if(j == 0)
                    dp[i][j] = grid[i][j] + dp[i - 1][j];
                else
                    dp[i][j] = grid[i][j] + min(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        
              for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
               cout<<" "<<dp[i][j];
               cout<<endl;
        }
        return dp[m - 1][n - 1];
        
        
    }
};