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 1:



Input: grid = [[1,3,1],[1,5,1],[4,2,1]]

Output: 7

Explanation: Because the path 1 → 3 → 1 → 1 → 1 minimizes the sum.

Example 2:


Input: grid = [[1,2,3],[4,5,6]]

Output: 12

 


Constraints:


m == grid.length

n == grid[i].length

1 <= m, n <= 200

0 <= grid[i][j] <= 200


class Solution {

       public int solve(int i,int j,int m,int n, int [][] dp,int[][] Grid){

        if(i==m-1 && j==n-1)
        {
           return Grid[i][j];
        }

       
       
        if(dp[i][j] !=-1)
        {
            return dp[i][j];
        }
        if(i==m-1)
        {
            return Grid[i][j]+solve(i,j+1,m,n,dp,Grid);//right move only
        }
        else if(j==n-1)
        {
            return Grid[i][j]+solve(i+1,j,m,n,dp,Grid);//down move only
        }
        else
        {
         int right=solve(i,j+1,m,n,dp,Grid);
        int down=solve(i+1,j,m,n,dp,Grid);
        dp[i][j]=Grid[i][j]+Math.min(right,down);
        return  dp[i][j];
        }
       
    }
    public int minPathSum(int[][] grid) {
       
        int row=grid.length;
        int col=grid[0].length;
         int [][] dp= new int[row][col];

          // Initialize all cells to -1 (uncomputed)
        for (int m = 0; m < row; m++) {
            for (int n = 0; n < col; n++) {
                dp[m][n] = -1;
            }
        }
        return solve(0,0,row,col,dp,grid);
    }
}

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