You are given an m x n matrix grid consisting of positive integers. You can move from a cell in the matrix to any other cell that is either to the bottom or to the right (not necessarily adjacent). The score of a move from a cell with the value c1 to a cell with the value c2 is c2 - c1.

You can start at any cell, and you have to make at least one move.

Return the maximum total score you can achieve.

class Solution {
public:
     int solve(vector<vector<int>>& grid,int n,int m,int i,int j,int mi,vector<vector<int>>& dp){
        if(i>=n||j>=m) return INT_MIN;
        int x = grid[i][j]-mi;
        dp[i][j] = mi;
        if(mi>grid[i][j]) mi = grid[i][j];      
        return   max({x,solve(grid,n,m,i+1,j,mi,dp),solve(grid,n,m,i,j+1,mi,dp)});
    }
    
    int maxScore(vector<vector<int>>& grid) {
        int n = grid.size();
        int m  = grid[0].size();
        vector<vector<int>>dp(n,vector<int>(m,-1));
        int mi = INT_MAX;
        int ma = solve(grid,n,m,0,0,mi,dp);
        return ma;
    }
};

how to memoize it properly ,here the dp[i] stores the min element traverse till now.