Silly doubt in today's potd

I tried to solve it using the 2D dp approach by taking row no. and col nom as my states but I got the wrong answer.Whn I looked into the solution it said we need to take lives left as the third state.

I tried to solve it using the 2D dp approach by taking row no. and col nom as my states but I got the wrong answer.Whn I looked into the solution it said we need to take lives left as the third state.I am not able to understand why so .Can someone explain why lives left should be taken as the third state For reference this was my approach: class Solution { public: int solve(vector<vector<int>>& coins,vector<vector<int>>& dp,int live,int i ,int j){ int n = coins.size(); int m = coins[0].size(); if(i>=n||j>=m){return INT_MIN;} if(i==n-1&&j==m-1){ if(coins[i][j]>=0){return dp[i][j]=coins[i][j];} else{ if(live>0){return dp[i][j]=0;} else{ dp[i][j]=coins[i][j]; } } } if(dp[i][j]!=INT_MIN){return dp[i][j];} if(coins[i][j]>=0){return dp[i][j]=coins[i][j]+max(solve(coins,dp,live,i+1,j),solve(coins,dp,live,i,j+1));} else{ if(live>0){return dp[i][j]=max(solve(coins,dp,live-1,i+1,j),solve(coins,dp,live-1,i,j+1));} else{ dp[i][j]=coins[i][j]+max(solve(coins,dp,live,i+1,j),solve(coins,dp,live,i,j+1)); } } return INT_MIN; } int maximumAmount(vector<vector<int>>& coins) { int n = coins.size(); int m = coins[0].size(); vector<vector<int>>dp(n,vector<int>(m,INT_MIN)); return solve(coins,dp,2,0,0); } };