Q- Recursion | Memoization | Tabulization in 2d dp

public class Solution {
//https://www.youtube.com/watch?v=AE39gJYuRog&list=PLgUwDviBIf0qUlt5H_kiKYaNSqJ81PMMY&index=8&ab_channel=takeUforward
    public static int ninjaTraining(int n, int points[][]) {
        // return solve(n-1 , points , 3 , new Integer[n][4]);
        int[][]dp = new int[n][4];
        
        /* while writing tabulization go from base case to n in recursion solution 
        1) handle for base case 
        2)also include loops for changing states 
        3) rest same code as recursion just replace recursion calls by dp
        */
        for(int i = 0 ;i < n ;i++){
            for(int last = 0 ; last < 4 ; last++){
                
                for(int j = 0 ; j < 3 ; j++){
                    if(j != last){
                        int point = points[i][j] + (i == 0 ?0 :dp[i-1][j]); //handling for base case
                        dp[i][last] = Math.max(dp[i][last] , point);
                    }
                }
            }
        }
        return dp[n-1][3];  //0-2 denotes activity , 3 denotes no last activity exist
    }
    
    //recursion + memoization
    public static int solve(int idx , int points[][] , int last , Integer[][] dp){
        if(idx == 0){
            int base = 0;
            
            for(int i = 0 ;i < 3 ; i++){
                if(i != last)
                base = Math.max(base , points[0][i]);
            }
            return base;
        }
        if(dp[idx][last] != null)
        return dp[idx][last];
        
        int max = 0;
        for(int i = 0 ;i < 3 ; i++){
            if(i != last){
                //we choose nth activity and send calls for rest of the ans;
                int point = points[idx][i] + solve(idx-1 , points , i , dp);
                max = Math.max(max , point);
            }
        }

        return dp[idx][last] = max;
    }

}