Grid Game

Problem

TC: O(n)
SC: O(n)

class Solution {
    public long gridGame(int[][] grid) {
        long prefix1[] = new long[grid[0].length];
        long prefix2[] = new long[grid[0].length];

        long sum  =0;
        for(int i=0;i<grid[0].length;i++){
            sum+=grid[0][i];
            prefix1[i]=  sum;
        }
        sum  =0;
        for(int i=0;i<grid[0].length;i++){
            sum+=grid[1][i];
            prefix2[i]=  sum;
        }
        long res = Long.MAX_VALUE;
        for(int i=0;i<grid[0].length;i++){
            //if robot 1 at 2 cross at index i
            //in this row if the robot 1 and 2 cross then robot 2 will get all the values after the from index i to n-1 th index (in top row) 
            //or values till i from the start 0 to i in the bottom row, as once in bottom row robot 2 can not move up

            long top  = prefix1[grid[0].length-1] - prefix1[i];
            long bottom = i>0 ? prefix2[i-1] : 0;
            long max = Math.max(top,bottom);// robot 2 will try to maximize the values it collects
            res  = Math.min(res,max); //robot 1 will try to minimize the value robot 2 collects
        }
        return res;
    }
}