Largest sum of any of the paths till row N-1 of a two-dimensional array

Approach explained pictorially:

Initial matrix


After first iteration


After second iteration

public class Matrix {
 public static void main(String[] args) {
  //Taking N = 3, for 3x3 matrix.
  int[][] A = {{1,2,3},
               {6,4,5},
               {9,8,7}};
  
  //Dynamic programming approach.
  //Logic is to compute the current maximum sum at any given point in the two-dimensional array.
  //Time complexity: O(N^2).
  for(int i = 1;i < A.length;i++) {
   for(int j = 0;j < A.length; j++) {
    if(j == 0) {
     A[i][j] = max((A[i][j] + A[i-1][j]),(A[i][j] + A[i-1][j+1]));
    }
    else if(j==A.length - 1) {
     A[i][j] = max((A[i][j] + A[i-1][j]),(A[i][j] + A[i-1][j-1]));
    }
    else {
     A[i][j] = max((A[i][j] + A[i-1][j-1]),(A[i][j] + A[i-1][j]),(A[i][j] + A[i-1][j+1]));
    }
   }
  }
  
  //The maximum sum is present in the last (N-1)th row. Traverse it to find the maximum sum in O(N) time complexity.
  int maxSum = Integer.MIN_VALUE;
  for(int i=0;i<A.length;i++) {
   if(maxSum < A[A.length - 1][i]) {
    maxSum = A[A.length - 1][i];
   }
  }
  
  System.out.println(maxSum);
 }
 
 //Returns the maximum of the two numbers passed as arguments.
 public static int max(int a, int b) {
  return a >= b ? a:b;
 }
 
 //Returns the maximum of the three numbers passed as arguments.
 public static int max(int a, int b, int c) {
  int x = (a >= b ? a:b);
  return c >= x ? c:x;
 }
}