Check if the given chessboard is valid or not

Given an NXN chessboard. The task is to check if the given chessboard is valid or not. A chess board is considered valid if every 2 adjacent cells are painted with different color. Two cells are considered adjacent if they share a boundary. 

First chessboard is valid whereas the second is invalid.

Examples: 

Input : N = 2
C = { 
      { 1, 0 },
      { 0, 1 }
    }
Output : Yes

Input : N = 2
C = { 
      { 0, 0 },
      { 0, 0 }
    }
Output : No

Observe, on a chess board, every adjacent pair of cells is painted in a different color. 
So, for each cell (i, j), check whether the adjacent cells i.e 
(i + 1, j), (i -1, j), (i, j + 1), (i, j – 1) is painted with different color than (i, j) or not.

Pseudocode: 

int X[] = {0, -1, 0, 1};
int Y[] = {1, 0, -1, 0};

bool validateConfiguration(int M[N][N]) 
{
  bool isValid = true;
  for (int i = 0; i < N; i++)
  {
    for (int j = 0; j < N; j++)
    {
      for (int k = 0; k < 4; k++)
      {
         int newX = i + X[k];
         int newY = j + Y[k];
         if (newX < N && newY < N && newX >= 0 &&
             newY >= 0 && M[newX][newY] == M[i][j])
         {
              isValid = false;
         } 
       }
     }
  }
  return isValid;
}

Implementation:

Yes

Time complexity: O(N^2) for given an N*N chessboard
Auxiliary space: O(1)

Implementation: A simpler approach would be to check if the cells with even sums are of the same color. 

true
true
false

Time complexity: O(N^2) for given an N*N chessboard
Auxiliary space: O(1)