Check If a Given Matrix is Good or Not in C++

Suppose we have one n x n matrix. The matrix is said to be a good matrix where every number not equal to 1 can be expressed as the sum of a number in the same row and a number in the same column. We have to check whether given matrix is good or not.

So, if the input is like

Then the output will be True, because the 6 in the bottom left corner is valid because when the sum of the 2 above it and the 4 on the right. The same holds for every number not equal to 1 in this matrix.

Steps

To solve this, we will follow these steps −

n := size of M
for initialize i := 0, when i < n, update (increase i by 1), do:
   for initialize j := 0, when j < n, update (increase j by 1), do:
      ok := 0
      if M[i, j] is not equal to 1, then:
         c := M[i, j]
      for initialize h := 0, when h < n, update (increase h by 1), do:
         for initialize k := 0, when k < n, update (increase k by 1), do:
            if c is same as M[i, h] + M[k, j], then:
               ok := 1
      if ok is same as 0 and M[i, j] is not equal to 1, then:
         return false
return true

Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;
bool solve(vector<vector<int>> M){
   int n = M.size();
   int c;
   bool ok;
   for (int i = 0; i < n; i++){
      for (int j = 0; j < n; j++){
         ok = 0;
         if (M[i][j] != 1)
            c = M[i][j];
         for (int h = 0; h < n; h++){
            for (int k = 0; k < n; k++)
               if (c == M[i][h] + M[k][j])
                  ok = 1;
         }
         if (ok == 0 && M[i][j] != 1){
            return false;
         }
      }
   }
   return true;
}
int main(){
   vector<vector<int>> matrix = { { 1, 1, 2 }, { 2, 3, 1 }, { 6, 4, 1 } };
   cout << solve(matrix) << endl;
}

Input

{ { 1, 1, 2 }, { 2, 3, 1 }, { 6, 4, 1 } }

Output

1