**PROBLEM LINK:** Practice contest

Author: salik_anwer

Tester: prasoonjain006

Editorialist: salik_anwer

**DIFFICULTY:**

Cake walk-Easy

**PREREQUISITES:**

Array, Matrix

**PROBLEM:**

Given a square matrix (N × N). Determine if the leading diagonal is *Strictly Increasing* or not. If it is strictly increasing then give output 1, otherwise print 0.

**EXPLANATION:**

- The leading diagonal of a square matrix is given by traversing the entries of the given matrix from top-left corner to bottom-right corner.
- It will be strictly increasing if the next element of the diagonal is greater than the current element when moving top to bottom in the diagonal.
- The coordinates of the leading diagonal elements are always (i, i) for a matrix. So, we have to use a for loop to traverse the (i, i) index and incrementing the index with every iteration.
- Meanwhile we will check if the current index entry is less than the entry on the next index in the leading diagonal.
- If the above condition is true for every element in the leading diagonal, the matrix is strictly increasing, else, it is not.

**TIME COMPLEXITY:**

**O(N²)**

**SOLUTION:**

#include <bits/stdc++.h>

using namespace std;

int main() {

int t;

cin>>t;

while(t–){

int n;

cin>>n;

int matrix[n][n];

for(int i=0;i<n;i++){

for(int j=0;j<n;j++){

cin>>matrix[i][j];

}

}

int chk;

for(int i=0;i<n-1;i++){

if(matrix[i+1][i+1]>matrix[i][i]){

chk=1;

}

else{

chk=0;

break;

}

}

cout<<chk<<endl;

}

return 0;

}

**Thanks for reading and do share your own solutions here **