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Day 2 : Find K-th Row

Last Updated: 4 Dec, 2020
Difficulty: Easy

PROBLEM STATEMENT

Given a square binary matrix ‘mat[n][n]’, find ‘K’ such that all elements in the Kth row are ‘0’ and all elements in the Kth column are ‘1’. The value of mat[k][k] can be anything (either ‘0’ or ‘1’). If no such k exists, return ‘-1’.

For example
``````Consider the following matrix :
0 1 1
0 1 0
1 1 0

You can see that row 1 (0-based) contains all 0’s except mat[1][1] and column 1 contains all 1’s. Hence the answer for the above case is 1.
``````
Input format :
``````The first line contains a single integer ‘T’ denoting the number of test cases to be run. Then the test cases follow.

The first line of each test case contains an integer N representing the number in the problem statement.

Then N lines follow, each of which contains N space-separated integers denoting the elements of the matrix.
``````
Output Format :
`````` For each test case print an integer denoting the value of K if such K exists else print -1.

Output for each test case will be printed in a separate line.
``````
Note:
``````You are not required to print anything; it has already been taken care of. Just implement the function.
``````
Constraints:
``````1 <= T <= 5
1 <= N <= 1000
0 <= Aij <= 1
Time Limit: 1sec
``````

Approach 1

Approach:

For each row, we assume this to be the answer. For each number except mat[k][k] we check if it is 1 and for every number in the kth column we check if it is 1. If both the conditions match, we return this row. If no rows match we return -1.

Algorithm:

1. Start K from 0 to n - 1.
2. Check if all numbers in kth row except mat[k][k] are 0.
3. Check if all numbers in kth column except mat[k][k] are 1.
4. If the current K satisfies the above conditions return it.
5. Return -1 if no such k is found.