# Check if Every Row and Column Contains All Numbers

Published: Sep 27, 2022

Easy Matrix Array

## Problem Description

An `n x n` matrix is valid if every row and every column contains all the integers from 1 to n (inclusive).

Given an `n x n` integer matrix `matrix`, return `true` if the matrix is valid. Otherwise, return `false`.

Constraints:

• `n == matrix.length == matrix[i].length`
• `1 <= n <= 100`
• `1 <= matrix[i][j] <= n`

https://leetcode.com/problems/check-if-every-row-and-column-contains-all-numbers/

## Examples

``````Example 1
Input: matrix = [[1,2,3],[3,1,2],[2,3,1]]
Output: true
``````
``````Example 2
Input: matrix = [[1,1,1],[1,2,3],[1,2,3]]
Output: false
``````

## How to Solve

The problem is about n x n matrix and 1 to n (inclusive) values. Using a set data structure, check the cell value of each row and column doesn’t appear more than once.

The solution uses a set for horizontal and array of sets for vertical. When a row is scanned, it checks the horizontal set. When all rows are scanned, it checks the array of vertical sets. If all are length n, return True.

## Solution

``````class CheckIfEveryRowAndColumnContainsAllNumbers {
public:
bool checkValid(vector<vector<int>>& matrix) {
int n = matrix.size();
// check rows
for (int r = 0; r < n; ++r) {
vector<int> counts(n + 1, 0);
for (int c = 0; c < n; ++c) {
if (counts[matrix[r][c]] > 0) { return false; }
counts[matrix[r][c]]++;
}
}
// check cols
for (int c = 0; c < n; ++c) {
vector<int> counts(n + 1, 0);
for (int r = 0; r < n; ++r) {
if (counts[matrix[r][c]] > 0) { return false; }
counts[matrix[r][c]]++;
}
}
return true;
}
};
``````
``````class CheckIfEveryRowAndColumnContainsAllNumbers {
public boolean checkValid(int[][] matrix) {
int n = matrix.length;
// check rows
for (int r = 0; r < n; ++r) {
int[] counts = new int[n + 1];
for (int c = 0; c < n; ++c) {
if (counts[matrix[r][c]] > 0) { return false; }
counts[matrix[r][c]]++;
}
}
// check cols
for (int c = 0; c < n; ++c) {
int[] counts = new int[n + 1];
for (int r = 0; r < n; ++r) {
if (counts[matrix[r][c]] > 0) { return false; }
counts[matrix[r][c]]++;
}
}
return true;
}
}
``````
``````/**
* @param {number[][]} matrix
* @return {boolean}
*/
var checkValid = function(matrix) {
const n = matrix.length;
// check rows
for (let r = 0; r < n; r++) {
let counts = new Array(n + 1);
counts.fill(0);
for (let c = 0; c < n; c++) {
if (counts[matrix[r][c]] > 0) { return false; }
counts[matrix[r][c]]++;
}
}
//check cols
for (let c = 0; c < n; c++) {
let counts = new Array(n + 1);
counts.fill(0);
for (let r = 0; r < n; r++) {
if (counts[matrix[r][c]] > 0) { return false; }
counts[matrix[r][c]]++;
}
}
return true;
};
``````
``````class CheckIfEveryRowAndColumnContainsAllNumbers:
def checkValid(self, matrix: List[List[int]]) -> bool:
n = len(matrix)
# check rows
for r in range(n):
counts = [0 for _ in range(n + 1)]
for c in range(n):
if counts[matrix[r][c]] > 0:
return False
counts[matrix[r][c]] += 1
# check cols
for c in range(n):
counts = [0 for _ in range(n + 1)]
for r in range(n):
if counts[matrix[r][c]] > 0:
return False
counts[matrix[r][c]] += 1
return True
``````
``````# @param {Integer[][]} matrix
# @return {Boolean}
def check_valid(matrix)
n = matrix.size
# check rows
(0...n).each do |r|
counts = Array.new(n + 1, 0)
(0...n).each do |c|
if counts[matrix[r][c]] > 0
return false
end
counts[matrix[r][c]] += 1
end
end
# check cols
(0...n).each do |c|
counts = Array.new(n + 1, 0)
(0...n).each do |r|
if counts[matrix[r][c]] > 0
return false
end
counts[matrix[r][c]] += 1
end
end
return true
end
``````

## Complexities

• Time: `O(n^2)`
• Space: `O(n^2)`