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2133. Check if Every Row and Column Contains All Numbers

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.

 

Example 1:

Input: matrix = [[1,2,3],[3,1,2],[2,3,1]]
Output: true
Explanation: In this case, n = 3, and every row and column contains the numbers 1, 2, and 3.
Hence, we return true.

Example 2:

Input: matrix = [[1,1,1],[1,2,3],[1,2,3]]
Output: false
Explanation: In this case, n = 3, but the first row and the first column do not contain the numbers 2 or 3.
Hence, we return false.

 

Constraints:

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

Solutions

Solution 1: Hash Table

Traverse each row and column of the matrix, using a hash table to record whether each number has appeared. If any number appears more than once in a row or column, return false; otherwise, return true

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the size of the matrix.

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class Solution:
    def checkValid(self, matrix: List[List[int]]) -> bool:
        n = len(matrix)
        return all(len(set(row)) == n for row in chain(matrix, zip(*matrix)))

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class Solution {
    public boolean checkValid(int[][] matrix) {
        int n = matrix.length;
        boolean[] vis = new boolean[n + 1];
        for (var row : matrix) {
            Arrays.fill(vis, false);
            for (int x : row) {
                if (vis[x]) {
                    return false;
                }
                vis[x] = true;
            }
        }
        for (int j = 0; j < n; ++j) {
            Arrays.fill(vis, false);
            for (int i = 0; i < n; ++i) {
                if (vis[matrix[i][j]]) {
                    return false;
                }
                vis[matrix[i][j]] = true;
            }
        }
        return true;
    }
}

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class Solution {
public:
    bool checkValid(vector<vector<int>>& matrix) {
        int n = matrix.size();
        bool vis[n + 1];
        for (const auto& row : matrix) {
            memset(vis, false, sizeof(vis));
            for (int x : row) {
                if (vis[x]) {
                    return false;
                }
                vis[x] = true;
            }
        }
        for (int j = 0; j < n; ++j) {
            memset(vis, false, sizeof(vis));
            for (int i = 0; i < n; ++i) {
                if (vis[matrix[i][j]]) {
                    return false;
                }
                vis[matrix[i][j]] = true;
            }
        }
        return true;
    }
};

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func checkValid(matrix [][]int) bool {
    n := len(matrix)
    for _, row := range matrix {
        vis := make([]bool, n+1)
        for _, x := range row {
            if vis[x] {
                return false
            }
            vis[x] = true
        }
    }
    for j := 0; j < n; j++ {
        vis := make([]bool, n+1)
        for i := 0; i < n; i++ {
            if vis[matrix[i][j]] {
                return false
            }
            vis[matrix[i][j]] = true
        }
    }
    return true
}

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function checkValid(matrix: number[][]): boolean {
    const n = matrix.length;
    const vis: boolean[] = Array(n + 1).fill(false);
    for (const row of matrix) {
        vis.fill(false);
        for (const x of row) {
            if (vis[x]) {
                return false;
            }
            vis[x] = true;
        }
    }
    for (let j = 0; j < n; ++j) {
        vis.fill(false);
        for (let i = 0; i < n; ++i) {
            if (vis[matrix[i][j]]) {
                return false;
            }
            vis[matrix[i][j]] = true;
        }
    }
    return true;
}

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