README_EN.md

1380. Lucky Numbers in a Matrix

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Description

Given an m x n matrix of distinct numbers, return all lucky numbers in the matrix in any order.

A lucky number is an element of the matrix such that it is the minimum element in its row and maximum in its column.

 

Example 1:

Input: matrix = [[3,7,8],[9,11,13],[15,16,17]]
Output: [15]
Explanation: 15 is the only lucky number since it is the minimum in its row and the maximum in its column.

Example 2:

Input: matrix = [[1,10,4,2],[9,3,8,7],[15,16,17,12]]
Output: [12]
Explanation: 12 is the only lucky number since it is the minimum in its row and the maximum in its column.

Example 3:

Input: matrix = [[7,8],[1,2]]
Output: [7]
Explanation: 7 is the only lucky number since it is the minimum in its row and the maximum in its column.

 

Constraints:

• m == mat.length
• n == mat[i].length
• 1 <= n, m <= 50
• 1 <= matrix[i][j] <= 105.
• All elements in the matrix are distinct.

Solutions

Python3

class Solution:
    def luckyNumbers(self, matrix: List[List[int]]) -> List[int]:
        rows = {min(row) for row in matrix}
        cols = {max(col) for col in zip(*matrix)}
        return list(rows & cols)

Java

class Solution {
    public List<Integer> luckyNumbers(int[][] matrix) {
        int m = matrix.length, n = matrix[0].length;
        int[] rows = new int[m];
        int[] cols = new int[n];
        Arrays.fill(rows, 1 << 30);
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                rows[i] = Math.min(rows[i], matrix[i][j]);
                cols[j] = Math.max(cols[j], matrix[i][j]);
            }
        }
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (rows[i] == cols[j]) {
                    ans.add(rows[i]);
                }
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> luckyNumbers(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        int rows[m];
        int cols[n];
        memset(rows, 0x3f, sizeof(rows));
        memset(cols, 0, sizeof(cols));
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                rows[i] = min(rows[i], matrix[i][j]);
                cols[j] = max(cols[j], matrix[i][j]);
            }
        }
        vector<int> ans;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (rows[i] == cols[j]) {
                    ans.push_back(rows[i]);
                }
            }
        }
        return ans;
    }
};

Go

func luckyNumbers(matrix [][]int) (ans []int) {
	m, n := len(matrix), len(matrix[0])
	rows, cols := make([]int, m), make([]int, n)
	for i := range rows {
		rows[i] = 1 << 30
	}
	for i, row := range matrix {
		for j, x := range row {
			rows[i] = min(rows[i], x)
			cols[j] = max(cols[j], x)
		}
	}
	for i, row := range matrix {
		for j, x := range row {
			if rows[i] == cols[j] {
				ans = append(ans, x)
			}
		}
	}
	return
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

TypeScript

function luckyNumbers(matrix: number[][]): number[] {
    const m = matrix.length;
    const n = matrix[0].length;
    const rows: number[] = new Array(m).fill(1 << 30);
    const cols: number[] = new Array(n).fill(0);
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; j++) {
            rows[i] = Math.min(rows[i], matrix[i][j]);
            cols[j] = Math.max(cols[j], matrix[i][j]);
        }
    }
    const ans: number[] = [];
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; j++) {
            if (rows[i] === cols[j]) {
                ans.push(rows[i]);
            }
        }
    }
    return ans;
}

Rust

impl Solution {
    pub fn lucky_numbers(matrix: Vec<Vec<i32>>) -> Vec<i32> {
        let m = matrix.len();
        let n = matrix[0].len();
        let mut res = vec![];
        let mut col = vec![0; n];
        for j in 0..n {
            for i in 0..m {
                col[j] = col[j].max(matrix[i][j]);
            }
        }
        for x in 0..m {
            let mut i = 0;
            for y in 1..n {
                if matrix[x][y] < matrix[x][i] {
                    i = y;
                }
            }
            if matrix[x][i] == col[i] {
                res.push(col[i]);
            }
        }
        res
    }
}

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