Lucky Numbers in a Matrix - LeetCode solutions

Problem

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] <= 10^5.
All elements in the matrix are distinct.

Solution (Java)

class Solution {
    public List<Integer> luckyNumbers(int[][] matrix) {
        List<Integer> mini = new ArrayList<>();
        List<Integer> maxi = new ArrayList<>();
        for (int[] arr : matrix) {
            int min = Integer.MAX_VALUE;
            for (int j : arr) {
                if (min > j) {
                    min = j;
                }
            }
            mini.add(min);
        }
        int cols = matrix[0].length;
        for (int c = 0; c < cols; ++c) {
            int max = Integer.MIN_VALUE;
            for (int[] ints : matrix) {
                if (ints[c] > max) {
                    max = ints[c];
                }
            }
            maxi.add(max);
        }
        List<Integer> res = new ArrayList<>();
        for (int value : mini) {
            if (maxi.contains(value)) {
                res.add(value);
            }
        }
        return res;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).