Problem

Given a m * 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]

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.

Pre analysis

Will need to iterate through each element and store corresponding rowMin and ColMax array. Eventually find the common number between both rowMin and ColMax.

Max(O(mn), O(n^2))

Another solution

Another way would be to find row mins and keep search in col, if you encounter the same number return it.

var luckyNumbers  = function(matrix) {
    for (let row = 0; row < matrix.length; row++) {
        // find the smallest number
        let smallest = matrix[row][0];
        let smallestIndex = 0;
        for (let col = 0; col < matrix[row].length; col++) {
            if (matrix[row][col] < smallest) {
                smallest = matrix[row][col];
                smallestIndex = col;
            }
        }

        // find if the number is the largest in the row
        let found = true;
        for (let r = 0; r < matrix.length; r++) {
            if (smallest < matrix[r][smallestIndex]) {
            found = false;
            break;
            };
        }

        if (found) {
        return [smallest];
        }
    }

    return [];
};