LeetCode 1380. Lucky Numbers in a Matrix

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Question

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

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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

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.

Solution

一开始想到到的就是暴力解法，搜索每一个数是否符合这样的限制，但是发现这样应该是m^2 * n^2的时间复杂度，因为需要遍历每个元素并且同时遍历当前行和列。但是根据观察发现，我们只需要每行的最小值以及每列的最大值，这样就想到可以先求其中一个然后看交集，如果存在则符合这样的要求。这样只需要遍历matrix每个元素一遍。

Code

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// time:O(mn) space:O(m)
    // set intersection
    public List<Integer> luckyNumbers (int[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0] == null || matrix[0].length == 0) return new ArrayList<>();
        HashSet<Integer> minOfRows = new HashSet<>();
        for (int[] row : matrix) {
            int min = row[0];
            for (int n : row) {
                min = Math.min(min, n);
            }
            minOfRows.add(min);
        } 
        int m = matrix.length;
        int n = matrix[0].length;
        List<Integer> res = new ArrayList<>();
        for (int j = 0; j < n; j++) {
            int max = matrix[0][j];
            for (int i = 0; i < m; i++) {
                max = Math.max(max, matrix[i][j]);
            }
            if (minOfRows.contains(max)) res.add(max);
        }
        return res;
    }