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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.

Related Topics:
Array

Solution 1.

// OJ: https://leetcode.com/problems/lucky-numbers-in-a-matrix
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(M + N)
class Solution {
public:
    vector<int> luckyNumbers (vector<vector<int>>& A) {
        int M = A.size(), N = A[0].size();
        vector<int> row(M, INT_MAX), col(N, INT_MIN);
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                row[i] = min(row[i], A[i][j]);
            }
        }
        for (int i = 0; i < N; ++i) {
            for (int j = 0; j < M; ++j) {
                col[i] = max(col[i], A[j][i]);
            }
        }
        vector<int> ans;
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (A[i][j] == row[i] && A[i][j] == col[j]) ans.push_back(A[i][j]);
            }
        }
        return ans;
    }
};