Problem

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

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

Examples

Example 1:

$$ matrix = \begin{bmatrix} 3 & 7 & 8 \\ 9 & 11 & 13 \\ \colorbox{green} {15} & 16 & 17 \end{bmatrix} $$

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:

$$ matrix = \begin{bmatrix} 1 & 10 & 4 & 2 \\ 9 & 3 & 8 & 7 \\ 15 & 16 & 17 & \colorbox{green} {12} \end{bmatrix} $$

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:

$$ matrix = \begin{bmatrix} \colorbox{green} {7} & 8 \\ 1 & 2 \end{bmatrix} $$

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.

Solution

Method 1 - Two Pass

Here is the step by step approach:

  1. Find Minimum Elements in Each Row: Traverse each row and keep track of the minimum element in that row.
  2. Find Maximum Elements in Each Column: Traverse each column and keep track of the maximum element in that column.
  3. Identify Lucky Numbers: For each row, check if the minimum element in that row is also the maximum element in its column. If it is, it’s a lucky number.

Pseudocode

function findLuckyNumbers(matrix):
    initialize rowMins as empty list
    initialize colMaxs as empty list with length of n and initialized to very small values
 
    for each row in matrix:
        find the minimum element in the row and append to rowMins
 
    for each column in matrix:
        find the maximum element in the column and store in colMaxs
 
    initialize luckyNumbers as empty list
    for each element in rowMins:
        if element is in colMaxs:
            append element to luckyNumbers
 
    return luckyNumbers

Video

Here is the video explanation:

Code

Java
public class Solution {

	public List<Integer> luckyNumbers(int[][] matrix) {
		int m = matrix.length;
		int n = matrix[0].length;

		int[] rowMins = new int[m];
		int[] colMaxs = new int[n];

		// Initialize rowMins with maximum values and colMaxs with minimum values
		Arrays.fill(rowMins, Integer.MAX_VALUE);
		Arrays.fill(colMaxs, Integer.MIN_VALUE);

		// Step 1 and Step 2 combined: Find the minimum element in each row and
		// the maximum element in each column in a single pass
		for (int i = 0; i < m; i++) {
			for (int j = 0; j < n; j++) {
				rowMins[i] = Math.min(rowMins[i], matrix[i][j]);
				colMaxs[j] = Math.max(colMaxs[j], matrix[i][j]);
			}
		}

		// Step 3: Find all lucky numbers
		List<Integer> luckyNumbers = new ArrayList<>();

		for (int i = 0; i < m; i++) {
			for (int j = 0; j < n; j++) {
				if (matrix[i][j] == rowMins[i] && matrix[i][j] == colMaxs[j]) {
					luckyNumbers.add(matrix[i][j]);
				}
			}
		}

		return luckyNumbers;
	}
}
Python
def findLuckyNumbers(matrix):
    if not matrix or not matrix[0]:
        return []

    m, n = len(matrix), len(matrix[0])

    # Step 1: Find the minimum element in each row
    rowMins = [min(row) for row in matrix]

    # Step 2: Find the maximum element in each column
    colMaxs = [max(column) for column in zip(*matrix)]

    # Step 3: Find all lucky numbers
    luckyNumbers = []
    for i in range(m):
        for j in range(n):
            if matrix[i][j] == rowMins[i] and matrix[i][j] == colMaxs[j]:
                luckyNumbers.append(matrix[i][j])

    return luckyNumbers

Complexity

  • ⏰ Time complexity: O(m*n) where m is number of rows, and n is number of columns.
  • 🧺 Space complexity: O(m + n)

Dry Run

Lets take eg. 1, with matrix: $$ matrix = \begin{bmatrix} 3 & 7 & 8 \\ 9 & 11 & 13 \\ 15 & 16 & 17 \end{bmatrix} $$ In first pass, we get, rowMins = [3, 9, 15] and colMaxs = 15, 16, 17. In second pass, we get, only 15 as lucky number, as rowMins[2] == 15 and colMaxs[0] == 15.