LeetCode #1380 — EASY

Lucky Numbers in a Matrix

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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] <= 10⁵.
All elements in the matrix are distinct.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[3,7,8],[9,11,13],[15,16,17]]

Example 2

[[1,10,4,2],[9,3,8,7],[15,16,17,12]]

Example 3

[[7,8],[1,2]]

Step 02

Core Insight

What unlocks the optimal approach

Find out and save the minimum of each row and maximum of each column in two lists.
Then scan through the whole matrix to identify the elements that satisfy the criteria.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1380: Lucky Numbers in a Matrix
class Solution {
    public List<Integer> luckyNumbers(int[][] matrix) {
        int m = matrix.length, n = matrix[0].length;
        int[] rows = new int[m];
        int[] cols = new int[n];
        Arrays.fill(rows, 1 << 30);
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                rows[i] = Math.min(rows[i], matrix[i][j]);
                cols[j] = Math.max(cols[j], matrix[i][j]);
            }
        }
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (rows[i] == cols[j]) {
                    ans.add(rows[i]);
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1380: Lucky Numbers in a Matrix
func luckyNumbers(matrix [][]int) (ans []int) {
	m, n := len(matrix), len(matrix[0])
	rows, cols := make([]int, m), make([]int, n)
	for i := range rows {
		rows[i] = 1 << 30
	}
	for i, row := range matrix {
		for j, x := range row {
			rows[i] = min(rows[i], x)
			cols[j] = max(cols[j], x)
		}
	}
	for i, row := range matrix {
		for j, x := range row {
			if rows[i] == cols[j] {
				ans = append(ans, x)
			}
		}
	}
	return
}

# Accepted solution for LeetCode #1380: Lucky Numbers in a Matrix
class Solution:
    def luckyNumbers(self, matrix: List[List[int]]) -> List[int]:
        rows = {min(row) for row in matrix}
        cols = {max(col) for col in zip(*matrix)}
        return list(rows & cols)

// Accepted solution for LeetCode #1380: Lucky Numbers in a Matrix
impl Solution {
    pub fn lucky_numbers(matrix: Vec<Vec<i32>>) -> Vec<i32> {
        let m = matrix.len();
        let n = matrix[0].len();
        let mut res = vec![];
        let mut col = vec![0; n];
        for j in 0..n {
            for i in 0..m {
                col[j] = col[j].max(matrix[i][j]);
            }
        }
        for x in 0..m {
            let mut i = 0;
            for y in 1..n {
                if matrix[x][y] < matrix[x][i] {
                    i = y;
                }
            }
            if matrix[x][i] == col[i] {
                res.push(col[i]);
            }
        }
        res
    }
}

// Accepted solution for LeetCode #1380: Lucky Numbers in a Matrix
function luckyNumbers(matrix: number[][]): number[] {
    const m = matrix.length;
    const n = matrix[0].length;
    const rows: number[] = new Array(m).fill(1 << 30);
    const cols: number[] = new Array(n).fill(0);
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; j++) {
            rows[i] = Math.min(rows[i], matrix[i][j]);
            cols[j] = Math.max(cols[j], matrix[i][j]);
        }
    }
    const ans: number[] = [];
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; j++) {
            if (rows[i] === cols[j]) {
                ans.push(rows[i]);
            }
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n)

Space

O(m + n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.