How do you solve Lucky Numbers in a Matrix on LeetCode?

Lucky Numbers in a Matrix (LeetCode #1380) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Lucky numbers must satisfy two conditions: min in row and max in column.

What is the time complexity of Lucky Numbers in a Matrix?

The optimal solution for Lucky Numbers in a Matrix runs in O(n) time and uses O(n) space. We perform linear scans to find minimums and maximums, leading to O(n) time complexity overall.

What is the brute force solution for Lucky Numbers in a Matrix?

The brute force approach for Lucky Numbers in a Matrix is Brute Force, with O(n²) time complexity and O(1) space complexity. We will check each element in the matrix to see if it is the minimum in its row and the maximum in its column. This is straightforward but can be slow for larger matrices. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Lucky Numbers in a Matrix?

Lucky Numbers in a Matrix uses the following concepts: Array, Matrix. The recommended patterns to study are: Array, Matrix.

What are the interview tips for Lucky Numbers in a Matrix?

Always clarify the problem statement and constraints before jumping into coding. Discuss your thought process and approach with the interviewer. Consider edge cases, such as the smallest or largest possible matrices.

What are common mistakes when solving Lucky Numbers in a Matrix?

Not checking both conditions properly, leading to incorrect results. Confusing row and column indices when checking max/min.

#1380

Lucky Numbers in a Matrix

Easy

ArrayMatrixArrayMatrix

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of checking each element repeatedly, we can first find the minimum of each row and the maximum of each column, then check for lucky numbers in a single pass.

⚙️

Algorithm

3 steps

1Step 1: Create a list to store the minimum of each row.
2Step 2: Create a list to store the maximum of each column.
3Step 3: Iterate through the matrix and check if each minimum is also the maximum in its column.

solution.py4 lines

1def luckyNumbers(matrix):
2    min_row = [min(row) for row in matrix]
3    max_col = [max(matrix[i][j] for i in range(len(matrix))) for j in range(len(matrix[0]))]
4    return [min_row[i] for i in range(len(min_row)) if min_row[i] == max_col[matrix[i].index(min_row[i])]]

ℹ

Complexity note: We perform linear scans to find minimums and maximums, leading to O(n) time complexity overall.

1Lucky numbers must satisfy two conditions: min in row and max in column.
2Using pre-computed lists for row mins and column maxes reduces redundant checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.