How do you solve Delete Greatest Value in Each Row on LeetCode?

Delete Greatest Value in Each Row (LeetCode #2500) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Sorting allows us to efficiently access maximum values.

What is the brute force solution for Delete Greatest Value in Each Row?

The brute force approach for Delete Greatest Value in Each Row is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will repeatedly find and remove the maximum value from each row of the matrix until the matrix is empty. This method is straightforward but inefficient as it involves multiple scans of the matrix. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Delete Greatest Value in Each Row?

Delete Greatest Value in Each Row uses the following concepts: Array, Sorting, Heap (Priority Queue), Matrix, Simulation. The recommended patterns to study are: Sorting, Array.

What are the interview tips for Delete Greatest Value in Each Row?

Always clarify the problem constraints and edge cases. Think about the efficiency of your approach and mention it during the interview. Practice explaining your thought process clearly and concisely.

What are common mistakes when solving Delete Greatest Value in Each Row?

Not considering edge cases like single-row or single-column matrices. Failing to update the matrix correctly after removing elements.

#2500

Delete Greatest Value in Each Row

Easy

ArraySortingHeap (Priority Queue)MatrixSimulationSortingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(1)

This approach leverages sorting to efficiently find the maximum values in each row and eliminates the need for repeated scans. By sorting each row once, we can access the maximum values directly in each iteration.

⚙️

Algorithm

4 steps

1Step 1: Sort each row of the matrix in descending order.
2Step 2: Initialize a variable 'total' to 0.
3Step 3: For each column index from 0 to n-1, find the maximum value across all rows at that column index, add it to 'total'.
4Step 4: Return 'total' after processing all columns.

solution.py10 lines

1# Full working Python code
2
3def deleteGreatestValue(grid):
4    for row in grid:
5        row.sort(reverse=True)
6    total = 0
7    for col in range(len(grid[0])):
8        max_val = max(row[col] for row in grid)
9        total += max_val
10    return total

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting of each row, where n is the number of columns. The space complexity is O(1) since we are modifying the matrix in place without using additional data structures.

1Sorting allows us to efficiently access maximum values.
2Iterating through columns after sorting simplifies the process.

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