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Leetcode 2500. Delete Greatest Value in Each Row

Problem Statement

You are given a matrix grid of integers where every row and every column is sorted in increasing order. Your task is to find the greatest value in each row, delete it, and return the sum of these values. The greatest value in a sorted row will always be the last element of that row.

Clarifying Questions

  1. What is the size constraint for the grid?
    • Assume 1 <= grid.length <= 1000 and 1 <= grid[0].length <= 1000.
  2. Can the grid contain negative integers?
    • No, the grid contains only non-negative integers.
  3. What happens if the grid is empty?
    • The grid will always have at least one element.
  4. Are there any edge cases to consider?
    • Not particularly, since every row and column is sorted, the last element of each row will always be the greatest.

Strategy

  1. Initialize a variable sum to 0, which will store the sum of the greatest values.
  2. Iterate over each row in the grid.
  3. For each row, find the last element (since the rows are sorted in increasing order, the last element is the greatest).
  4. Add this greatest value to the sum.
  5. Return the sum.

Code

public class Solution {
    public int deleteGreatestValue(int[][] grid) {
        int sum = 0;
        
        // Iterate over each row in the grid
        for (int[] row : grid) {
            // The greatest value in each sorted row is the last element
            int greatestValue = row[row.length - 1];
            // Add the greatest value to the sum
            sum += greatestValue;
        }
        
        return sum;
    }
}

Time Complexity

Therefore, the overall time complexity is: [ O(n) ]

Where n is the number of rows in the grid. Considering each row can also have columns, the complexity in terms of the grid size is efficient. The constant time operations dominate for each row.

Summary

This approach effectively reduces the problem to a linear scan of the rows, leveraging the sorted property of the grid to directly access the greatest value in each row. The method is efficient and clean, focusing only on the necessary elements to achieve the solution.

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