Leetcode 2500. Delete Greatest Value in Each Row
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.
1 <= grid.length <= 1000
and 1 <= grid[0].length <= 1000
.sum
to 0, which will store the sum of the greatest values.sum
.sum
.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;
}
}
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.
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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