LeetCode 2500: Delete Greatest Value in Each Row
LeetCode 2500 Solution Explanation
Explanation:
To solve this problem, we iterate over each row of the matrix and find the maximum value in each row. We remove the maximum value from each row and add it to the answer. We repeat this process until the matrix becomes empty. The key idea is to keep track of the maximum value in each row and update the answer accordingly.
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Algorithm:
- Initialize answer to 0.
- While the number of columns is greater than 0:
- For each row, find the maximum value and remove it.
- Add the maximum value to the answer.
- Reduce the number of columns by 1.
- Return the final answer.
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Time Complexity: O(m * n^2), where m is the number of rows and n is the number of columns in the matrix.
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Space Complexity: O(1) as we are not using any extra space.
:
LeetCode 2500 Solutions in Java, C++, Python
class Solution {
public int maximumMatrixSum(int[][] grid) {
int answer = 0;
int m = grid.length;
int n = grid[0].length;
while (n > 0) {
int maxVal = Integer.MIN_VALUE;
int maxRow = -1;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] > maxVal) {
maxVal = grid[i][j];
maxRow = i;
}
}
}
answer += maxVal;
for (int j = 0; j < n; j++) {
if (grid[maxRow][j] == maxVal) {
grid[maxRow][j] = 0;
}
}
n--;
}
return answer;
}
}Interactive Code Editor for LeetCode 2500
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