3402. Minimum Operations to Make Columns Strictly Increasing
You are given a m x n
matrix grid
consisting of non-negative integers.
In one operation, you can increment the value of any grid[i][j]
by 1.
Return the minimum number of operations needed to make all columns of grid
strictly increasing.
Example 1:
Input: grid = [[3,2],[1,3],[3,4],[0,1]]
Output: 15
Explanation:
- To make the
0th
column strictly increasing, we can apply 3 operations ongrid[1][0]
, 2 operations ongrid[2][0]
, and 6 operations ongrid[3][0]
. - To make the
1st
column strictly increasing, we can apply 4 operations ongrid[3][1]
.
Example 2:
Input: grid = [[3,2,1],[2,1,0],[1,2,3]]
Output: 12
Explanation:
- To make the
0th
column strictly increasing, we can apply 2 operations ongrid[1][0]
, and 4 operations ongrid[2][0]
. - To make the
1st
column strictly increasing, we can apply 2 operations ongrid[1][1]
, and 2 operations ongrid[2][1]
. - To make the
2nd
column strictly increasing, we can apply 2 operations ongrid[1][2]
.
Constraints:
m == grid.length
n == grid[i].length
1 <= m, n <= 50
0 <= grid[i][j] < 2500
class Solution {
public int minimumOperations(int[][] grid) {
int row = grid.length;
int col = grid[0].length;
int result = 0;
for (int i = 0; i < col; i++){// 0
for (int j = 1; j < row; j++){// 1
int pre = grid[j - 1][i]; // [0,0] = 3
int need = pre + 1; // 4
int cur = grid[j][i]; // 1
if (cur > pre){
continue;
}else{
grid[j][i] = need;
result = result + (need - cur);
// 0 + (4 - 1 )= 3
}
}
}
return result;
}
}
// TC: O(n * m)
// SC: O(1)