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Range Addition II
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Range Addition II
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/*
LeetCode 598:
Given an m * n matrix M initialized with all 0's and several update operations.
Operations are represented by a 2D array, and each operation is represented by an array with two positive integers a and b, which means M[i][j] should be added by one for all 0 <= i < a and 0 <= j < b.
You need to count and return the number of maximum integers in the matrix after performing all the operations.
Example 1:
Input:
m = 3, n = 3
operations = [[2,2],[3,3]]
Output: 4
Explanation:
Initially, M =
[[0, 0, 0],
[0, 0, 0],
[0, 0, 0]]
After performing [2,2], M =
[[1, 1, 0],
[1, 1, 0],
[0, 0, 0]]
After performing [3,3], M =
[[2, 2, 1],
[2, 2, 1],
[1, 1, 1]]
So the maximum integer in M is 2, and there are four of it in M. So return 4.
Note:
The range of m and n is [1,40000].
The range of a is [1,m], and the range of b is [1,n].
The range of operations size won't exceed 10,000.
Solution:
Language: C++
Tag:
*/
class Solution {
public:
#define min(a, b) (a < b ? a : b)
int maxCount(int m, int n, vector<vector<int>>& ops) {
/*
// 1. brute force -- O(size*mn) time + O(mn) size -- memory limit exceed
vector<vector<int>> matrix(m, vector<int>(n, 0));
for (int i = 0; i < ops.size(); ++i) {
int rowSize = ops[i][0], colSize = ops[i][1];
for (int j = 0; j < rowSize; ++j) {
for (int k = 0; k < colSize; ++k) {
++matrix[j][k];
}
}
}
int res = 0, tar = matrix[0][0];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (matrix[i][j] == tar) {
++res;
}
}
}
return res;
*/
// 2. O(size) time + O(1) space
int row = m, col = n;
for (int i = 0; i < ops.size(); ++i) {
row = min(row, ops[i][0]);
col = min(col, ops[i][1]);
}
return row * col;
}
};