Zero Array Transformation IV
Time complexity: $O(mnq)$
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
|
#include <iostream>
#include <vector>
class Solution {
public:
int minZeroArray(std::vector<int>& nums, std::vector<std::vector<int>>& queries) {
int res = 0;
for (int i = 0; i < nums.size(); ++i) {
int num = nums[i];
if (!num) {
continue;
}
std::vector<int> dp(num + 1);
dp[0] = 1;
for (int k = 0; k < queries.size(); ++k) {
if (i < queries[k][0] || i > queries[k][1]) {
continue;
}
for (int j = num; j >= queries[k][2]; --j) {
dp[j] |= dp[j - queries[k][2]];
}
if (dp.back()) {
res = std::max(res, k + 1);
break;
}
}
if (!dp.back()) {
return -1;
}
}
return res;
}
};
|
