Leetcode-3489

Zero Array Transformation IV

Time complexity: $O(mnq)$

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#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;
    }
};

Leetcode-3489

Licensed under CC BY-NC-SA 4.0