3356. Zero Array Transformation II

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3356. Zero Array Transformation II

Description

You are given an integer array nums of length n and a 2D array queries where queries[i] = [li, ri, vali].

Each queries[i] represents the following action on nums:

  • Decrement the value at each index in the range [li, ri] in nums by at most vali.
  • The amount by which each value is decremented can be chosen independently for each index.

A Zero Array is an array with all its elements equal to 0.

Return the minimum possible non-negative value of k, such that after processing the first k queries in sequence , nums becomes a Zero Array . If no such k exists, return -1.

Example 1:

Input: nums = [2,0,2], queries = [[0,2,1],[0,2,1],[1,1,3]]

Output: 2

Explanation:

  • For i = 0 (l = 0, r = 2, val = 1):

  • Decrement values at indices [0, 1, 2] by [1, 0, 1] respectively.

  • The array will become [1, 0, 1].

  • For i = 1 (l = 0, r = 2, val = 1):

  • Decrement values at indices [0, 1, 2] by [1, 0, 1] respectively.

  • The array will become [0, 0, 0], which is a Zero Array. Therefore, the minimum value of k is 2.

Example 2:

Input: nums = [4,3,2,1], queries = [[1,3,2],[0,2,1]]

Output: -1

Explanation:

  • For i = 0 (l = 1, r = 3, val = 2):

  • Decrement values at indices [1, 2, 3] by [2, 2, 1] respectively.

  • The array will become [4, 1, 0, 0].

  • For i = 1 (l = 0, r = 2, val = 1):

  • Decrement values at indices [0, 1, 2] by [1, 1, 0] respectively.

  • The array will become [3, 0, 0, 0], which is not a Zero Array.

Constraints:

  • 1 <= nums.length <= 10^5
  • 0 <= nums[i] <= 5 * 10^5
  • 1 <= queries.length <= 10^5
  • queries[i].length == 3
  • 0 <= li <= ri < nums.length
  • 1 <= vali <= 5

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    int minZeroArray(vector<int>& nums, vector<vector<int>>& queries) {
        int m = queries.size();
        int left = 0, right = m + 1;
        // [left, right), ending point: left = right
        // for values < left the return value is false
        // for values >= right the return value is true
        while (left < right) {
            int mid = (left + right) / 2;
            if (check(mid, nums, queries)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return right == m + 1 ? -1 : left;
    }

    bool check(int k, vector<int> nums, vector<vector<int>>& queries) {
        int n = nums.size();
        vector<int> diffs(n + 1, 0);
        for (int i = 0; i < k; i++) {
            auto& q = queries[i];
            int l = q[0], r = q[1], v = q[2];
            diffs[l] -= v;
            diffs[r + 1] += v;
        }
        partial_sum(diffs.begin(), diffs.end(), diffs.begin());
        for (int i = 0; i < n; i++) {
            nums[i] = nums[i] + diffs[i];
            if (nums[i] > 0) {
                return false;
            }
        }
        return true;
    }
};