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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 <= 105
  • 0 <= nums[i] <= 5 * 105
  • 1 <= queries.length <= 105
  • queries[i].length == 3
  • 0 <= li <= ri < nums.length
  • 1 <= vali <= 5

Solutions

Solution 1

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class Solution:
    def minZeroArray(self, nums: List[int], queries: List[List[int]]) -> int:
        def check(k: int) -> bool:
            d = [0] * (len(nums) + 1)
            for l, r, val in queries[:k]:
                d[l] += val
                d[r + 1] -= val
            s = 0
            for x, y in zip(nums, d):
                s += y
                if x > s:
                    return False
            return True

        m = len(queries)
        l = bisect_left(range(m + 1), True, key=check)
        return -1 if l > m else l

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