3356. Zero Array Transformation II

Problem Link

Description

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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):

  <ul>
  	<li>Decrement values at indices <code>[0, 1, 2]</code> by <code>[1, 0, 1]</code> respectively.</li>
  	<li>The array will become <code>[1, 0, 1]</code>.</li>
  </ul>
  </li>
  <li><strong>For i = 1 (l = 0, r = 2, val = 1):</strong>
  <ul>
  	<li>Decrement values at indices <code>[0, 1, 2]</code> by <code>[1, 0, 1]</code> respectively.</li>
  	<li>The array will become <code>[0, 0, 0]</code>, which is a Zero Array. Therefore, the minimum value of <code>k</code> is 2.</li>
  </ul>
  </li>
  

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):

  <ul>
  	<li>Decrement values at indices <code>[1, 2, 3]</code> by <code>[2, 2, 1]</code> respectively.</li>
  	<li>The array will become <code>[4, 1, 0, 0]</code>.</li>
  </ul>
  </li>
  <li><strong>For i = 1 (l = 0, r = 2, val<span style="font-size: 13.3333px;"> </span>= 1):</strong>
  <ul>
  	<li>Decrement values at indices <code>[0, 1, 2]</code> by <code>[1, 1, 0]</code> respectively.</li>
  	<li>The array will become <code>[3, 0, 0, 0]</code>, which is not a Zero Array.</li>
  </ul>
  </li>
  

 

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

Solution

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Python3

class Solution:
    def minZeroArray(self, nums: List[int], queries: List[List[int]]) -> int:
        n = len(nums)
        cnt = [0 for _ in range(n + 1)]
        s = k = 0
        for i in range(n):
            while s + cnt[i] < nums[i]:
                k += 1
                if k - 1 >= len(queries): return -1
                l, r, val = queries[k - 1]
                if r < i: continue
                cnt[max(l, i)] += val
                cnt[r + 1] -= val
            s += cnt[i]
        return k