3643 Zero Array Transformation Ii

Zero Array Transformation II

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

class Solution:
    def minZeroArray(self, nums: List[int], queries: List[List[int]]) -> int:
        n = len(nums)
        left, right = 0, len(queries)

        # Zero array isn't formed after all queries are processed
        if not self.can_form_zero_array(nums, queries, right):
            return -1

        # Binary Search
        while left <= right:
            middle = left + (right - left) // 2
            if self.can_form_zero_array(nums, queries, middle):
                right = middle - 1
            else:
                left = middle + 1

        # Return earliest query that zero array can be formed
        return left

    def can_form_zero_array(
        self, nums: List[int], queries: List[List[int]], k: int
    ) -> bool:
        n = len(nums)
        total_sum = 0
        difference_array = [0] * (n + 1)

        # Process query
        for query_index in range(k):
            start, end, val = queries[query_index]

            # Process start and end of range
            difference_array[start] += val
            difference_array[end + 1] -= val

        # Check if zero array can be formed
        for num_index in range(n):
            total_sum += difference_array[num_index]
            if total_sum < nums[num_index]:
                return False
        return True