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Author: iamAntimPal

3356. Zero Array Transformation II.py

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+/*
+
+3356. 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:
+        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