Watch 10 video solutions for Zero Array Transformation II, a medium level problem involving Array, Binary Search, Prefix Sum. This walkthrough by codestorywithMIK has 20,566 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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:
[li, ri] in nums by at most vali.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:
[0, 1, 2] by [1, 0, 1] respectively.[1, 0, 1].[0, 1, 2] by [1, 0, 1] respectively.[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:
[1, 2, 3] by [2, 2, 1] respectively.[4, 1, 0, 0].[0, 1, 2] by [1, 1, 0] respectively.[3, 0, 0, 0], which is not a Zero Array.
Constraints:
1 <= nums.length <= 1050 <= nums[i] <= 5 * 1051 <= queries.length <= 105queries[i].length == 30 <= li <= ri < nums.length1 <= vali <= 5Problem Overview: You are given an integer array nums and a list of range queries. Each query allows decreasing every element in a subarray by a certain value. The task is to determine the minimum number of queries (in order) required so that every element in nums becomes zero or negative. If it is impossible even after all queries, return -1.
Approach 1: Brute Force Simulation (O(n * m) time, O(1) space)
The direct approach applies queries one by one and updates the affected range in the array. For each query [l, r, val], iterate from l to r and subtract val from each element. After applying a query, scan the array to check if all values are <= 0. If so, return the number of queries used so far. This approach clearly shows the mechanics of the problem but becomes slow when both the array length and number of queries are large because every query touches up to n elements.
Approach 2: Binary Search + Difference Array (O((n + m) log m) time, O(n) space)
The key observation is that queries are applied in order, and we only care about the earliest prefix of queries that can reduce every index enough. This makes the problem monotonic: if the first k queries can zero the array, any k+1 queries will also work. That property allows binary search over the number of queries.
To efficiently evaluate whether the first k queries are sufficient, use a difference array for range updates. For each query [l, r, val], add val at diff[l] and subtract it at diff[r+1]. After processing the first k queries, compute the prefix sum to obtain the total decrement available for each index. If the accumulated decrement at index i is at least nums[i], that element can be reduced to zero. Otherwise, the first k queries are insufficient.
This technique converts repeated range updates into O(1) operations and a single prefix sum sweep. It leverages ideas from prefix sums and efficient range updates commonly used in array problems.
Recommended for interviews: The binary search + difference array approach is the expected solution. The brute force simulation demonstrates understanding of the operations, but it does not scale. Interviewers typically look for recognition of the monotonic condition and the use of a difference array to process range updates efficiently.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Simulation | O(n * m) | O(1) | Useful for understanding the mechanics of the queries or when constraints are very small. |
| Binary Search + Difference Array | O((n + m) log m) | O(n) | Optimal solution for large inputs where many range updates must be processed efficiently. |