Zero Array Transformation IV solution and explanation
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The problem
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:
- Select a subset of indices in the range
[li, ri]fromnums. - Decrement the value at each selected index by exactly
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:
- For query 0 (l = 0, r = 2, val = 1):
- Decrement the values at indices
[0, 2]by 1. - The array will become
[1, 0, 1].
- Decrement the values at indices
- For query 1 (l = 0, r = 2, val = 1):
- Decrement the values at indices
[0, 2]by 1. - The array will become
[0, 0, 0], which is a Zero Array. Therefore, the minimum value ofkis 2.
- Decrement the values at indices
Example 2:
Input: nums = [4,3,2,1], queries = [[1,3,2],[0,2,1]]
Output: -1
Explanation:
It is impossible to make nums a Zero Array even after all the queries.
Example 3:
Input: nums = [1,2,3,2,1], queries = [[0,1,1],[1,2,1],[2,3,2],[3,4,1],[4,4,1]]
Output: 4
Explanation:
- For query 0 (l = 0, r = 1, val = 1):
- Decrement the values at indices
[0, 1]by1. - The array will become
[0, 1, 3, 2, 1].
- Decrement the values at indices
- For query 1 (l = 1, r = 2, val = 1):
- Decrement the values at indices
[1, 2]by 1. - The array will become
[0, 0, 2, 2, 1].
- Decrement the values at indices
- For query 2 (l = 2, r = 3, val = 2):
- Decrement the values at indices
[2, 3]by 2. - The array will become
[0, 0, 0, 0, 1].
- Decrement the values at indices
- For query 3 (l = 3, r = 4, val = 1):
- Decrement the value at index 4 by 1.
- The array will become
[0, 0, 0, 0, 0]. Therefore, the minimum value ofkis 4.
Example 4:
Input: nums = [1,2,3,2,6], queries = [[0,1,1],[0,2,1],[1,4,2],[4,4,4],[3,4,1],[4,4,5]]
Output: 4
Constraints:
1 <= nums.length <= 100 <= nums[i] <= 10001 <= queries.length <= 1000queries[i] = [li, ri, vali]0 <= li <= ri < nums.length1 <= vali <= 10
Approach overview
Problem Overview: You are given an integer array and a sequence of allowed transformations. Each transformation affects a range or position and reduces values according to specific rules. The goal is to determine whether you can apply these operations so every element in the array becomes exactly zero.
Approach 1: Brute Force Simulation (Exponential Time, O(2^m) time, O(m) space)
The most direct strategy is to try every possible combination of operations. For each subset of operations, simulate how the array changes and check if all values become zero. This approach repeatedly copies the array and applies reductions step by step. While this demonstrates the mechanics of the problem, the number of combinations grows exponentially with the number of operations, making it impractical for large inputs.
Approach 2: Dynamic Programming Over Operations (O(n * m) time, O(n) space)
A better strategy models the process using dynamic programming. Iterate through operations while maintaining the effective reduction applied to each index. The DP state tracks how much reduction has been accumulated up to a certain position. When processing an operation, update the affected range and propagate its impact forward. This avoids recomputing transformations from scratch and reduces the complexity to O(n * m), where n is the array size and m is the number of operations.
Approach 3: Prefix Accumulation + Greedy DP (O(n + m) time, O(n) space)
The optimized solution combines prefix accumulation with a lightweight DP decision process. Instead of applying each operation directly to the array, track the net effect using a difference array. As you scan from left to right, maintain the current cumulative reduction and decide whether additional operations must be activated to keep the value non‑negative and eventually reach zero. This technique relies on fast prefix updates, a common pattern in array problems, and turns repeated range updates into constant‑time adjustments. The result is a linear pass across the array with O(n + m) time and O(n) auxiliary space.
Recommended for interviews: Interviewers typically expect the prefix‑based dynamic programming approach. Brute force shows you understand the transformation rules, but the optimized method demonstrates stronger problem‑solving skills by converting repeated range updates into prefix effects and processing the array in a single pass. Problems like this frequently combine ideas from dynamic programming and prefix techniques, so recognizing that pattern is key.
The written solution for this problem is still being prepared. Try it yourself in the meantime.
Try solving it yourselfComplexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Simulation | O(2^m) | O(m) | Useful only for very small inputs or understanding the transformation rules |
| Dynamic Programming Over Operations | O(n * m) | O(n) | General case when operations must be evaluated sequentially |
| Prefix Accumulation + Greedy DP | O(n + m) | O(n) | Optimal approach for large arrays and many operations |
Frequently asked questions
Is Zero Array Transformation IV easy or hard?
Zero Array Transformation IV is considered a medium difficulty problem. The challenge comes from recognizing that direct simulation is inefficient and that prefix accumulation with dynamic programming can reduce the complexity significantly.
How to solve Zero Array Transformation IV in O(n)?
Use a prefix or difference array to track cumulative reductions. While scanning the array, maintain the current applied reduction and activate operations only when necessary to ensure each value can be reduced to zero. This converts range updates into constant‑time adjustments and results in a linear pass.
Zero Array Transformation IV Python or Java solution?
The algorithm can be implemented in Python, Java, C++, or Go using the same idea: maintain a prefix difference array, iterate through the main array, update cumulative reductions, and apply operations when needed. The logic remains identical across languages.
What is the best approach for Zero Array Transformation IV?
The most efficient approach uses prefix accumulation combined with dynamic programming. Instead of applying every operation directly, track the net effect using a difference or prefix array and process the array from left to right. This reduces repeated updates and achieves O(n + m) time with O(n) extra space.
What data structure is used in Zero Array Transformation IV?
The optimal solution primarily uses arrays with prefix or difference tracking. Dynamic programming logic determines when operations must be applied, while the prefix structure efficiently accumulates the effects of range updates.
What is the time complexity of Zero Array Transformation IV?
The optimized solution runs in O(n + m) time, where n is the array length and m is the number of operations. Earlier dynamic programming approaches may take O(n * m) time because each operation potentially affects multiple indices.
Is Zero Array Transformation IV asked at Google, Amazon, or Meta?
Problems combining range updates, prefix techniques, and dynamic programming frequently appear in interviews at companies like Amazon, Google, and Meta. Variants of transformation or prefix‑update array problems are common in coding rounds because they test optimization and data‑structure knowledge.
