Watch 10 video solutions for Zero Array Transformation I, a medium level problem involving Array, Prefix Sum. This walkthrough by Techdose has 7,132 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].
For each queries[i]:
[li, ri] in nums.A Zero Array is an array where all elements are equal to 0.
Return true if it is possible to transform nums into a Zero Array after processing all the queries sequentially, otherwise return false.
Example 1:
Input: nums = [1,0,1], queries = [[0,2]]
Output: true
Explanation:
[0, 2] and decrement the values at these indices by 1.[0, 0, 0], which is a Zero Array.Example 2:
Input: nums = [4,3,2,1], queries = [[1,3],[0,2]]
Output: false
Explanation:
[1, 2, 3] and decrement the values at these indices by 1.[4, 2, 1, 0].[0, 1, 2] and decrement the values at these indices by 1.[3, 1, 0, 0], which is not a Zero Array.
Constraints:
1 <= nums.length <= 1050 <= nums[i] <= 1051 <= queries.length <= 105queries[i].length == 20 <= li <= ri < nums.lengthProblem Overview: You are given an integer array nums and several range queries. Each query allows decreasing every element in the interval [l, r] by 1 exactly once. The goal is to determine whether these operations provide enough decrements to reduce every value in nums to zero without running out of operations for any index.
Approach 1: Direct Simulation of Operations (O(n * q) time, O(1) space)
The straightforward idea is to simulate every query exactly as described. For each query [l, r], iterate through the range and decrement the corresponding elements in nums. After processing all queries, check whether every element became zero. This approach mirrors the problem statement and is useful for validating logic on small inputs.
The drawback is performance. If the array size is n and the number of queries is q, repeatedly iterating across ranges leads to O(n * q) time in the worst case. Large inputs quickly make this approach too slow. It also mutates the original array, which may require copying if the input must be preserved.
Approach 2: Efficient Interval Update using Difference Array (O(n + q) time, O(n) space)
A more scalable solution focuses on counting how many decrement operations are available for each index instead of applying them directly. Each query contributes one potential decrement to every element in its range. Using a difference array, mark +1 at l and -1 at r + 1. After processing all queries, compute a prefix sum to determine how many operations cover each index.
This converts multiple range updates into constant‑time operations. The prefix sum pass reconstructs the total number of decrements available at every position. If the coverage count at index i is less than nums[i], the element cannot reach zero because it lacks enough decrement operations. If every index has sufficient coverage, the transformation is possible.
This technique relies on the classic prefix sum pattern combined with interval marking from the array toolkit. It avoids repeated range iteration and processes the entire input in linear time.
Recommended for interviews: The difference array approach is the expected solution. Interviewers look for the observation that queries only contribute counts of operations, not the actual decrements themselves. Implementing the prefix accumulation demonstrates strong understanding of prefix sums and efficient range updates. Starting with the simulation approach can help explain the intuition before optimizing to the linear-time solution.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Direct Simulation of Operations | O(n * q) | O(1) | Good for understanding the problem or when constraints are very small |
| Difference Array + Prefix Sum | O(n + q) | O(n) | Best for large inputs with many range queries; converts range updates to constant time |