#3355

Medium

LeetCode Problem Workspace

Zero Array Transformation I

Transform the given integer array into a zero array using range operations, focusing on prefix sum optimization and careful index handling.

array prefix sum

Problem Statement

You are given an integer array nums of length n and a list of queries, where each query specifies a subarray [li, ri]. Each query represents an operation that can increment or decrement elements in that subarray. The goal is to determine if after performing all queries, every element in nums can be zero.

A Zero Array is defined as an array where every element is equal to 0. For each query, you must evaluate whether applying the operation maintains the possibility of transforming the array into a zero array. Return true if it is possible and false otherwise.

Examples

Example 1

Input: nums = [1,0,1], queries = [[0,2]]

Output: true

Example 2

Input: nums = [4,3,2,1], queries = [[1,3],[0,2]]

Output: false

Constraints

1 <= nums.length <= 105
0 <= nums[i] <= 105
1 <= queries.length <= 105
queries[i].length == 2
0 <= li <= ri < nums.length

Solution Approach

Use a Difference Array

Initialize a difference array diff of length n+1. For each query [l, r], update diff[l] and diff[r+1] to track the net effect of increment/decrement operations efficiently without modifying the original array directly.

Compute Prefix Sum

Convert the difference array into cumulative values using prefix sums. This helps compute the net value at each index after all queries and allows for O(1) checks to see if the array element can be zero.

Validate Zero Array

Iterate over the prefix sum array and check if each position results in a zero after applying all queries. If any element cannot become zero, return false. Otherwise, return true.

Complexity Analysis

Metric	Value
Time	O(n + m)
Space	O(n)

Time complexity is O(n + m) where n is the length of nums and m is the number of queries, because each query is processed once and prefix sums are computed in a single pass. Space complexity is O(n) for the difference array.

What Interviewers Usually Probe

Check whether you are correctly using a difference array instead of modifying the original array per query.
Make sure your prefix sum computation correctly accumulates all range operations.
Consider edge cases where queries overlap or cover entire array sections.

Common Pitfalls or Variants

Common pitfalls

Updating the original array directly for each query leads to TLE for large n or m.
Incorrect handling of diff[r+1] can miscalculate net effects at array boundaries.
Forgetting to validate that all elements are zero after prefix sum accumulation.

Follow-up variants

Allowing queries that increment or decrement by different values, not just ±1.
Checking for partial zero arrays where only a subset must be zero.
Transforming using non-contiguous index sets instead of continuous ranges.

FAQ

What is the key pattern for Zero Array Transformation I?

The main pattern is using a difference array combined with prefix sums to track range operations efficiently.

Can I modify the original array directly for each query?

No, directly modifying the array per query is inefficient and may lead to timeouts for large inputs.

Why do we use a prefix sum after the difference array?

Prefix sums convert the difference array into actual net values at each index to check if they can become zero.

What should I watch out for with overlapping queries?

Ensure the difference array correctly accumulates overlapping range effects; otherwise, zero validation can fail.

What is the expected time complexity for this problem?

The solution runs in O(n + m) time, where n is nums length and m is the number of queries, due to efficient difference array and prefix sum usage.

terminal

Solution

Solution 1: Difference Array

We can use a difference array to solve this problem.

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class Solution:
    def isZeroArray(self, nums: List[int], queries: List[List[int]]) -> bool:
        d = [0] * (len(nums) + 1)
        for l, r in queries:
            d[l] += 1
            d[r + 1] -= 1
        s = 0
        for x, y in zip(nums, d):
            s += y
            if x > s:
                return False
        return True

Continue Practicing

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