How do you solve Zero Array Transformation II on LeetCode?

Zero Array Transformation II (LeetCode #3356) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log m) time complexity and O(n) space complexity. Key insight: Using a difference array allows for efficient range updates.

What is the time complexity of Zero Array Transformation II?

The optimal solution for Zero Array Transformation II runs in O(n log m) time and uses O(n) space. The binary search runs in O(log m) where m is the number of queries, and for each mid, we apply the difference array in O(n).

What is the brute force solution for Zero Array Transformation II?

The brute force approach for Zero Array Transformation II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves processing each query one by one and checking if the array can be transformed into a zero array after each query. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log m).

What algorithm or data structure is used to solve Zero Array Transformation II?

Zero Array Transformation II uses the following concepts: Array, Two Pointers, Binary Search, Prefix Sum. The recommended patterns to study are: Binary Search, Difference Array.

What are the interview tips for Zero Array Transformation II?

Always think about how to optimize your brute-force solution. Consider edge cases, like when no queries are needed or when the array is already zero. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Zero Array Transformation II?

Not considering the case where some queries may not be needed to zero out the array. Failing to reset the temporary array correctly between checks.

#3356

Zero Array Transformation II

Medium

ArrayTwo PointersBinary SearchPrefix SumBinary SearchDifference Array

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log m)
Space	O(1)	O(n)

💡

Intuition

Time O(n log m)Space O(n)

We can use a binary search combined with a difference array to efficiently determine the minimum number of queries needed to transform nums into a zero array. This approach reduces unnecessary checks and speeds up the process.

⚙️

Algorithm

3 steps

1Step 1: Use a binary search to find the minimum k such that applying the first k queries can zero out the nums array.
2Step 2: For each mid value in the binary search, use a difference array to apply the decrements efficiently.
3Step 3: Check if the resulting nums array can be zeroed out after applying the decrements from the first mid queries.

solution.py22 lines

1def min_queries_to_zero(nums, queries):
2    def can_zero(k):
3        diff = [0] * (len(nums) + 1)
4        for i in range(k):
5            l, r, val = queries[i]
6            diff[l] += val
7            if r + 1 < len(diff):
8                diff[r + 1] -= val
9        current = 0
10        for i in range(len(nums)):
11            current += diff[i]
12            if nums[i] > current:
13                return False
14        return True
15    left, right = 0, len(queries)
16    while left < right:
17        mid = (left + right) // 2
18        if can_zero(mid + 1):
19            right = mid
20        else:
21            left = mid + 1
22    return left if left < len(queries) else -1

ℹ

Complexity note: The binary search runs in O(log m) where m is the number of queries, and for each mid, we apply the difference array in O(n).

1Using a difference array allows for efficient range updates.
2Binary search helps in efficiently finding the minimum number of queries needed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.