How do you solve Zero Array Transformation III on LeetCode?

Zero Array Transformation III (LeetCode #3362) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Greedy approach maximizes efficiency.

What is the time complexity of Zero Array Transformation III?

The optimal solution for Zero Array Transformation III runs in O(n log n) time and uses O(n) space. The sorting step dominates the complexity, making it efficient compared to brute force.

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

The brute force approach for Zero Array Transformation III is Brute Force, with O(n²) time complexity and O(1) space complexity. Try every combination of removing queries and check if the remaining can zero out the array. This is inefficient as it explores all possibilities. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Zero Array Transformation III?

Explain your thought process clearly. Discuss edge cases and how to handle them. Practice similar problems to build confidence.

What are common mistakes when solving Zero Array Transformation III?

Not considering all ranges when decrementing. Ignoring edge cases where nums cannot be zeroed.

#3362

Zero Array Transformation III

Medium

ArrayTwo PointersGreedySortingHeap (Priority Queue)Prefix SumGreedyPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

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

Sort the queries and use a greedy approach to maximize the number of removable queries while ensuring the remaining can still zero out the array.

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Algorithm

3 steps

1Step 1: Sort queries based on their end index.
2Step 2: Use a prefix sum to track the total decrements needed for each index.
3Step 3: Iterate through queries, checking if removing the last query still allows nums to be zeroed out.

solution.py15 lines

1def maxRemovable(nums, queries):
2    queries.sort(key=lambda x: x[1])
3    total_decrements = [0] * (len(nums) + 1)
4    for l, r in queries:
5        total_decrements[l] += 1
6        if r + 1 < len(total_decrements):
7            total_decrements[r + 1] -= 1
8    for i in range(1, len(total_decrements)):
9        total_decrements[i] += total_decrements[i - 1]
10    max_removable = 0
11    for i in range(len(nums)):
12        if total_decrements[i] > nums[i]:
13            return -1
14        max_removable += total_decrements[i]
15    return len(queries) - max_removable

ℹ

Complexity note: The sorting step dominates the complexity, making it efficient compared to brute force.

1Greedy approach maximizes efficiency.
2Sorting helps in managing overlapping ranges.

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