How do you solve Zero Array Transformation I on LeetCode?

Zero Array Transformation I (LeetCode #3355) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + q) time complexity and O(n) space complexity. Key insight: Using a difference array allows for efficient range updates, which is crucial for handling multiple queries.

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

The brute force approach for Zero Array Transformation I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves directly simulating the decrement operations for each query on the array. We check if we can decrement the values in the specified ranges to zero sequentially. The optimal Optimal Solution approach improves this to O(n + q).

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

Zero Array Transformation I uses the following concepts: Array, Prefix Sum. The recommended patterns to study are: Difference Array, Prefix Sum.

What are the interview tips for Zero Array Transformation I?

Always clarify the problem constraints and edge cases before diving into coding. Think about how to optimize your solution from the start, especially with large input sizes. Explain your thought process clearly as you code, as this shows your understanding and can help in debugging.

What are common mistakes when solving Zero Array Transformation I?

Not considering the edge cases where queries overlap or do not cover all indices. Failing to reset the array or difference array between test cases in a coding interview.

#3355

Zero Array Transformation I

Medium

ArrayPrefix SumDifference ArrayPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n + q)Space O(n)

Using a difference array allows us to efficiently apply the decrement operations for each query. We can then compute the prefix sum to determine the final values in the nums array.

⚙️

Algorithm

4 steps

1Step 1: Create a difference array of the same length as nums initialized to zero.
2Step 2: For each query [l, r], increment the difference array at index l and decrement at index r + 1.
3Step 3: Compute the prefix sum of the difference array to get the final decrements for each index in nums.
4Step 4: Check if all values in nums are zero after applying the decrements.

solution.py9 lines

1def canTransformToZeroArray(nums, queries):
2    diff = [0] * (len(nums) + 1)
3    for l, r in queries:
4        diff[l] += 1
5        if r + 1 < len(diff):
6            diff[r + 1] -= 1
7    for i in range(1, len(diff)):
8        diff[i] += diff[i - 1]
9    return all(nums[i] - diff[i] == 0 for i in range(len(nums)))

ℹ

Complexity note: This complexity is efficient because we only make a single pass through the queries and then a single pass through the nums array, leading to linear time complexity.

1Using a difference array allows for efficient range updates, which is crucial for handling multiple queries.
2The final check for zeros can be done in a single pass after applying all decrements.

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