How do you solve Minimum Operations to Reduce X to Zero on LeetCode?

Minimum Operations to Reduce X to Zero (LeetCode #1658) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Thinking in reverse can simplify the problem — instead of removing elements, focus on the subarray that can be kept.

What is the brute force solution for Minimum Operations to Reduce X to Zero?

The brute force approach for Minimum Operations to Reduce X to Zero is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying every possible combination of removing elements from both ends of the array to see if we can reach the target value x. This is straightforward but inefficient as it checks all possible combinations. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Minimum Operations to Reduce X to Zero?

Always clarify the problem statement and constraints before jumping to solutions. Think about different approaches and their trade-offs; don't settle for the first solution. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Minimum Operations to Reduce X to Zero?

Failing to account for cases where x is larger than the total sum of the array. Not considering edge cases where the entire array might need to be removed.

#1658

Minimum Operations to Reduce X to Zero

Medium

ArrayHash TableBinary SearchSliding WindowPrefix SumSliding WindowTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

Instead of removing elements from both ends, we can think in reverse. We want to find a subarray whose sum equals the total sum of the array minus x. This allows us to minimize the number of operations by focusing on the maximum length of the subarray that can be kept.

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Algorithm

4 steps

1Step 1: Calculate the total sum of the array.
2Step 2: Determine the target sum we need to find, which is total sum - x.
3Step 3: Use a sliding window approach to find the longest subarray that sums to the target sum.
4Step 4: Calculate the minimum operations as the total length of the array minus the length of the found subarray.

solution.py16 lines

1def minOperations(nums, x):
2    total = sum(nums)
3    target = total - x
4    if target < 0:
5        return -1
6    left = 0
7    max_length = -1
8    current_sum = 0
9    for right in range(len(nums)):
10        current_sum += nums[right]
11        while current_sum > target:
12            current_sum -= nums[left]
13            left += 1
14        if current_sum == target:
15            max_length = max(max_length, right - left + 1)
16    return len(nums) - max_length if max_length != -1 else -1

ℹ

Complexity note: This complexity is linear because we traverse the array with a sliding window approach, adjusting the window size dynamically without needing nested loops.

1Thinking in reverse can simplify the problem — instead of removing elements, focus on the subarray that can be kept.
2The sliding window technique is powerful for finding subarrays with specific sum properties.

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