LeetCode 1658: Minimum Operations to Reduce X to Zero

Meta Data

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Difficulty: medium First Attempt: 2025-08-03

• Total time: 00:00.00

Intuition

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We can solve this problem by thinking about it in reverse. Instead of finding the minimum number of operations to reduce x to zero, we can find the maximum length of a subarray whose sum equals sum(nums) - x. This is because if we remove elements from both ends to reduce x to zero, the remaining elements must sum to sum(nums) - x. The minimum operations needed will be len(nums) - max_subarray_length.

Approach

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Reverse Thinking with Sliding Window

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class Solution:
    def minOperations(self, nums: List[int], x: int) -> int:
        
        target_sum = sum(nums) - x

        # Handle edge cases
        if target_sum < 0:
            return -1
        if target_sum == 0:
            return len(nums)
    
        max_len = 0
        left = sum_now = 0

        # Find maximum subarray with sum equal to target_sum
        for right in range(len(nums)):
            sum_now += nums[right]

            # Shrink window if sum exceeds target
            while sum_now > target_sum:
                sum_now -= nums[left]
                left += 1
            
            # Update maximum length if sum equals target
            if sum_now == target_sum:
                max_len = max(max_len, right - left + 1)

        return len(nums) - max_len if max_len != 0 else -1

Algorithm Analysis

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Time Complexity

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• Time: O(n) where n is the length of the input array
• Space: O(1) since we only use a constant amount of extra space

Key Insights

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1. Reverse Thinking: Transform the problem to find maximum subarray with specific sum
2. Sliding Window: Use sliding window to find the longest subarray with target sum
3. Edge Case Handling: Handle cases where target sum is negative or zero

Findings

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1. Reverse Problem Transformation: This problem demonstrates how thinking about a problem in reverse can simplify the solution. Instead of finding minimum operations, we find the maximum subarray length.

2. Sliding Window Application: The transformed problem becomes a classic sliding window problem - finding the longest subarray with a specific sum.

3. Target Sum Calculation: The key insight is that target_sum = sum(nums) - x represents the sum of elements we want to keep in the middle.

4. Window Management: We maintain a sliding window and adjust its size based on whether the current sum equals, exceeds, or is less than the target sum.

5. Edge Case Handling: We need to handle several edge cases: when target sum is negative (impossible), when target sum is zero (remove all elements), and when no valid subarray is found.

6. Result Calculation: The final result is len(nums) - max_len, representing the number of elements we need to remove from both ends.

7. Single Pass Solution: The solution requires only one pass through the array, making it very efficient for large inputs.

8. Memory Efficiency: O(1) space complexity makes this solution very memory-efficient, as we only need a few variables to track the current state.

Encountered Problems

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1. Initial Approach Confusion: I initially tried to solve this directly by removing elements from both ends, which was complex and inefficient. The key insight was the reverse thinking approach.

2. Target Sum Understanding: Understanding that target_sum = sum(nums) - x was crucial. I had to think carefully about what this represents in the context of the original problem.

3. Edge Case Management: I had to handle multiple edge cases: negative target sum, zero target sum, and cases where no valid subarray exists.