Minimum Operations to Reduce X to Zero (Leetcode 1658)

Pattern:

Idea:

Variations :

• part of Subarray with Given Sum — Important Interview Variations (Solutions)
• part of Complementary Counting

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💻 Code

def isPalindrome(x):
    if x < 0:
        return False

    original = x
    rev = 0

    while x > 0:
        digit = x % 10
        rev = rev * 10 + digit
        x //= 10

    return original == rev

Time complexity - O(D) , D is no of digits

Aux. Space complexity - O(1)

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Minimum Operations to Reduce X to Zero (Leetcode 1658)

Tags: #SlidingWindow #TwoPointers #PrefixSum #Arrays #Optimization #Interview-Pattern #LeetCode #FAANG

Problem Statement

Given an integer array nums and an integer x, you may remove elements only from the left or the right end of the array.

Each removed element is subtracted from x.

Return the minimum number of operations needed to reduce x to exactly 0. If impossible, return -1.

Example

• nums = [1,1,4,2,3]

• x = 5

• Answer: 2 (2 + 3 removed from the right)

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Key Idea

Instead of deciding what to remove, find what to keep.

Let:

• total = sum(nums)

• Removed sum = x

Then the kept middle subarray must have sum:

So the problem becomes:

Find the longest subarray whose sum is total - x.

If that subarray has length L, then:

This transformation is the entire interview trick.

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Intuition (The WHY)

Removing only from both ends always leaves one contiguous middle subarray.

Example:

[1, 1, 4, 2, 3]
 ↑           ↑
remove     remove

Remaining:

[1, 1, 4]

Instead of exploring every left/right combination, maximize the length of the kept middle segment.

Minimum removed = Maximum kept.

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Approach — Longest Subarray with Sum = Target

Since all numbers are positive in this problem, the target-sum subarray can be found using a variable-size sliding window.

Algorithm

1. Compute target = sum(nums) - x.

2. Edge cases:

   • target < 0 → impossible

   • target == 0 → remove every element

3. Use sliding window to find the longest subarray with sum = target.

4. Return n - longest.

Python Solution

def minOperations(nums, x):
    target = sum(nums) - x

    if target < 0:
        return -1

    if target == 0:
        return len(nums)

    left = 0
    curr = 0
    longest = -1

    for right in range(len(nums)):
        curr += nums[right]

        while curr > target:
            curr -= nums[left]
            left += 1

        if curr == target:
            longest = max(longest, right - left + 1)

    return -1 if longest == -1 else len(nums) - longest

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Dry Run

nums = [1,1,4,2,3]

x = 5

Total sum:

Target:

Find the longest subarray with sum 6.

Window	Sum
[1]	1
[1,1]	2
[1,1,4]	6 ✅
[1,1,4,2]	8
[4,2]	6 ✅

Longest valid window:

[1, 1, 4]

Length = 3

Operations:

Answer = 2

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Why Sliding Window Works Here

Sliding window requires the sum to behave monotonically.

Since every element is positive:

• Expand → sum increases

• Shrink → sum decreases

This guarantees every pointer moves only forward, giving linear time.

If negative numbers were allowed, this approach would fail and we’d use Prefix Sum + HashMap instead.

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Edge Cases

Case 1 — Impossible

nums = [5,6,7]
x = 30

Target:

Negative target is impossible.

Return -1.

Case 2 — Remove Everything

nums = [3,2,5]
x = 10

Target:

Keep nothing.

Answer = 3.

Case 3 — No Valid Middle

nums = [1,1,1]
x = 2

Target:

Longest window = [1]

Operations = 3 - 1 = 2.

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Complexity

Metric	Value
Time	O(n)
Auxiliary Space	O(1)

The window pointers each traverse the array at most once.

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Relationship to Previous Problems

Original Problem	Transformed Into
Remove ends to make sum = x	Longest subarray with sum = total-x
Longest Subarray with Sum K	Same sliding-window pattern
Count Subarrays with Sum K	Prefix Sum + HashMap

This is a classic problem reversal interview pattern.

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Common Mistakes

1. Solving the removal problem directly

Many candidates try DFS or two-pointer removal from both ends, leading to exponential or quadratic solutions.

Instead, transform it into a kept subarray problem.

2. Forgetting target == 0

If:

the answer is simply:

return len(nums)

3. Using Prefix Sum Unnecessarily

Because all numbers are positive, sliding window is both simpler and more space-efficient.

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Key Takeaways / Pattern Recognition

• Remove from both ends almost always means keep one contiguous middle subarray.

• Convert:

  • Removed Sum = x

  • Kept Sum = total - x

• Then maximize the kept length.

• This is one of the most important examples of reframing an optimization problem into a familiar sliding-window pattern.