Minimum Operations to Reduce X to Zero: Sliding Window Complement

LeetCode 1658, Minimum Operations to Reduce X to Zero, gives you an integer array nums and an integer x. In one operation, you can remove an element from either the left end or the right end of the array and subtract its value from x. Return the minimum number of operations to reduce x to exactly zero. If it is not possible, return -1.

For example, given nums = [1,1,4,2,3] and x = 5, the answer is 2. You can remove 2 and 3 from the right end (2 + 3 = 5) in just two operations.

Why this problem matters

This problem teaches one of the most powerful reframing tricks in algorithm design: the complement approach. Instead of directly solving "minimize elements removed from both ends," you flip it into "maximize a contiguous middle subarray." This inversion transforms a tricky two-sided removal problem into a classic sliding window problem.

Once you see this pattern, it shows up everywhere. Any time a problem asks you to minimize what you take from the edges of an array, consider maximizing what you keep in the middle. It is a mental shortcut that simplifies a whole family of problems.

The key insight

Every set of elements you remove comes from a prefix (left end) and a suffix (right end). What remains in between is a contiguous subarray. If the removed elements sum to x, then the remaining middle subarray sums to total_sum - x. Minimizing the number of removed elements is the same as maximizing the length of that middle subarray.

So the problem reduces to: find the longest contiguous subarray with sum equal to total_sum - x. If you find one with length L, the answer is n - L. If no such subarray exists, return -1.

The solution

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

    if target < 0:
        return -1
    if target == 0:
        return len(nums)

    best = -1
    window_sum = 0
    left = 0

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

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

        if window_sum == target:
            best = max(best, right - left + 1)

    return len(nums) - best if best != -1 else -1

The code first computes the target sum for the middle subarray. Then it uses a sliding window: expand right to grow the window, and when the window sum exceeds the target, shrink from left until it no longer exceeds. Each time the window sum exactly equals the target, record the window length if it is the longest seen so far.

Visual walkthrough

The window slides across the array, expanding to the right and shrinking from the left whenever the sum exceeds the target. The longest window that hits exactly target = 6 has length 3 (the subarray [1, 1, 4]), giving us an answer of 5 - 3 = 2 operations.

Complexity analysis

Time: O(n). Each element is visited at most twice, once when right passes over it and once when left passes over it. Both pointers traverse the array at most once.

Space: O(1). We only use a few variables: left, window_sum, best, and target. No additional data structures are needed.

The building blocks

1. Complement/inversion trick

The core reframing: instead of minimizing removals from both ends, maximize the contiguous middle. This reduces to computing a target and searching for it.

target = sum(nums) - x

If target is negative, no solution exists because even keeping nothing in the middle (removing everything) would not remove enough. If target is zero, you need to remove everything, so the answer is n.

2. Sliding window for exact target sum

With all positive integers, the sliding window works perfectly. As you expand, the sum grows. As you shrink, the sum decreases. This monotonic behavior guarantees you never miss a valid window.

for right in range(len(nums)):
    window_sum += nums[right]
    while window_sum > target:
        window_sum -= nums[left]
        left += 1
    if window_sum == target:
        best = max(best, right - left + 1)

The key difference from "minimum length subarray" problems is that here you want the longest window that hits the target exactly, not the shortest that exceeds it. So you record the length when the sum equals the target, not when it first becomes valid.

Edge cases

• target is negative: sum(nums) is less than x, so it is impossible to reduce x to zero even by removing every element. Return -1.
• target is zero: The entire array must be removed. Return n.
• No subarray sums to target: The sliding window never finds a match, best stays -1, and we return -1.
• Single element equals x: The target is total - x. If removing one element from either end gives x, the complement window has length n - 1.
• Entire array sums to x: Same as target = 0. Every element is removed, answer is n.
• Multiple valid windows of the same length: The algorithm naturally picks the longest, so ties do not matter.

From understanding to recall

You now understand the complement trick and how it turns edge removals into a sliding window search. But the details matter in an interview: remembering to check for target < 0, knowing to maximize window length rather than minimize it, and correctly returning n - best instead of best itself. Spaced repetition drills these specifics at increasing intervals so that the full solution, including edge cases and the return logic, becomes automatic recall rather than something you have to re-derive under pressure.

Related posts

• Minimum Size Subarray Sum - Classic sliding window finding shortest subarray with sum at least k
• Subarray Sum Equals K - Using prefix sums to find subarrays with a target sum
• Maximum Subarray - Another subarray optimization using a different technique

Mastering the complement reframing and sliding window mechanics prepares you for a wide range of subarray problems. CodeBricks helps you build this kind of deep pattern recognition through structured practice and spaced repetition.