Minimum Operations to Reduce X to Zero — LC 1658 Complement Sliding Window

Problem Statement#

Remove leftmost or rightmost elements from nums, subtracting each from x. Return the minimum number of removals to make x exactly zero, or -1 if impossible.

Constraints:

• 1 less than or equal to nums.length less than or equal to 10 to the 5
• 1 less than or equal to nums[i] less than or equal to 10 to the 4
• 1 less than or equal to x less than or equal to 10 to the 9

Input:  nums = [1, 1, 4, 2, 3], x = 5
Output: 2

Input:  nums = [5, 6, 7, 8, 9], x = 4
Output: -1

Why This Problem Matters#

LeetCode 1658 — Minimum Operations to Reduce X to Zero is the trickiest two-pointer problem in the medium tier because the obvious DFS-from-both-ends solution times out, and the elegant reformulation requires a leap of insight. Amazon, Google, and Meta use it as a 30-minute mid-loop screen specifically to test problem reformulation.

The reformulation is identical in spirit to LC 1423 (Maximum Points You Can Obtain from Cards): if you remove a prefix and a suffix that sum to x, the remaining middle is a contiguous subarray that sums to total - x. Maximizing the middle length minimizes the number of removals. Once you spot this, the problem collapses to LC 325 (Maximum Size Subarray Sum Equals K) with a clean sliding window.

In production, the same reframe surfaces in budget-trimming algorithms, where you cut from both ends until you hit a quota.

The Core Insight#

Direct simulation is exponential because at each step you choose either end. Reframe instead:

• Total sum S is fixed.
• The elements you keep form a contiguous middle subarray.
• If their sum equals S - x, then the removed prefix and suffix together sum to x.
• Minimum removals equals n minus the maximum length of a contiguous subarray with sum equal to S - x.

So the problem becomes: find the longest subarray with sum exactly target = S - x. If target less than 0, no answer; if target equals 0, the longest such subarray is empty and the answer is n.

Because all values are positive, prefix sums are strictly increasing, which lets us use a sliding window. As right advances, shrink left while current greater than target, then check equality.

Visual Dry Run#

Input: nums = [1, 1, 4, 2, 3], x = 5, S = 11, target = 6.

Best length 3, answer 5 - 3 = 2.

Solution (Optimal)#

class Solution:
    def minOperations(self, nums, x):
        target = sum(nums) - x
        if target < 0:
            return -1
        if target == 0:
            return len(nums)
        left, current, best = 0, 0, -1
        for right, value in enumerate(nums):
            current += value
            while current > target and left <= right:
                current -= nums[left]
                left += 1
            if current == target:
                if right - left + 1 > best:
                    best = right - left + 1
        return -1 if best == -1 else len(nums) - best

var minOperations = function(nums, x) {
    let total = 0;
    for (const v of nums) total += v;
    const target = total - x;
    if (target < 0) return -1;
    if (target === 0) return nums.length;
    let left = 0, current = 0, best = -1;
    for (let right = 0; right < nums.length; right++) {
        current += nums[right];
        while (current > target && left <= right) {
            current -= nums[left];
            left++;
        }
        if (current === target) {
            const len = right - left + 1;
            if (len > best) best = len;
        }
    }
    return best === -1 ? -1 : nums.length - best;
};

Time: O(n) — single linear sliding window. Space: O(1).

Common Mistakes#

• Trying DFS or DP on both ends. Exponential and times out.
• Forgetting target less than 0 short circuit when x greater than total.
• Forgetting target equals 0 case where the entire array must be removed.
• Comparing current less than or equal to target after the shrink instead of strict equality.
• Returning the best length instead of n minus best.

Interview Tips#

• Pitch the inversion explicitly: "Maximize the middle subarray, do not simulate the removals."
• Note the prerequisite: positive values are required for the sliding window.
• Walk through the target equals 0 and target less than 0 cases before coding.
• Mention prefix sums plus hash map as a backup for negative inputs.
• Trace once on a small example so the interviewer sees the bookkeeping.

Follow-up Questions#

• What if values can be negative? Switch to prefix sum plus hashmap; sliding window breaks.
• Can you return the actual removed sequence? Track the boundary indices when best updates.
• What if you must minimize the suffix length specifically? Becomes a one-sided two-pointer over suffix sums.
• What if multiple test cases share the array but vary x? Precompute prefix sums once and binary search per query.
• Stream variant: process additional appended values. Maintain prefix sum hash map online.

Key Takeaways#

• LeetCode 1658 — Minimum Operations to Reduce X to Zero solves in O(n) time and O(1) space.
• Reframe: minimum removals equals n minus the longest middle subarray summing to total - x.
• Positive values make prefix sums monotonic, enabling the sliding window.
• Always handle target less than 0 and target equals 0 separately.
• Asked at Amazon, Google, Meta, and Microsoft as a mid-loop reformulation problem.
• Same complement insight applies to LC 1423 (Maximum Points from Cards).
• Without positivity, fall back to prefix sum plus hash map.