#1658

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Minimum Operations to Reduce X to Zero

Find the minimum number of operations to reduce a value x to zero by removing elements from an array.

array hash table binary search sliding window prefix sum

Problem Statement

You are given an integer array nums and a target integer x. In one operation, you can either remove the leftmost or rightmost element of nums and subtract its value from x. The goal is to determine the minimum number of operations required to reduce x to exactly 0. If it is impossible, return -1.

To solve this, you need to identify a subarray with a sum equal to the remaining value of x, and the length of that subarray will determine the number of operations required. The key to an efficient solution lies in finding the longest subarray with the sum equal to the target using a sliding window approach.

Examples

Example 1

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

Output: 2

The optimal solution is to remove the last two elements to reduce x to zero.

Example 2

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

Output: -1

Example details omitted.

Example 3

Input: nums = [3,2,20,1,1,3], x = 10

Output: 5

The optimal solution is to remove the last three elements and the first two elements (5 operations in total) to reduce x to zero.

Constraints

1 <= nums.length <= 105
1 <= nums[i] <= 104
1 <= x <= 109

Solution Approach

Sliding Window + Hash Lookup

By using a sliding window to find the largest subarray whose sum is equal to the remaining value of x, we can efficiently calculate the required number of operations. This approach utilizes a hash map to store sums of subarrays and their lengths to speed up the process.

Two-Pointer Technique

The two-pointer technique helps in dynamically adjusting the window of elements, ensuring that we explore every possible subarray that could sum to the remaining value of x. One pointer starts at the leftmost element, and the other at the rightmost, progressively shrinking or expanding the window as needed.

Binary Search Optimization

For further optimization, binary search can be used to quickly locate the largest possible subarray that sums to x. By sorting the array, we can efficiently search for subarrays with specific sums, improving the time complexity in some cases.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach chosen, with the sliding window and hash lookup method typically offering a time complexity of O(n). The space complexity also depends on the hash map's size, which can grow linearly with the input size.

What Interviewers Usually Probe

Understand how to efficiently find subarrays with a sum equal to a target value.
Can articulate the trade-offs between different sliding window techniques.
Demonstrates the ability to optimize solutions with binary search or hash-based lookups.

Common Pitfalls or Variants

Common pitfalls

Failing to handle edge cases where no subarray can sum to x, leading to a return value of -1.
Incorrectly managing the array's boundaries when adjusting the window, causing out-of-bound errors.
Overcomplicating the solution with brute-force methods that result in suboptimal time complexity.

Follow-up variants

Modify the problem to find the minimum number of operations to reduce x to any given target.
Introduce constraints that prevent removal from both ends, restricting operations to one side of the array.
Extend the problem to handle negative numbers or floating point values in the array.

FAQ

What is the primary pattern used in the Minimum Operations to Reduce X to Zero problem?

The primary pattern involves scanning the array using a sliding window technique combined with hash lookups to find the longest subarray with the required sum.

What approach is optimal for solving the problem Minimum Operations to Reduce X to Zero?

The optimal approach uses a sliding window with hash lookup to efficiently find the longest subarray that sums to x.

How do sliding window techniques help in solving array problems like Minimum Operations to Reduce X to Zero?

Sliding window techniques allow for dynamically adjusting the window of elements, which helps efficiently find subarrays with specific sums.

Why is binary search sometimes used to optimize the solution for this problem?

Binary search can optimize the solution by quickly locating subarrays that sum to the target value, reducing the search time compared to brute force methods.

What is the time complexity of solving Minimum Operations to Reduce X to Zero?

The time complexity typically ranges from O(n) to O(n log n), depending on the specific approach used, with sliding window offering the most efficient solution.

terminal

Solution

Solution 1: Hash Table + Prefix Sum

According to the problem description, we need to remove elements from both ends of the array $nums$ so that the sum of the removed elements equals $x$, and the number of removed elements is minimized. We can transform the problem into: find the longest consecutive subarray in the array $nums$ such that the sum of the subarray $s = \sum_{i=0}^{n} nums[i] - x$. In this way, we can transform the problem into finding the length $mx$ of the longest consecutive subarray in the array $nums$ with a sum of $s$, and the answer is $n - mx$.

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class Solution:
    def minOperations(self, nums: List[int], x: int) -> int:
        s = sum(nums) - x
        vis = {0: -1}
        mx, t = -1, 0
        for i, v in enumerate(nums):
            t += v
            if t not in vis:
                vis[t] = i
            if t - s in vis:
                mx = max(mx, i - vis[t - s])
        return -1 if mx == -1 else len(nums) - mx

Solution 2: Two Pointers

Based on the analysis of Solution 1, we need to find the length $mx$ of the longest consecutive subarray in the array $nums$ with a sum of $s$. Since all elements in the array $nums$ are positive integers, the prefix sum of the array will only increase monotonically, so we can use two pointers to solve this problem.

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class Solution:
    def minOperations(self, nums: List[int], x: int) -> int:
        s = sum(nums) - x
        vis = {0: -1}
        mx, t = -1, 0
        for i, v in enumerate(nums):
            t += v
            if t not in vis:
                vis[t] = i
            if t - s in vis:
                mx = max(mx, i - vis[t - s])
        return -1 if mx == -1 else len(nums) - mx

Continue Practicing

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