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.

#1658

Medium

auto_awesome数组·哈希·扫描

LeetCode 题解工作台

将 x 减到 0 的最小操作数

给你一个整数数组 nums 和一个整数 x 。每一次操作时，你应当移除数组 nums 最左边或最右边的元素，然后从 x 中减去该元素的值。请注意，需要修改数组以供接下来的操作使用。如果可以将 x 恰好减到 0 ，返回最小操作数；否则，返回 -1 。示例 1：输入： nums = [1…

数组哈希表二分查找滑动窗口前缀和

题目描述

给你一个整数数组 nums 和一个整数 x 。每一次操作时，你应当移除数组 nums 最左边或最右边的元素，然后从 x 中减去该元素的值。请注意，需要修改数组以供接下来的操作使用。

如果可以将 x 恰好减到 0 ，返回 最小操作数 ；否则，返回 -1 。

示例 1：

输入：nums = [1,1,4,2,3], x = 5
输出：2
解释：最佳解决方案是移除后两个元素，将 x 减到 0 。

示例 2：

输入：nums = [5,6,7,8,9], x = 4
输出：-1

示例 3：

输入：nums = [3,2,20,1,1,3], x = 10
输出：5
解释：最佳解决方案是移除后三个元素和前两个元素（总共 5 次操作），将 x 减到 0 。

提示：

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

lightbulb

解题思路

方法一：哈希表 + 前缀和

根据题目描述，我们需要移除数组 $nums$ 左右两端的元素，使得移除的元素和等于 $x$ ，且移除的元素个数最少。我们可以将问题转化为：找到数组 $nums$ 中最长的连续子数组，使得子数组的和 $s = \sum_{i=0}^{n} nums[i] - x$ 。这样，我们就可以将问题转化为求解数组 $nums$ 中和为 $s$ 的最长连续子数组的长度 $mx$ ，答案即为 $n - mx$ 。

我们初始化 $mx = -1$ ，然后使用哈希表 $vis$ 来存储前缀和，键为前缀和，值为前缀和对应的下标。

遍历数组 $nums$ ，对于当前元素 $nums[i]$ ，计算前缀和 $t$ ，如果 $t$ 不在哈希表中，则将 $t$ 加入哈希表；如果 $t - s$ 在哈希表中，则更新 $mx = \max(mx, i - vis[t - s])$ 。

最后，如果 $mx = -1$ ，则返回 $-1$ ，否则返回 $n - mx$ 。

时间复杂度 $O(n)$ ，空间复杂度 $O(n)$ 。其中 $n$ 为数组 $nums$ 的长度。

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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

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

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

warning

常见陷阱

外企场景

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

swap_horiz

进阶变体

外企场景

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

help

常见问题

外企场景

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