What is the best way to approach the Minimum Absolute Difference Between Elements With Constraint problem?

A binary search approach is often the most efficient, especially when combined with ordered sets to maintain values in a constrained index range.

How can I handle the index separation constraint efficiently?

You can utilize binary search or a sliding window approach to ensure the indices are separated by at least x, while minimizing the absolute difference.

What is the time complexity of solving this problem with binary search?

The time complexity is generally O(n log(max(nums))) when using binary search on the difference, where n is the number of elements and max(nums) is the range of values.

Are there alternative methods to solve the Minimum Absolute Difference Between Elements With Constraint?

Yes, you can also solve this problem using sliding window, two-pointer techniques, or ordered sets for more efficient searching of minimum differences.

How does GhostInterview assist with solving the Minimum Absolute Difference problem?

GhostInterview provides step-by-step assistance with understanding the binary search approach, optimizing for constraints, and refining the solution for large inputs.

#2817

Medium

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LeetCode 题解工作台

限制条件下元素之间的最小绝对差

给你一个下标从 0 开始的整数数组 nums 和一个整数 x 。请你找到数组中下标距离至少为 x 的两个元素的差值绝对值的最小值。换言之，请你找到两个下标 i 和 j ，满足 abs(i - j) >= x 且 abs(nums[i] - nums[j]) 的值最小。请你返回一个整数，…

数组二分查找有序集合

题目描述

给你一个下标从 0 开始的整数数组 nums 和一个整数 x 。

请你找到数组中下标距离至少为 x 的两个元素的 差值绝对值 的 最小值 。

换言之，请你找到两个下标 i 和 j ，满足 abs(i - j) >= x 且 abs(nums[i] - nums[j]) 的值最小。

请你返回一个整数，表示下标距离至少为 x 的两个元素之间的差值绝对值的 最小值 。

示例 1：

输入：nums = [4,3,2,4], x = 2
输出：0
解释：我们选择 nums[0] = 4 和 nums[3] = 4 。
它们下标距离满足至少为 2 ，差值绝对值为最小值 0 。
0 是最优解。

示例 2：

输入：nums = [5,3,2,10,15], x = 1
输出：1
解释：我们选择 nums[1] = 3 和 nums[2] = 2 。
它们下标距离满足至少为 1 ，差值绝对值为最小值 1 。
1 是最优解。

示例 3：

输入：nums = [1,2,3,4], x = 3
输出：3
解释：我们选择 nums[0] = 1 和 nums[3] = 4 。
它们下标距离满足至少为 3 ，差值绝对值为最小值 3 。
3 是最优解。

提示：

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

lightbulb

解题思路

方法一：有序集合

我们创建一个有序集合，用于存储距离当前下标至少为 $x$ 的元素。

接下来，我们从下标 $i = x$ 开始枚举，每次将 $nums[i - x]$ 加入到有序集合中。然后找出有序集合中与 $nums[i]$ 最接近的两个元素，它们的差值绝对值的最小值就是答案。

时间复杂度 $O(n \times \log n)$ ，空间复杂度 $O(n)$ 。其中 $n$ 是数组 $nums$ 的长度。

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class Solution:
    def minAbsoluteDifference(self, nums: List[int], x: int) -> int:
        sl = SortedList()
        ans = inf
        for i in range(x, len(nums)):
            sl.add(nums[i - x])
            j = bisect_left(sl, nums[i])
            if j < len(sl):
                ans = min(ans, sl[j] - nums[i])
            if j:
                ans = min(ans, nums[i] - sl[j - 1])
        return ans

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
Evaluate the candidate's understanding of binary search and its application to constrained problems.
question_mark
Look for clarity in the explanation of how to handle the index separation constraint efficiently.
question_mark
Check for familiarity with ordered set or tree-based data structures for optimizing search operations.

warning

常见陷阱

外企场景

error
Failing to efficiently manage the index separation constraint and checking all pairs directly can lead to time-limit exceeded errors.
error
Overlooking the impact of large input sizes and failing to optimize the solution with binary search or an ordered set.
error
Confusing the problem with simpler pair difference problems that do not have the index separation requirement.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Adjusting the index separation constraint (e.g., making it larger) can change the approach and affect performance.
arrow_right_alt
Solving the problem for an array with a fixed length or pre-sorted array may require different optimizations.
arrow_right_alt
Extending the problem to multidimensional arrays or higher-dimensional constraints would introduce added complexity.

help

常见问题

外企场景

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