What is the easiest way to solve Minimum Distance to the Target Element?

Scan the full array, and every time nums[i] equals target, compute abs(i - start) and keep the minimum. That is the most direct and reliable solution for this problem.

Can I search outward from start instead of scanning everything?

Yes. Check start first, then move one step left and right, then two steps, and so on. The first target you hit gives the answer, but you must handle array boundaries carefully.

Why is the Array-driven solution strategy enough here?

The problem only asks for the closest matching index in one array, so a single traversal already compares every valid candidate. There is no need for dynamic programming, hashing, or sorting.

What mistake causes wrong answers most often?

A common bug is returning when you see the first target during a normal left-to-right pass. That can miss another target index that is closer to start.

What are the time and space costs for this problem?

The standard approach is O(n) time and O(1) space. You read each element once and store only the best distance found so far.

#1848

Easy

auto_awesome数组·driven

LeetCode 题解工作台

到目标元素的最小距离

给你一个整数数组 nums （下标从 0 开始计数）以及两个整数 target 和 start ，请你找出一个下标 i ，满足 nums[i] == target 且 abs(i - start) 最小化。注意： abs(x) 表示 x 的绝对值。返回 abs(i - start) 。题目…

数组

题目描述

给你一个整数数组 nums （下标 从 0 开始 计数）以及两个整数 target 和 start ，请你找出一个下标 i ，满足 nums[i] == target 且 abs(i - start) 最小化 。注意：abs(x) 表示 x 的绝对值。

返回 abs(i - start) 。

题目数据保证 target 存在于 nums 中。

示例 1：

输入：nums = [1,2,3,4,5], target = 5, start = 3
输出：1
解释：nums[4] = 5 是唯一一个等于 target 的值，所以答案是 abs(4 - 3) = 1 。

示例 2：

输入：nums = [1], target = 1, start = 0
输出：0
解释：nums[0] = 1 是唯一一个等于 target 的值，所以答案是 abs(0 - 0) = 0 。

示例 3：

输入：nums = [1,1,1,1,1,1,1,1,1,1], target = 1, start = 0
输出：0
解释：nums 中的每个值都是 1 ，但 nums[0] 使 abs(i - start) 的结果得以最小化，所以答案是 abs(0 - 0) = 0 。

提示：

1 <= nums.length <= 1000
1 <= nums[i] <= 10⁴
0 <= start < nums.length
target 存在于 nums 中

lightbulb

解题思路

方法一：一次遍历

遍历数组，找到所有等于 $target$ 的下标，然后计算 $|i - start|$ ，取最小值即可。

时间复杂度 $O(n)$ ，其中 $n$ 为数组 $nums$ 的长度。空间复杂度 $O(1)$ 。

1

2

3

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class Solution:
    def getMinDistance(self, nums: List[int], target: int, start: int) -> int:
        return min(abs(i - start) for i, x in enumerate(nums) if x == target)

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
They want you to notice that only indices with nums[i] == target matter, not the values around them.
question_mark
A strong answer mentions that target is guaranteed to exist, so no fallback case is needed.
question_mark
If they hint at checking both directions from start, they are steering you toward an outward array search with early stop.

warning

常见陷阱

外企场景

error
Stopping at the first target seen in a left-to-right scan can fail when a later target is closer to start.
error
Forgetting absolute value leads to wrong comparisons when the target index is left of start.
error
Overengineering with sorting or extra maps misses that this problem is just a direct array distance check.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Return the closest target index instead of the distance, with a clear tie rule if both sides match.
arrow_right_alt
Remove the guarantee that target exists and return -1 when no matching value appears.
arrow_right_alt
Process many queries on the same nums, where preprocessing target indices could help repeated lookups.

help

常见问题

外企场景

继续练习

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