Formatted question description: https://leetcode.ca/all/1848.html

# 1848. Minimum Distance to the Target Element

## Level

Easy

## Description

Given an integer array `nums`

**(0-indexed)** and two integers `target`

and `start`

, find an index `i`

such that `nums[i] == target`

and `abs(i - start)`

is **minimized**. Note that `abs(x)`

is the absolute value of `x`

.

Return `abs(i - start)`

.

It is **guaranteed** that `target`

exists in `nums`

.

**Example 1:**

**Input:** nums = [1,2,3,4,5], target = 5, start = 3

**Output:** 1

**Explanation:** nums[4] = 5 is the only value equal to target, so the answer is abs(4 - 3) = 1.

**Example 2:**

**Input:** nums = [1], target = 1, start = 0

**Output:** 0

**Explanation:** nums[0] = 1 is the only value equal to target, so the answer is abs(0 - 0) = 0.

**Example 3:**

**Input:** nums = [1,1,1,1,1,1,1,1,1,1], target = 1, start = 0

**Output:** 0

**Explanation:** Every value of nums is 1, but nums[0] minimizes abs(i - start), which is abs(0 - 0) = 0.

**Constraints:**

`1 <= nums.length <= 1000`

`1 <= nums[i] <= 10^4`

`0 <= start < nums.length`

`target`

is in`nums`

.

## Solution

Loop over `nums`

. For each `i`

such that `nums[i] == target`

, calculate `Math.abs(i - start)`

and maintain the minimum distance. Finally, return the minimum distance.

```
class Solution {
public int getMinDistance(int[] nums, int target, int start) {
int minDistance = Integer.MAX_VALUE;
int length = nums.length;
for (int i = 0; i < length; i++) {
if (nums[i] == target) {
int distance = Math.abs(i - start);
minDistance = Math.min(minDistance, distance);
}
}
return minDistance;
}
}
```