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

# 1848. Minimum Distance to the Target Element

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 = 5 is the only value equal to target, so the answer is abs(4 - 3) = 1.

Example 2:

Input: nums = , target = 1, start = 0

Output: 0

Explanation: nums = 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 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;
}
}