# LeetCode 1848. Minimum Distance to the Target Element

## Description

https://leetcode.com/problems/minimum-distance-to-the-target-element/

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) = 1.
```

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] <= 104`
• `0 <= start < nums.length`
• `target` is in `nums`.

## Explanation

Iterate the list and calculate the min distance.

## Python Solution

``````class Solution:
def getMinDistance(self, nums: List[int], target: int, start: int) -> int:
min_distance = sys.maxsize

for i in range(0, len(nums)):
if nums[i] == target:
min_distance = min(min_distance, abs(i - start))

return min_distance``````
• Time Complexity: O(N).
• Space Complexity: O(1).