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# 1848. Minimum Distance to the Target Element

### 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`.

### Constraints

* `1 <= nums.length <= 1000`
* `1 <= nums[i] <= 104`
* `0 <= start < nums.length`
* `target` is in `nums`.

### Approach

### Links

* GeeksforGeeks
* [Leetcode](https://leetcode.com/problems/minimum-distance-to-the-target-element/)
* ProgramCreek
* YouTube

### **Examples**

{% tabs %}
{% tab title="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.
{% endtab %}

{% tab title="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.
{% endtab %}

{% tab title="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.
{% endtab %}
{% endtabs %}

### **Solutions**

{% tabs %}
{% tab title="Solution 1" %}

```java
/**
 * Time complexity : 
 * Space complexity : 
 */

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

{% endtab %}
{% endtabs %}

### **Follow up**

*