1848. 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) = 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] <= 104`
• `0 <= start < nums.length`
• `target` is in `nums`.

1848. Minimum Distance to the Target Element
``````struct Solution;

impl Solution {
fn get_min_distance(nums: Vec<i32>, target: i32, start: i32) -> i32 {
let n = nums.len();
let mut res = std::i32::MAX;
for i in 0..n {
if nums[i] == target {
res = res.min((i as i32 - start).abs());
}
}
res
}
}

#[test]
fn test() {
let nums = vec![1, 2, 3, 4, 5];
let target = 5;
let start = 3;
let res = 1;
assert_eq!(Solution::get_min_distance(nums, target, start), res);
let nums = vec![1];
let target = 1;
let start = 0;
let res = 0;
assert_eq!(Solution::get_min_distance(nums, target, start), res);
let nums = vec![1, 1, 1, 1, 1, 1, 1, 1, 1, 1];
let target = 1;
let start = 0;
let res = 0;
assert_eq!(Solution::get_min_distance(nums, target, start), res);
}
``````