1848. Minimum Distance to the Target Element

#### QUESTION:

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

#### SOLUTION:

``````class Solution {
func getMinDistance(_ nums: [Int], _ target: Int, _ start: Int) -> Int {
var result = Int.max
for index in nums.indices {
if nums[index] == target {
result = min(result, abs(index - start))
}
}
return result
}
}
``````