# 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.

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.

## Solutions

Solution 1: Single Pass

 Traverse the array, find all indices equal to $target$, then calculate $i - start$, and take the minimum value.

The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.

• class Solution {
public int getMinDistance(int[] nums, int target, int start) {
int n = nums.length;
int ans = n;
for (int i = 0; i < n; ++i) {
if (nums[i] == target) {
ans = Math.min(ans, Math.abs(i - start));
}
}
return ans;
}
}

• class Solution {
public:
int getMinDistance(vector<int>& nums, int target, int start) {
int n = nums.size();
int ans = n;
for (int i = 0; i < n; ++i) {
if (nums[i] == target) {
ans = min(ans, abs(i - start));
}
}
return ans;
}
};

• class Solution:
def getMinDistance(self, nums: List[int], target: int, start: int) -> int:
return min(abs(i - start) for i, x in enumerate(nums) if x == target)


• func getMinDistance(nums []int, target int, start int) int {
ans := 1 << 30
for i, x := range nums {
if t := abs(i - start); x == target && t < ans {
ans = t
}
}
return ans
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}

• function getMinDistance(nums: number[], target: number, start: number): number {
let ans = Infinity;
for (let i = 0; i < nums.length; ++i) {
if (nums[i] === target) {
ans = Math.min(ans, Math.abs(i - start));
}
}
return ans;
}


• impl Solution {
pub fn get_min_distance(nums: Vec<i32>, target: i32, start: i32) -> i32 {
nums.iter()
.enumerate()
.filter(|&(_, &x)| x == target)
.map(|(i, _)| ((i as i32) - start).abs())
.min()
.unwrap_or_default()
}
}