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.

Approach

C++

#include <bits/stdc++.h>
using namespace std;

int getMinDistance(vector<int> &nums, int target, int start)
{

    int ans;
    for (int i = start; i < nums.size(); i++)
    {
        if (nums[i] == target)
        {
            ans = abs(i - start);
            break;
        }
    }
    for (int i = 0; i < start; i++)
    {
        if (nums[i] == target)
        {
            ans = min(abs(i - start), ans);
        }
    }
    return ans;
}

int main()
{
    vector<int> nums = {1, 2, 3, 4, 5};
    int target = 5, start = 3;

    cout << getMinDistance(nums, target, start) << "\n";

    return 0;
}