main.cpp

// Source: https://leetcode.com/problems/minimum-distance-to-the-target-element
// Title: Minimum Distance to the Target Element
// Difficulty: Easy
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// 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] <= 10^4`
// - `0 <= start < nums.length`
// - `target` is in `nums`.
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

#include <vector>

using namespace std;

// Linear Scan
class Solution {
 public:
  int getMinDistance(const vector<int>& nums, int target, int start) {
    const int n = nums.size();

    int minDist = n;
    for (int i = 0; i < n; ++i) {
      if (nums[i] == target) {
        minDist = min(minDist, abs(i - start));
      }
    }

    return minDist;
  }
};