1848

1848. Minimum Distance to the Target Element (Easy)

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) = 1.

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.

Related Topics:
Array

Solution 1.

// OJ: https://leetcode.com/problems/minimum-distance-to-the-target-element/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int getMinDistance(vector<int>& A, int target, int start) {
        if (A[start] == target) return 0;
        int i = 1;
        for (int N = A.size(); start + i < N || start - i >= 0; ++i) {
            if ((start + i < N && A[start + i] == target)
               || (start - i >= 0 && A[start - i] == target)) break;
        }
        return i;
    }
};