README_EN.md

1848. Minimum Distance to the Target Element

中文文档

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) = 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.

Solutions

Python3

class Solution:
    def getMinDistance(self, nums: List[int], target: int, start: int) -> int:
        res = float('inf')
        for i, num in enumerate(nums):
            if num == target:
                res = min(res, abs(i - start))
        return res

Java

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

C++

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

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