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
- targetis in- nums.
Solution: Brute Force
Time complexity: O(n)
Space complexity: O(1)
C++
| 1 2 3 4 5 6 7 8 9 10 11 | // Author: Huahua class Solution { public:   int getMinDistance(vector<int>& nums, int target, int start) {     int ans = INT_MAX;     for (int i = 0; i < nums.size(); ++i)       if (nums[i] == target)         ans = min(ans, abs(i - start));         return ans;   } }; | 
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