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 innums
.
Solution: Brute Force
Time complexity: O(n)
Space complexity: O(1)
C++
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// 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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