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Formatted question description: https://leetcode.ca/all/1848.html
1848. Minimum Distance to the Target Element
Level
Easy
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) = 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 innums
.
Solution
Loop over nums
. For each i
such that nums[i] == target
, calculate Math.abs(i - start)
and maintain the minimum distance. Finally, return the minimum distance.
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class Solution { public int getMinDistance(int[] nums, int target, int start) { int minDistance = Integer.MAX_VALUE; int length = nums.length; for (int i = 0; i < length; i++) { if (nums[i] == target) { int distance = Math.abs(i - start); minDistance = Math.min(minDistance, distance); } } return minDistance; } } ############ 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; } }
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// OJ: https://leetcode.com/problems/minimum-distance-to-the-target-element/ // 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; } };
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class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: res = inf for i, num in enumerate(nums): if num == target: res = min(res, abs(i - start)) return res ############ # 1848. Minimum Distance to the Target Element # https://leetcode.com/problems/minimum-distance-to-the-target-element/ class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: res = float('inf') for i,x in enumerate(nums): if x == target: res = min(res, abs(i - start)) return res
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func getMinDistance(nums []int, target int, start int) int { ans := 1 << 30 for i, x := range nums { if t := abs(i - start); x == target && t < ans { ans = t } } return ans } func abs(x int) int { if x < 0 { return -x } return x }
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function getMinDistance(nums: number[], target: number, start: number): number { let ans = Infinity; for (let i = 0; i < nums.length; ++i) { if (nums[i] === target) { ans = Math.min(ans, Math.abs(i - start)); } } return ans; }
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impl Solution { pub fn get_min_distance(nums: Vec<i32>, target: i32, start: i32) -> i32 { nums.iter() .enumerate() .filter(|&(_, &x)| x == target) .map(|(i, _)| ((i as i32) - start).abs()) .min() .unwrap_or_default() } }