LeetCode 1848. Minimum Distance to the Target Element Solution in Java, C++, Python & More | Explanation + Code

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] <= 104
• 0 <= start < nums.length
• target is in nums.

Solutions

Solution 1: Single Pass

Traverse the array, find all indices equal to target, then calculate |i - start|, and take the minimum value.

The time complexity is O(n), where n is the length of the array nums. The space complexity is O(1).

PythonJavaC++GoTypeScriptRust

class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: return min(abs(i - start) for i, x in enumerate(nums) if x == target)(code-box)

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

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

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 }(code-box)

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; }(code-box)

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() } }(code-box)