Skip to content

1848. Minimum Distance to the Target Element

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 $O(1)$.

1
2
3
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)

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
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;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
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;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
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
}

1
2
3
4
5
6
7
8
9
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;
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
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()
    }
}

Comments