1848. Minimum Distance to the Target Element

Array

TIME

O(n)

SPACE

O(1)

Intuition

Simple linear scan to find the target value closest to the start index.

Solution

Iterate through the array. For each index where nums[i] == target, compute |i - start| and track the minimum.

min_dist = infinity
for i in range(len(nums)):
    if nums[i] == target:
        min_dist = min(min_dist, abs(i - start))

Optimization (not needed for small n)

Could expand outward from start and return immediately when target is found, but linear scan is sufficient for n ≤ 1000.

Complexity

Time: O(n)
Space: O(1)

Edge Cases

Target at start index → return 0
Single element array → return 0
Multiple occurrences of target → pick closest
Target only at one end

Code

from typing import List


class Solution:
    def getMinDistance(self, nums: List[int], target: int, start: int) -> int:
        """
        1848. Minimum Distance to the Target Element
        Time: O(n)
        Space: O(1)
        """
        min_dist = float('inf')

        for i, num in enumerate(nums):
            if num == target:
                min_dist = min(min_dist, abs(i - start))

        return min_dist

/**
 * 1848. Minimum Distance to the Target Element
 * Time: O(n)
 * Space: O(1)
 */
class Solution {
    public int getMinDistance(int[] nums, int target, int start) {
        int minDist = Integer.MAX_VALUE;

        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == target) {
                minDist = Math.min(minDist, Math.abs(i - start));
            }
        }

        return minDist;
    }
}