Minimum Distance to the Target Element

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.

SOLUTION:

import heapq

class Solution:
    def getMinDistance(self, nums: List[int], target: int, start: int) -> int:
        heap = []
        for i, num in enumerate(nums):
            heapq.heappush(heap, (abs(num - target), abs(i - start)))
        return heap[0][1]