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 innums
.
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]
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