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Tested Python solution for LeetCode 961 with 18 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 961 # by problem number
lcpy gen -s n_repeated_element_in_size_2n_array # by problem name
nums with the following properties:
nums.length == 2 * n.nums contains n + 1 unique values, n of which occur exactly once in the array.nums is repeated n times.n times.
Input: nums = [1,2,3,3]
Output: 3
Input: nums = [2,1,2,5,3,2]
Output: 2
Input: nums = [5,1,5,2,5,3,5,4]
Output: 5
class Solution:
# Time: O(n) worst case, returns after a handful of checks in practice
# Space: O(1)
def repeated_n_times(self, nums: list[int]) -> int:
# Pigeonhole: the n copies of the answer cannot all sit 3+ slots apart,
# so some pair at distance 1 or 2 must match. The only way no such pair
# exists is n == 2 with the answer at both ends, hence the fallback.
for i in range(2, len(nums)):
if nums[i] == nums[i - 1] or nums[i] == nums[i - 2]:
return nums[i]
return nums[0]
| Time | Space |
|---|---|
| O(n) worst case, returns after a handful of checks in practice | O(1) |