Find the Duplicate Number
Created: October 27, 2018 by [lek-tin]
Last updated: October 27, 2018
Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.
Example 1
Input: [1,3,4,2,2]
Output: 2
Example 2
Input: [3,1,3,4,2]
Output: 3
Note
You must not modify the array (assume the array is read only).
You must use only constant, O(1) extra space.
Your runtime complexity should be less than O(n2).
There is only one duplicate number in the array, but it could be repeated more than once.
Solution
class Solution:
def findDuplicate(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
newNums = list(nums)
newNums.sort()
print(nums)
for i in range(1, len(newNums)):
if newNums[i] == newNums[i-1]:
return newNums[i]
Complexity Analysis
Time complexity : O(nlgn)
The sort invocation costs O(nlgn) time in Python and Java, so it dominates the subsequent linear scan.
Space complexity : O(1) (or O(n))
Here, we sort nums in place, so the memory footprint is constant. If we cannot modify the input array, then we must allocate linear space for a copy of nums and sort that instead.