Problem Decription 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:

  1. You must not modify the array (assume the array is read only).
  2. You must use only constant, O(1) extra space.
  3. Your runtime complexity should be less than O(n2).
  4. There is only one duplicate number in the array, but it could be repeated more than once.

brute force solution

class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        ordered = sorted(nums)
        i,j = 0,1
        while j <len(ordered):
            if ordered[i] == ordered[j]:
                return ordered[i]
            i += 1
            j += 1

bisection solution

class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        low = 0
        high = len(nums)-1
        
        while low < high:
            mid = (low+high)>>1
            count = 0
            for i in nums:
                if i <= mid:
                    count +=1
            if count > mid:
                # duplicated number is between low and mid
                high = mid
            else:
                #  duplicated number is between mid and high
                low = mid+1
        return low