Find the Duplicate Number

Given an array of integers nums containing n + 1 integers where each integer is in the range [1, n] inclusive.

There is only one repeated number in nums, return this repeated number.

You must solve the problem without modifying the array nums and using only constant extra space.

Example 1:

Input: nums = [1,3,4,2,2]
Output: 2

Example 2:

Input: nums = [3,1,3,4,2]
Output: 3

Example 3:

Input: nums = [3,3,3,3,3]
Output: 3

Constraints:

• 1 <= n <= 105
• nums.length == n + 1
• 1 <= nums[i] <= n
• All the integers in nums appear only once except for precisely one integer which appears two or more times.

Follow up:

• How can we prove that at least one duplicate number must exist in nums?
• Can you solve the problem in linear runtime complexity?

Approach - Hare Tortoise

• slow fast pointer like the linked list if both meet then it is a loop but doesn’t necessarily mean that is the duplicate, the duplicate we will find by finding the entrance of the of the loop
• Time: O(n)
  • Each pointer moves at most O(n) steps across both phases.
• Space: O(1) extra
  • Only a handful of pointers and counters, no arrays or recursion.

class Solution {
    public int findDuplicate(int[] nums) {
        int slow = nums[0], fast = nums[0];
        do {
            slow = nums[slow];
            fast = nums[nums[fast]];
        } while (slow != fast);
 
        slow = nums[0];
        while (slow != fast) {
            slow = nums[slow];
            fast = nums[fast];
        }
 
        return slow;
    }
}