Find the Duplicate Number

Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate element must exist. Assume that there is only one duplicate number, find the duplicate one.

Note:

1. You must not modify the array (assume the array is read only).
2. You must use only constant extra space.
3. Your runtime complexity should be less than O(n2).
4. 
   

The most straightforward way is to utilize pigeon hole theorem: keep swapping elements so that nums[i] = [i] until we find nums[nums[i]] = nums[i], then nums[i] is the duplicate one. If everything is in place, return nums[0].

BUT, the question requires no modification to the array, and no extra space. So we cannot use swap or construct an external array. There is a very interesting solution online that utilizes the solution in find cycle in linkedlist problem. So we consider each index as a list node, and the next node is its element, i.e., i -> nums[i]. Now we can construct a linked list in our mind. Since there is a duplicate node, we know there must exist a cycle. Now we can use the solution of Linked list Cycle II to solve the problem.

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