We treat the array as a linked list where nums[i] points to the index nums[i]. Since there's a duplicate, there must be a cycle in this linked list. We can use Floyd's Cycle-Finding Algorithm (Tortoise and Hare).

1. Initialize two pointers, slow and fast, both at nums[0].
2. Move slow one step (slow = nums[slow]) and fast two steps (fast = nums[nums[fast]]) until they meet.
3. Once they meet, move slow back to nums[0].
4. Move both slow and fast one step at a time. The point where they meet again is the start of the cycle, which corresponds to the duplicate number.

• Time Complexity: O(N), as we traverse the "list" a small number of times.
• Space Complexity: O(1), no extra space besides pointers.

def find_duplicate(nums):
    # Phase 1: Finding the intersection point in the cycle
    slow = nums[0]
    fast = nums[0]
    
    while True:
        slow = nums[slow]
        fast = nums[nums[fast]]
        if slow == fast:
            break
            
    # Phase 2: Finding the entrance to the cycle
    slow = nums[0]
    while slow != fast:
        slow = nums[slow]
        fast = nums[fast]
        
    return slow