Find the Duplicate Number

 

1. Clarify the problem: Before diving into solving the problem, let's clarify the requirements:

  • We are given an array of integers where each integer is in the range of 1 to n, inclusive.
  • The array has a length of n+1.
  • There is exactly one duplicate number in the array, and it can appear more than once.
  • We need to find the duplicate number.

2. Analyze the problem: To solve this problem, we can use the Floyd's Tortoise and Hare algorithm, also known as the "Hare and Tortoise" algorithm or the "Cycle Detection" algorithm.

  • This algorithm is based on the principle that if we have a cycle in a linked list, we can find it using two pointers: one slow pointer and one fast pointer.
  • We can apply this principle to our problem by treating the array as a linked list, where the array elements represent the indices to jump to.
  • By initializing two pointers, slow and fast, we can detect the cycle in the array.
  • Once the cycle is detected, we can find the entry point of the cycle, which represents the duplicate number.

3. Design an algorithm: Here is the algorithm to solve the problem:

  1. Initialize two pointers, slow and fast, to the first element of the array.
  2. Move slow one step at a time and fast two steps at a time until they meet.
  3. Set the slow pointer back to the first element of the array and keep the fast pointer at the meeting point.
  4. Move both pointers one step at a time until they meet again. The meeting point is the entry point of the cycle and represents the duplicate number.
  5. Return the duplicate number.

4. Explain your approach: The approach involves using the Floyd's Tortoise and Hare algorithm to detect the cycle in the array. By treating the array as a linked list and using two pointers, slow and fast, we can find the entry point of the cycle, which represents the duplicate number.

5. Write clean and readable code:

python
def findDuplicate(nums):
    slow = fast = nums[0]
    
    while True:
        slow = nums[slow]
        fast = nums[nums[fast]]
        if slow == fast:
            break
    
    slow = nums[0]
    while slow != fast:
        slow = nums[slow]
        fast = nums[fast]
    
    return slow

6. Test your code: Let's test the code with different test cases:

  1. nums = [1, 3, 4, 2, 2]
    • Expected output: 2
    • Explanation: The array represents a linked list with a cycle: 1 -> 3 -> 2 -> 4 -> 2. The duplicate number is 2, which is the entry point of the cycle.
  2. nums = [3, 1, 3, 4, 2]
    • Expected output: 3
    • Explanation: The array represents a linked list with a cycle: 3 -> 4 -> 2 -> 3. The duplicate number is 3, which is the entry point of the cycle.
  3. nums = [1, 1, 2]
    • Expected output: 1
    • Explanation: The array represents a linked list with a cycle: 1 -> 1. The duplicate number is 1, which is the entry point of the cycle.

7. Optimize if necessary: The solution already follows an optimal approach using the Floyd's Tortoise and Hare algorithm. There are no further optimizations required.

8. Handle error cases: The code assumes that the input array nums is valid and always contains a duplicate number. If the input array is empty or does not have a duplicate number, the behavior of the code is undefined.

9. Discuss complexity analysis:

  • The time complexity of this solution is O(n), where n is the length of the input array. The Floyd's Tortoise and Hare algorithm guarantees to find the duplicate number in linear time.
  • The space complexity is O(1) since we are not using any additional data structures.
  • The solution has an optimal time and space complexity given the problem requirements.
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