Introduction

In this blog post, we will explore how to solve the LeetCode problem "287" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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?

Explanation

Here's a solution to find the duplicate number in an array, adhering to the given constraints and follow-up questions:

• Treat the array as a linked list: The array indices can be considered nodes, and the values at each index can be treated as pointers to the next node. Because there's a duplicate, the linked list will have a cycle.
• Floyd's Cycle-Finding Algorithm: Use the "tortoise and hare" approach to detect the cycle. One pointer (tortoise) moves one step at a time, and the other (hare) moves two steps at a time.

• Find the entrance to the cycle: Once the two pointers meet, reset one pointer to the beginning of the array and move both pointers one step at a time until they meet again. This meeting point is the duplicate number.

• Runtime & Storage Complexity: O(n) time complexity, O(1) space complexity.

Code

    def findDuplicate(nums):
    """
    Finds the duplicate number in an array using Floyd's Cycle-Finding Algorithm.

    Args:
      nums: A list of integers where each integer is in the range [1, n] inclusive,
            and there is exactly one repeated number.

    Returns:
      The repeated number in the array.
    """

    # Phase 1: Detect the cycle
    slow = nums[0]
    fast = nums[nums[0]]

    while slow != fast:
        slow = nums[slow]
        fast = nums[nums[fast]]

    # Phase 2: Find the entrance to the cycle
    slow = 0
    while slow != fast:
        slow = nums[slow]
        fast = nums[fast]

    return slow