Find The Duplicate Number Leetcode Problem 287 [Python Solution]

In this blog post, we will discuss the LeetCode problem "Find the Duplicate Number," which falls under the "Linked List" category and is rated as a medium difficulty problem.

The problem is particularly interesting as it challenges us to find a duplicate number in an array of integers, given certain constraints.

Problem Overview
The problem statement is as follows: 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, and the task is to return this repeated number.

However, there are some critical constraints:

1. You must solve the problem without modifying the array nums.
2. You can use only constant extra space.

Let's illustrate this with an example:

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

In this example, the duplicate number is 2, and we need to return it.

The challenge here is to solve this problem efficiently while adhering to the constraints mentioned above.

Understanding the Constraints

Before we delve into the solution, it's important to understand the constraints of the problem, which make it more challenging.

1. Constant Extra Space: This means you cannot create any additional data structures like sets or dictionaries to store values.

You must find the duplicate using only a few variables.

1. No Modification of the Input Array: You cannot sort the array or alter it in any way.

This constraint eliminates many common techniques for solving such problems.

Now, let's move on to the solution.

Find The Duplicate Number LeetCode Problem Solution

We'll solve this problem using a well-known algorithm called Floyd's Tortoise and Hare (Cycle Detection).

The idea is to consider the given array as a linked list, where each element is treated as a pointer to the next element.

1. Floyd's Algorithm

Here's how we apply Floyd's algorithm to find the duplicate number:

def findDuplicate(nums):
    # Initialize slow and fast pointers
    slow, fast = 0, 0

    # Phase 1: Detect the intersection point of the two pointers
    while True:
        slow = nums[slow]
        fast = nums[nums[fast]
        if slow == fast:
            break

    # Phase 2: Find the "entrance" to the cycle
    slow2 = 0
    while slow != slow2:
        slow = nums[slow]
        slow2 = nums[slow2]

    return slow

This algorithm consists of two phases.

Let's break down each phase:

1.1 Phase 1

In this phase, we initialize two pointers, slow and fast, at the beginning of the array.

We move the slow pointer one step at a time, and the fast pointer two steps at a time.

If there's a cycle in the array (which is guaranteed in this problem), they will eventually meet at the same position, indicating the start of the cycle.

1.2 Phase 2

Once we find the intersection point in Phase 1, we reset one of the pointers (in this case, slow) to the beginning of the array.

The other pointer (fast) stays at the intersection point.

We then advance both pointers one step at a time until they meet again.

The meeting point will be the start of the cycle, which corresponds to the duplicate number in the array.

The reason this algorithm works is based on mathematical analysis, and it ensures that the two pointers will meet at the start of the cycle.

It's a clever way of finding the duplicate without extra space or modifying the array.

Time and Space Complexity

Now, let's discuss the time and space complexity of this solution:

• Time Complexity: The time complexity of this algorithm is O(n), where n is the length of the input array.

The algorithm's time complexity is linear because the two pointers move through the array at different speeds, but eventually, they will meet after at most two traversals.

• Space Complexity: The space complexity is O(1) because we are using a constant amount of extra space to maintain the pointers, regardless of the input array's size.

Reasoning Behind Our Approach

The reasoning behind this approach is based on understanding the problem as a linked list cycle detection problem.

By applying Floyd's Tortoise and Hare algorithm, we can efficiently find the duplicate number while satisfying the constraints of not modifying the array and using constant extra space.

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• Find The Duplicate Number LeetCode
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Conclusion

In this blog post, we tackled the Find The Duplicate Number problem from LeetCode, which falls under the "Linked List" category.

We discussed the problem's constraints, provided an efficient Python solution using Floyd's algorithm, and explained the reasoning behind our approach.

This problem is a great example of how mathematical concepts can be applied to solve real-world coding challenges.

We hope this explanation has been helpful, especially for beginners, and that it equips you with the knowledge and approach needed to tackle similar problems.

If you have any questions, suggestions, or insights, please feel free to comment and share your thoughts.

You can access the original problem on LeetCode here.

Happy coding!