Find Duplicate Number Leetcode

[Quincey Homer]

Idid see binary search being listed as a solution, probably to reduce time complexity by half(so O(log n)) with constant space complexity. But I was more interested in another solution and that was a bit challenging. Somehow, it took me about a day to understand this.

The idea is to have two pointers - slow and fast. These would move through the list by using the current number as the index to the next element to iterate through. And as always, fast moves two steps ahead of slow. The iterations would create a linked list that contains a cycle due to the duplicate element.

Constructing our linked list

We would now find the sequence of numbers by iterating through nums(given above) and creating a linked list. The duplicate would have the same value and link back to the original value thus creating the cycle. The explanation of how each next number is identified is mentioned in () next to the number.

Find if there's a loop - while iterating through the list, if at some point, both the slow and fast pointers are pointing to the same node, then yes, a loop is present. Since fast is moving two nodes ahead, fast and slow would point to the same node only if there is a loop.

Why?

The intersection node is not necessarily the duplicate element/node as this intersection node is a node in the cycle where the fast caught up with the slow or where the fast and slow pointers randomly met - this just proves that there's a cycle in the linked list.

To find this origin, we need to push the slow pointer to the start of the list and move both fast and slow pointers at the same rate, one at a time, and when they now meet at the same node, that is the start of the cycle/loop.

Some math-ish stuff

Assuming D = the number of steps/distance from the start of the linked list to the start of the cycle and K = the number of steps/distance from the start of the cycle to the intersection point we determined in Step 1. Since slow moves to the start(0), its position after D steps is D. The fast pointer continues from the intersection point, so its position after D steps would be, nC(some number of cycles) - K, which is equal to D => nC - K = D.

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

Note:You must not modify the array (assume the array is read only).You must use only constant, O(1) extra space.Your runtime complexity should be less than O(n2).There is only one duplicate number in the array, but it could be repeated more than once.

The naive solution to this problem would be to use a nested loop and compare each element with every other element in the array. For an array of size n+1, this solution would have a time complexity of O(n^2). However, this solution is not practical for larger arrays.

A better solution is to use the pigeonhole principle and binary search. We know that there are n integers between 1 and n (inclusive), so at least one integer must be repeated. We can divide the array into two halves, with one half containing the numbers 1 to n/2 and the other half containing the numbers n/2 + 1 to n.

If the number of integers in the first half of the array is greater than n/2, then there must be a duplicate in that half of the array. Otherwise, there must be a duplicate in the second half of the array. We can repeat this process recursively until we find the duplicate.

To implement this solution, we can use a modified binary search algorithm. Initially, we set the lower and upper bounds to 1 and n, respectively. We then calculate the mid-point of the bounds. We count the number of integers in the array that are less than or equal to the mid-point value. If the number of integers is greater than the mid-point value (i.e. there are more integers in the first half of the array), we set the upper bound to the mid-point value. Otherwise, we set the lower bound to the mid-point value + 1. We repeat this process until we find the duplicate.

The array contains a lot of problems in the world of programming. One of the most interesting challenges is the identification of repeated numbers within an array. This article will discuss different approaches to finding the duplicate number present and implementing their code in C++, Java, and Python.

In the Find the Duplicate Number problem, we are given an array of integer nums containing n + 1 integers where each integer is inclusive in the range [1, n]. The array contains a duplicate number and our task is to return that duplicate number.

First, we will try to use the brute force approach, where we will use sorting to find the duplicate number. The initial step in this approach involves sorting the array in ascending order. After sorting duplicates numbers are adjacent to each other. So now we will simply compare the adjacent number using a loop.

The Time complexity is O(Nlogn + N) for this approach NlogN for sorting the array and O(N) for traversing through the array and checking if adjacent elements are equal. The Space complexity is O(1) as we are not using any extra space.

In conclusion, we explored three approaches to solve this leetcode problem of identifying the duplicate number in an array: the Sorting Approach, the Frequency Approach, and the optimal Linked List Cycle Detection. The Linked List Cycle Detection stands out with linear time complexity and minimal space usage, making it the optimal choice.

I was trying to solve this leetcode problem -the-duplicate-number/ using my own implementation of the tortoise and hare algorithm which resulted in an infinite loop when given the following array of integers:

Now, I'm sure someone who is skilled in discrete mathematics would have been able to intuitively know that this approach would not have lead to the correct solution without first having to trace through an example like I had to do.

What inferences or observations could I have made that would have lead me to see that this algorithm was not going to work? I'd like to know how one could intuitively identity a flaw in this logic through a series of logical statements. In other words, what's the explanation for why the two runners will never find the duplicates in this example? I feel like it may have something to do with counting, but I do not have a very strong background in discrete.

And to clarify, I have looked at the correct implementation so I do know what the correct way to solve it is. I just thought that this way would have worked too similar to applying it to linked lists, where you'd move the fast runner two nodes up and the slow runner one node up. Thank you for your help.

Floyd's tortoise algorithm works when you're detecting a cycle in a linked list. It relies on the fact that if both pointers are moving at a different pace, the gap between them will keep on increasing to a limit, after which it'll be reset if a cycle exists.

In this case, the algorithm does find a cycle, since both pointers converge to the index 0 after some iterations. However, you're not looking to detect a cycle here; you're trying to find a duplicate. That's why this gets stuck in infinite recursion: it is meant to detect a cycle (which it correctly does), but not detect duplicates in its basic implementation.

If you run Floyd's algorithm, you find that the cycle will get detected at index 0, since both pointers will converge there. It works by checking if fast and slow point to the same location and not if they have the same values of nodes (fast==slow isn't the same as fast.value==slow.value).

You are attempting to check duplicates by comparing the value on the nodes, and checking if the nodes don't point to the same location. That is actually the flaw, since Floyd's algorithm works to check if both pointers point to the same location in order to detect a cycle.

You can read this simple, informative proof to improve your intuition as to why the pointers will converge.

Your question is actually hard to answer. Typically when dealing with complexities, there's an assumed machine model. A standard model assumes that memory locations are of size log(n) bits when the input is of size n, and that arithmetic operations on numbers of size log(n) bits are O(1).

To solve your problem in O(1) space and O(n) time, you've got to make sure your values don't get too large. One approach is to xor all the numbers in the array, and then you'll get 1^2^3...^n ^ d where d is the duplicate. Thus you can xor 1^2^3^..^n from the total xor of the array, and find the duplicate value.

Now you converted this to linked list, you have to use Floyd's hare and tortoise algorithm. Basically you have two pointers, slow and fast. If there is cycle, slow and fast pointers are gonna meet at some point.

Notice none of the elements after first index ever points to value at index 0. because our range is 1-n. We are tracking where we point to by nums[value] but since no value will be 0, nothing will point to nums[0]

In this problem, we're given an array called nums which contains n + 1 integers, where every integer is within the range of 1 to n, both inclusive. There's an important constraint in the problem: there is exactly one number in the array nums that appears more than once, and our task is to find that one repeated number. The challenge is to solve this problem under the following two conditions: we are not permitted to alter the original array, and we have to solve it using only a constant amount of space, which eliminates the possibility of using additional data structures that can grow with the size of the input.

We can approach this problem with a binary search technique despite the fact that the array is not sorted. This might seem counterintuitive at first because binary search usually requires a sorted array. However, the key insight here is to use binary search not on the elements of the array itself, but rather on the range of numbers between 1 to n.

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