Find duplicates in constant array with elements 0 to N-1 in O(1) space

Given a constant array of n elements which contains elements from 1 to n-1, with any of these numbers appearing any number of times. Find any one of these repeating numbers in O(n) and using only constant memory space.

Examples:

Input : arr[] = {1, 2, 3, 4, 5, 6, 3}
Output : 3

As the given array is constant methods given in below articles will not work.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

So, here is an approach that is based on Floyd’s cycle finding algorithm. We use this to detect loop in a linked list.

The idea is to consider array items as linked list nodes. Any particular index is pointing to the value at that index. And you will see that there is loop as shown in the image below- In case of duplicate, two indexes will have same value and they will form a cycle just like in the image given below.

Linked list formed for above example would be :
1->2->3->4->5->6->3 So we can find the entry point of cycle in the linked list and that will be our duplicate element.

1. We maintain two pointers fast and slow
2. For each step fast will move to the index that is equal to arr[arr[fast]](two jumps at a time) and slow will move to the index arr[slow](one step at a time)
3. When fast==slow that means now we are in a cycle.
4. Fast and slow will meet in a circle and the entry point of that circle will be the duplicate element.
5. Now we need to find entry point so we will start with fast=0 and visit one step at a time for both fast and slow.
6. When fast==slow that will be entry point.
7. Return the entry point.

C++

 // CPP program to find a duplicate // element in an array with values in // range from 0 to n-1. #include using namespace std;    // function to find one duplicate int findduplicate(const int arr[], int n) {     // return -1 because in these cases      // there can not be any repeated element     if (n <= 1)         return -1;        // initialize fast and slow     int slow = arr;     int fast = arr[arr];        // loop to enter in the cycle     while (fast != slow) {            // move one step for slow         slow = arr[slow];            // move two step for fast         fast = arr[arr[fast]];     }        // loop to find entry point of the cycle     fast = 0;     while (slow != fast) {         slow = arr[slow];         fast = arr[fast];     }     return slow; }    int main() {     const int arr[] = { 1, 2, 3, 4, 5, 6, 3 };     int n = sizeof(arr) / sizeof(arr);     cout << findduplicate(arr, n);     return 0; }

Java

 // Java program to find a duplicate // element in an array with values in // range from 0 to n-1. import java.io.*; import java.util.*;    public class GFG {             // function to find one duplicate     static int findduplicate(int []arr, int n)     {                     // return -1 because in these cases          // there can not be any repeated element         if (n <= 1)             return -1;                 // initialize fast and slow         int slow = arr;         int fast = arr[arr];                 // loop to enter in the cycle         while (fast != slow)          {                     // move one step for slow             slow = arr[slow];                     // move two step for fast             fast = arr[arr[fast]];         }                 // loop to find entry         // point of the cycle         fast = 0;         while (slow != fast)         {             slow = arr[slow];             fast = arr[fast];         }         return slow;     }             // Driver Code     public static void main(String args[])     {         int []arr = {1, 2, 3, 4, 5, 6, 3};         int n = arr.length;         System.out.print(findduplicate(arr, n));     } }     // This code is contributed by // Manish Shaw (manishshaw1)

Python 3

 # Python 3 program to find a duplicate # element in an array with values in # range from 0 to n-1.    # function to find one duplicate def findduplicate(arr, n):        # return -1 because in these cases      # there can not be any repeated element     if (n <= 1):         return -1        # initialize fast and slow     slow = arr     fast = arr[arr]        # loop to enter in the cycle     while (fast != slow) :            # move one step for slow         slow = arr[slow]            # move two step for fast         fast = arr[arr[fast]]        # loop to find entry point of the cycle     fast = 0     while (slow != fast):         slow = arr[slow]         fast = arr[fast]     return slow    # Driver Code if __name__ == "__main__":            arr = [1, 2, 3, 4, 5, 6, 3 ]     n = len(arr)     print(findduplicate(arr, n))    # This code is contributed by ita_c

C#

 // C# program to find a duplicate // element in an array with values in // range from 0 to n-1. using System; using System.Collections.Generic; class GFG {            // function to find one duplicate     static int findduplicate(int []arr, int n)     {                    // return -1 because in these cases          // there can not be any repeated element         if (n <= 1)             return -1;                // initialize fast and slow         int slow = arr;         int fast = arr[arr];                // loop to enter in the cycle         while (fast != slow)          {                    // move one step for slow             slow = arr[slow];                    // move two step for fast             fast = arr[arr[fast]];         }                // loop to find entry         // point of the cycle         fast = 0;         while (slow != fast)         {             slow = arr[slow];             fast = arr[fast];         }         return slow;     }            // Driver Code     public static void Main()     {         int []arr = {1, 2, 3, 4, 5, 6, 3};         int n = arr.Length;         Console.Write(findduplicate(arr, n));     } }    // This code is contributed by // Manish Shaw (manishshaw1)

PHP



Output:

3

Time Complexity : O(n)
Auxiliary Space : O(1)

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

3

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.