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

Difficulty Level : Hard
Last Updated : 03 May, 2021

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.

So, here is an approach that is based on Floyd’s cycle finding algorithm. We use this to detect loops 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 a loop as shown in the image below- In case of a 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.

We maintain two pointers fast and slow
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)
When fast==slow that means now we are in a cycle.
Fast and slow will meet in a circle and the entry point of that circle will be the duplicate element.
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.
When fast==slow that will be entry point.
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 <bits/stdc++.h>
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[0];
    int fast = arr[arr[0]];
 
    // 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[0]);
    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[0];
        int fast = arr[arr[0]];
      
        // 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[0]
    fast = arr[arr[0]]
 
    # 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[0];
        int fast = arr[arr[0]];
     
        // 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

<?php
// PHP program to find a duplicate
// element in an array with values in
// range from 0 to n-1.
 
// function to find one duplicate
function 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[0];
    $fast = $arr[$arr[0]];
 
    // 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
$arr = array( 1, 2, 3, 4, 5, 6, 3 );
$n = sizeof($arr);
echo findduplicate($arr, $n);
 
// This code is contributed by Tushil
?>

Javascript

<script>
 
    // Javascript program to find a duplicate
    // element in an array with values in
    // range from 0 to n-1.
     
    // function to find one duplicate
    function 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
        let slow = arr[0];
        let fast = arr[arr[0]];
       
        // 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;
    }
     
    let arr = [1, 2, 3, 4, 5, 6, 3];
    let n = arr.length;
    document.write(findduplicate(arr, n));
     
</script>

Output:

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

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