Find duplicate in an array in O(n) and by using O(1) extra space

Given an array arr[] containing n + 1 integers where each integer is between 1 and n (inclusive). There is only one duplicate element, find the duplicate element in O(n) time complexity and O(1) space.
Examples :

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

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

Approach :
Firstly, the constraints of this problem imply that a cycle must exist. Because each number in an array arr[] is between 1 and n, it will necessarily point to an index that exists. Therefore, the list can be traversed infinitely, which implies that there is a cycle. Additionally, because 0 cannot appear as a value in an array arr[], arr[0] cannot be part of the cycle. Therefore, traversing the array in this manner from arr[0] is equivalent to traversing a cyclic linked list. The problem can be solved just like linked list cycle.
Below is the implementation of above approach: 

C++

// CPP code to find the repeated elements
// in the array where every other is present once
#include <iostream>
using namespace std;

// Function to find duplicate
int findDuplicate(int arr[])
{
    // Find the intersection point of the slow and fast.
    int slow = arr[0];
    int fast = arr[0];
    do {
        slow = arr[slow];
        fast = arr[arr[fast]];
    } while (slow != fast);
    // Find the "entrance" to the cycle.
    int ptr1 = arr[0];
    int ptr2 = slow;
    while (ptr1 != ptr2) {
        ptr1 = arr[ptr1];
        ptr2 = arr[ptr2];
    }
    return ptr1;
}

// Driver code
int main()
{
    int arr[] = { 1, 3, 2, 1 };
    cout << findDuplicate(arr) << endl;
    return 0;
}

// This code is contributed by Sania Kumari Gupta (kriSania804)

C

// C code to find the repeated elements
// in the array where every other is present once
#include <stdio.h>

// Function to find duplicate
int findDuplicate(int arr[])
{
    // Find the intersection point of the slow and fast.
    int slow = arr[0];
    int fast = arr[0];
    do {
        slow = arr[slow];
        fast = arr[arr[fast]];
    } while (slow != fast);

    // Find the "entrance" to the cycle.
    int ptr1 = arr[0];
    int ptr2 = slow;
    while (ptr1 != ptr2) {
        ptr1 = arr[ptr1];
        ptr2 = arr[ptr2];
    }
    return ptr1;
}

// Driver code
int main()
{
    int arr[] = { 1, 3, 2, 1 };
    printf("%d", findDuplicate(arr));
    return 0;
}

// This code is contributed by Sania Kumari Gupta (kriSania804)

Java

// Java code to find the repeated
// elements in the array where
// every other is present once
import java.util.*;

class GFG
{

// Function to find duplicate
public static int findDuplicate(int []arr)
{
    // Find the intersection
    // point of the slow and fast.
    int slow = arr[0];
    int fast = arr[0];
    do
    {
        slow = arr[slow];
        fast = arr[arr[fast]];
    } while (slow != fast);

    // Find the "entrance"
    // to the cycle.
    int ptr1 = arr[0];
    int ptr2 = slow;
    while (ptr1 != ptr2)
    {
        ptr1 = arr[ptr1];
        ptr2 = arr[ptr2];
    }

    return ptr1;
}

// Driver Code
public static void main(String[] args)
{
    int []arr = {1, 3, 2, 1};

    System.out.println("" +
               findDuplicate(arr));

    System.exit(0);
}
}

// This code is contributed
// by Harshit Saini

Python3

# Python code to find the
# repeated elements in the
# array where every other
# is present once

# Function to find duplicate
def findDuplicate(arr):

    # Find the intersection
    # point of the slow and fast.
    slow = arr[0]
    fast = arr[0]
    while True:
        slow = arr[slow]
        fast = arr[arr[fast]]
        if slow == fast:
            break

    # Find the "entrance"
    # to the cycle.
    ptr1 = arr[0]
    ptr2 = slow
    while ptr1 != ptr2:
        ptr1 = arr[ptr1]
        ptr2 = arr[ptr2]

    return ptr1

# Driver code
if __name__ == '__main__':


    arr = [ 1, 3, 2, 1 ]

    print(findDuplicate(arr))


# This code is contributed
# by Harshit Saini

C#

// C# code to find the repeated
// elements in the array where
// every other is present once
using System;

class GFG
{

// Function to find duplicate
public static int findDuplicate(int []arr)
{
    // Find the intersection
    // point of the slow and fast.
    int slow = arr[0];
    int fast = arr[0];
    do
    {
        slow = arr[slow];
        fast = arr[arr[fast]];
    } while (slow != fast);

    // Find the "entrance"
    // to the cycle.
    int ptr1 = arr[0];
    int ptr2 = slow;
    while (ptr1 != ptr2)
    {
        ptr1 = arr[ptr1];
        ptr2 = arr[ptr2];
    }

    return ptr1;
}

// Driver Code
public static void Main()
{
    int[] arr = {1, 3, 2, 1};

    Console.WriteLine("" +
            findDuplicate(arr));

}
}

// This code is contributed
// by Akanksha Rai(Abby_akku)

PHP

<?php
// PHP code to find the repeated
// elements in the array where
// every other is present once

// Function to find duplicate
function findDuplicate(&$arr)
{
    $slow = $arr[0];
    $fast = $arr[0];
    do
    {
        $slow = $arr[$slow];
        $fast = $arr[$arr[$fast]];
    } while ($slow != $fast);

    // Find the "entrance"
    // to the cycle.
    $ptr1 = $arr[0];
    $ptr2 = $slow;
    while ($ptr1 != $ptr2)
    {
        $ptr1 = $arr[$ptr1];
        $ptr2 = $arr[$ptr2];
    }

    return $ptr1;
}

// Driver code
$arr = array(1, 3, 2, 1);
echo " " . findDuplicate($arr);

// This code is contributed
// by Shivi_Aggarwal
?>

Javascript

<script>
// JavaScript code to find the repeated elements
// in the array where every other is present once

// Function to find duplicate
function findDuplicate(arr)
{

    // Find the intersection point of
    // the slow and fast.
    let slow = arr[0];
    let fast = arr[0];
    do
    {
        slow = arr[slow];
        fast = arr[arr[fast]];
    } while (slow != fast);

    // Find the "entrance" to the cycle.
    let ptr1 = arr[0];
    let ptr2 = slow;
    while (ptr1 != ptr2)
    {
        ptr1 = arr[ptr1];
        ptr2 = arr[ptr2];
    }
    return ptr1;
}

// Driver code
    let arr = [ 1, 3, 2, 1 ];
    document.write(findDuplicate(arr) + "<br>");

// This code is contributed by Surbhi Tyagi.
</script>

Output:

1

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