Find the only repetitive element between 1 to N-1

Given an array of size N filled with numbers from 1 to N-1 in random order. The array has only one repetitive element. The task is to find the repetitive element.

Examples:

Input: a[] = {1, 3, 2, 3, 4}
Output: 3
Explanation: The number 3 is the only repeating element.

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

Naive Approach:  To solve the problem follow the below idea:

Use two nested loops. The outer loop traverses through all elements and the inner loop check if the element picked by the outer loop appears anywhere else.

Below is the implementation of the above approach:

C++

// C++ program to find the only repeating
// element in an array where elements are
// from 1 to N-1.
#include <bits/stdc++.h>
using namespace std;

int findRepeating(int arr[], int N)
{
    for (int i = 0; i < N; i++) {
        for (int j = i + 1; j < N; j++) {
            if (arr[i] == arr[j])
                return arr[i];
        }
    }
}

// Driver's code
int main()
{
    int arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function call
    cout << findRepeating(arr, N);
    return 0;
}
// This code is added by Arpit Jain

Java

// Java program to find the only repeating
// element in an array where elements are
// from 1 to N-1.using System;
public class GFG {
    static int findRepeating(int[] arr)
    {
        for (int i = 0; i < arr.length; i++) {
            for (int j = i + 1; j < arr.length; j++) {
                if (arr[i] == arr[j])
                    return arr[i];
            }
        }
        return -1;
    }

    // Driver's code
    static public void main(String[] args)
    {

        // Code
        int[] arr
            = new int[] { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
        int repeatingNum = findRepeating(arr);

        // Function call
        System.out.println(repeatingNum);
    }
}

// This code is contributed by phasing17

Python3

# Python3 program to find the only
# repeating element in an array where
# elements are from 1 to N-1.


def findRepeating(arr, N):
    for i in range(N):
        for j in range(i + 1, N):
            if (arr[i] == arr[j]):
                return arr[i]


# Driver's Code
if __name__ == "__main__":
  arr = [9, 8, 2, 6, 1, 8, 5, 3, 4, 7]
  N = len(arr)

  # Function call
  print(findRepeating(arr, N))

# This code is contributed by Arpit Jain

C#

// C# program to find the only repeating
// element in an array where elements are
// from 1 to N-1.
using System;

public class GFG {
    static int findRepeating(int[] arr)
    {
        for (int i = 0; i < arr.Length; i++) {
            for (int j = i + 1; j < arr.Length; j++) {
                if (arr[i] == arr[j])
                    return arr[i];
            }
        }
        return -1;
    }

    // Driver's code
    static public void Main()
    {
        int[] arr
            = new int[] { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };

          // Function call
        int repeatingNum = findRepeating(arr);
        Console.WriteLine(repeatingNum);
    }
}

// This code is contributed by Mohd Nizam

Javascript

<script>
//Javascript program to find the only repeating
// element in an array where elements are
// from 1 to N-1.
function findRepeating(arr, N){
    for (let i = 0; i <N; i++) {
            for (let j = i + 1; j < N; j++) {
                if (arr[i] == arr[j])
                    return arr[i];
            }
}
}
             // Driver code
let arr= [ 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 ];
let N = arr.length;

// Function call
console.log(findRepeating(arr, N));

// This code is contributed by aarohirai2616.
</script>

Output

8

Time Complexity: O(N²)
Auxiliary Space: O(1)

Find the only repetitive element using sorting:

Sort the given input array. Traverse the array and if value of the ith element is not equal to i+1, then the current element is repetitive as value of elements is between 1 and N-1 and every element appears only once except one element.

Follow the below steps to solve the problem:

• Sort the given array.
• Traverse the array and  compare the array elements with its index 
  • if arr[i] != i+1, it means that arr[i] is repetitive, So Just return arr[i]. 
• Otherwise, the array does not contain duplicates from 1 to n-1, In this case, return -1  

Below is the implementation of the above approach:

C++

// CPP program to find the only repeating
// element in an array where elements are
// from 1 to N-1.
#include <bits/stdc++.h>
using namespace std;

int findRepeating(int arr[], int N)
{
    sort(arr, arr + N); // sort array

    for (int i = 0; i < N; i++) {

        // compare array element with its index
        if (arr[i] != i + 1) {
            return arr[i];
        }
    }
    return -1;
}

// driver's code
int main()
{
    int arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function call
    cout << findRepeating(arr, N);
    return 0;
}
// this code is contributed by devendra solunke

Java

// Java program to find the only repeating
// element in an array where elements are
// from 1 to N-1.

import java.util.Arrays;

public class GFG {
    static int findRepeating(int[] arr, int N)
    {
        Arrays.sort(arr); // sort array

        for (int i = 0; i < N; i++) {

            // compare array element with its index
            if (arr[i] != i + 1) {
                return arr[i];
            }
        }
        return -1;
    }

    // Driver's code
    static public void main(String[] args)
    {
        int[] arr
            = new int[] { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
        int N = arr.length;

          // Function call
        int repeatingNum = findRepeating(arr, N);
        System.out.println(repeatingNum);
    }
}

// This code is contributed by Abhijeet Kumar(abhijeet19403)

Python3

# Python3 program to find the only
# repeating element in an array where
# elements are from 1 to N-1.


def findRepeating(arr, N):
    arr.sort()
    for i in range(1, N):
        if(arr[i] != i+1):
            return arr[i]


# Driver's Code
if __name__ == "__main__":
  arr = [9, 8, 2, 6, 1, 8, 5, 3, 4, 7]
  N = len(arr)

  # Function call
  print(findRepeating(arr, N))

# This code is contributed by Arpit Jain

C#

// C# program to find the only repeating element in an array
// where elements are from 1 to N-1.

using System;

public class GFG {

    static int findRepeating(int[] arr, int N)
    {
        Array.Sort(arr); // sort array
        for (int i = 0; i < N; i++) {
            // compare array element with its index
            if (arr[i] != i + 1) {
                return arr[i];
            }
        }
        return -1;
    }

      // Driver's code
    static public void Main()
    {
        int[] arr = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
        int N = arr.Length;

          // Function call
        int repeatingNum = findRepeating(arr, N);
        Console.WriteLine(repeatingNum);
    }
}

// This code is contributed by lokesh (lokeshmvs21).

Javascript

<script>
//Javascript program to find the only repeating
// element in an array where elements are
// from 1 to N-1.
function findRepeating(arr, N){
    arr.sort();
    for (let i = 0; i < N; i++) {

    // compare array element with its index
        if (arr[i] != i + 1) {
              return arr[i];
            }
        }
        return -1;
}
           // Driver code
let arr= [ 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 ];
let N = arr.length;

// Function call
console.log(findRepeating(arr, N));

// This code is contributed by aarohirai2616.
</script>

Output

8

Time complexity: O(N * log N) 
Auxiliary Space: O(1)

Find the only repetitive element using a frequency array:

Use a array to store frequency of  elements appeared in array. If frequency of element is greater than one,then return it.

C++

#include <bits/stdc++.h>
using namespace std;

int findRepeating(int arr[], int N)
{
    int freq[N+1];// initializing array for all
       memset(freq, 0, sizeof(freq));// elements with frequency 0.
    for (int i = 0; i < N; i++) {
        freq[arr[i]]++;
    }
    for (int i = 0; i < N; i++) {
        if (freq[arr[i]] > 1) {
            return arr[i];
        }
    }

    return -1;
}

int main()
{
    int arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function call
    cout << findRepeating(arr, N);
    return 0;
}

Java

public class Main {
    static int findRepeating(int arr[], int N) {
        int[] freq = new int[N + 1];

        for (int i = 0; i < N; i++) {
            freq[arr[i]]++;
        }

        for (int i = 0; i < N; i++) {
            if (freq[arr[i]] > 1) {
                return arr[i];
            }
        }

        return -1;
    }

    public static void main(String[] args) {
        int arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
        int N = arr.length;

        // Function call
        System.out.println(findRepeating(arr, N));
    }
}

Python3

def findRepeating(arr, N):
    freq = [0]*(N+1)
    for i in range(N):
        freq[arr[i]] += 1
    for i in range(N):
        if freq[arr[i]] > 1:
            return arr[i]
    return -1

arr = [9, 8, 2, 6, 1, 8, 5, 3, 4, 7]
N = len(arr)

# Function call
print(findRepeating(arr, N))

C#

using System;
using System.Linq;

class MainClass {
    public static int findRepeating(int[] arr, int N) {
        int[] freq = new int[N + 1];
        for (int i = 0; i < N; i++) {
            freq[arr[i]]++;
        }
        for (int i = 0; i < N; i++) {
            if (freq[arr[i]] > 1) {
                return arr[i];
            }
        }
        return -1;
    }

    public static void Main (string[] args) {
        int[] arr = {9, 8, 2, 6, 1, 8, 5, 3, 4, 7};
        int N = arr.Length;

        // Function call
        Console.WriteLine(findRepeating(arr, N));
    }
}

Javascript

function findRepeating(arr, N) {
    var freq = new Array(N + 1).fill(0);
    for (var i = 0; i < N; i++) {
        freq[arr[i]]++;
    }
    for (var i = 0; i < N; i++) {
        if (freq[arr[i]] > 1) {
            return arr[i];
        }
    }
    return -1;
}

var arr = [9, 8, 2, 6, 1, 8, 5, 3, 4, 7];
var N = arr.length;

// Function call
console.log(findRepeating(arr, N));

Output

8

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

Find the only repetitive element using the hash set:

Use a hash table to store elements visited. If an already visited element appears again, return it.

Follow the below steps to solve the problem:

• Create a hash set to store the visited elements
• Traverse the array
  • If the given element is already present in the hash set then, return this element
  • else insert this element into the hash set
• Return -1, if no repeating is found

Below is the implementation of the above approach:

C++

// C++ program to find the only repeating
// element in an array where elements are
// from 1 to N-1.

#include <bits/stdc++.h>
using namespace std;

int findRepeating(int arr[], int N)
{
    unordered_set<int> s;
    for (int i = 0; i < N; i++) {
        if (s.find(arr[i]) != s.end())
            return arr[i];
        s.insert(arr[i]);
    }

    // If input is correct, we should
    // never reach here
    return -1;
}

// Driver's code
int main()
{
    int arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);

      // Function call
    cout << findRepeating(arr, N);
    return 0;
}

Java

import java.util.*;
// Java program to find the only repeating
// element in an array where elements are
// from 1 to N-1.

class GFG {

    static int findRepeating(int arr[], int N)
    {
        HashSet<Integer> s = new HashSet<Integer>();
        for (int i = 0; i < N; i++) {
            if (s.contains(arr[i]))
                return arr[i];
            s.add(arr[i]);
        }

        // If input is correct, we should
        // never reach here
        return -1;
    }

    // Driver's code
    public static void main(String[] args)
    {
        int arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
        int N = arr.length;

          // Function call
        System.out.println(findRepeating(arr, N));
    }
}

// This code is contributed by Rajput-Ji

Python3

# Python3 program to find the only
# repeating element in an array
# where elements are from 1 to n-1.


def findRepeating(arr, N):
    s = set()
    for i in range(N):
        if arr[i] in s:
            return arr[i]
        s.add(arr[i])

    # If input is correct, we should
    # never reach here
    return -1


# Driver code
if __name__ == "__main__":
  arr = [9, 8, 2, 6, 1, 8, 5, 3]
  N = len(arr)

  # Function call
  print(findRepeating(arr, N))

# This code is contributed
# by Shrikant13

C#

// C# program to find the only repeating
// element in an array where elements are
// from 1 to N-1.
using System;
using System.Collections.Generic;

class GFG {

    static int findRepeating(int[] arr, int N)
    {
        HashSet<int> s = new HashSet<int>();
        for (int i = 0; i < N; i++) {
            if (s.Contains(arr[i]))
                return arr[i];
            s.Add(arr[i]);
        }

        // If input is correct, we should
        // never reach here
        return -1;
    }

    // Driver code
    public static void Main(String[] args)
    {
        int[] arr = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
        int N = arr.Length;

          // Function call
        Console.WriteLine(findRepeating(arr, N));
    }
}

// This code has been contributed by 29AjayKumar

Javascript

// JavaScript program to find the only repeating
// element in an array where elements are
// from 1 to n-1.

function findRepeating(arr,n)
{
  s = new Set();
   for (let i = 0; i < n; i++)
   {
      if (s.has(arr[i]))
         return arr[i];
      s.add(arr[i]);
   }

   // If input is correct, we should
   // never reach here
   return -1;
}

// driver code
let arr = [ 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 ];
let n = arr.length;
document.write(findRepeating(arr, n));

// This code is contributed by shinjanpatra.

Output

8

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

Find the only repetitive element using the Sum of first N elements:

We know sum of first n-1 natural numbers is (N – 1)*N/2. We compute sum of array elements and subtract natural number sum from it to find the only missing element.

Follow the below steps to solve the problem:

• Calculate the sum of array elements and the sum of first (N-1) natural numbers
• Return (array sum) – ((N-1) natural numbers sum)

Below is the implementation of the above approach:

C++

// CPP program to find the only repeating
// element in an array where elements are
// from 1 to N-1.
#include <bits/stdc++.h>
using namespace std;

int findRepeating(int arr[], int N)
{
    // Find array sum and subtract sum
    // first N-1 natural numbers from it
    // to find the result.
    return accumulate(arr, arr + N, 0) - ((N - 1) * N / 2);
}

// Driver's code
int main()
{
    int arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);

      // Function call
    cout << findRepeating(arr, N);
    return 0;
}

Java

// Java program to find the only repeating
// element in an array where elements are
// from 1 to N-1.
import java.io.*;
import java.util.*;

class GFG {

    static int findRepeating(int[] arr, int N)
    {
        // Find array sum and subtract sum
        // first n-1 natural numbers from it
        // to find the result.

        int sum = 0;
        for (int i = 0; i < N; i++)
            sum += arr[i];
        return sum - (((N - 1) * N) / 2);
    }

    // Driver's code
    public static void main(String args[])
    {
        int[] arr = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
        int N = arr.length;

          // Function call
        System.out.println(findRepeating(arr, N));
    }
}

// This code is contributed by rachana soma

Python3

# Python3 program to find the only
# repeating element in an array where
# elements are from 1 to N-1.


def findRepeating(arr, N):

    # Find array sum and subtract sum
    # first n-1 natural numbers from it
    # to find the result.
    return sum(arr) - (((N - 1) * N) // 2)


# Driver's Code
if __name__ == "__main__":
  arr = [9, 8, 2, 6, 1, 8, 5, 3, 4, 7]
  N = len(arr)

  # Function call
  print(findRepeating(arr, N))

# This code is contributed
# by mohit kumar

C#

// C# program to find the only repeating
// element in an array where elements are
// from 1 to N-1.
using System;

class GFG {

    static int findRepeating(int[] arr, int N)
    {
        // Find array sum and subtract sum
        // first N-1 natural numbers from it
        // to find the result.

        int sum = 0;
        for (int i = 0; i < N; i++)
            sum += arr[i];
        return sum - (((N - 1) * N) / 2);
    }

    // Driver's code
    public static void Main(String[] args)
    {
        int[] arr = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
        int N = arr.Length;

          // Function call
        Console.WriteLine(findRepeating(arr, N));
    }
}

/* This code contributed by PrinciRaj1992 */

Javascript

// javascript program to find the only repeating
// element in an array where elements are
// from 1 to n-1.
function findRepeating(arr , n)
{

        // Find array sum and subtract sum
        // first n-1 natural numbers from it
        // to find the result.
        var sum = 0;
        for (i = 0; i < n; i++)
            sum += arr[i];
        return sum - (((n - 1) * n) / 2);
    }

    // Driver code
        var arr = [ 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 ];
        var n = arr.length;
        document.write(findRepeating(arr, n));

// This code is contributed by Rajput-Ji

Output

8

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

Note: This approach Causes overflow for large arrays.

Find the only repetitive element using XOR:

The idea is based on the fact that x ^ x = 0 and if x ^ y = z then x ^ z = y

Follow the below steps to solve the problem:

• Compute XOR of elements from 1 to n-1.
• Compute XOR of array elements.
• XOR of the above two would be our result.

Below is the implementation of the above approach:

C++

// CPP program to find the only repeating
// element in an array where elements are
// from 1 to N-1.
#include <bits/stdc++.h>
using namespace std;

int findRepeating(int arr[], int N)
{

    // res is going to store value of
    // 1 ^ 2 ^ 3 .. ^ (N-1) ^ arr[0] ^
    // arr[1] ^ .... arr[N-1]
    int res = 0;
    for (int i = 0; i < N - 1; i++)
        res = res ^ (i + 1) ^ arr[i];
    res = res ^ arr[N - 1];
    return res;
}

// Driver code
int main()
{
    int arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);

      // Function call
    cout << findRepeating(arr, N);
    return 0;
}

Java

// Java program to find the only repeating
// element in an array where elements are
// from 1 to N-1.
import java.io.*;
class GFG {

    static int findRepeating(int arr[], int N)
    {

        // res is going to store value of
        // 1 ^ 2 ^ 3 .. ^ (N-1) ^ arr[0] ^
        // arr[1] ^ .... arr[n-1]
        int res = 0;
        for (int i = 0; i < N - 1; i++)
            res = res ^ (i + 1) ^ arr[i];
        res = res ^ arr[N - 1];

        return res;
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
        int N = arr.length;

          // Function call
        System.out.println(findRepeating(arr, N));
    }
}

// This code is contributed by
// Smitha Dinesh Semwal.

Python3

# Python3 program to find the only
# repeating element in an array where
# elements are from 1 to N-1.


def findRepeating(arr, N):

    # res is going to store value of
    # 1 ^ 2 ^ 3 .. ^ (N-1) ^ arr[0] ^
    # arr[1] ^ .... arr[n-1]
    res = 0
    for i in range(0, N-1):
        res = res ^ (i+1) ^ arr[i]
    res = res ^ arr[N-1]

    return res


# Driver code
if __name__ == "__main__":
  arr = [9, 8, 2, 6, 1, 8, 5, 3, 4, 7]
  N = len(arr)

  # Function call
  print(findRepeating(arr, N))

# This code is contributed by Smitha Dinesh Semwal.

C#

// C# program to find the only repeating
// element in an array where elements are
// from 1 to N-1.
using System;
public class GFG {

    static int findRepeating(int[] arr, int N)
    {

        // res is going to store value of
        // 1 ^ 2 ^ 3 .. ^ (n-1) ^ arr[0] ^
        // arr[1] ^ .... arr[N-1]
        int res = 0;
        for (int i = 0; i < N - 1; i++)
            res = res ^ (i + 1) ^ arr[i];
        res = res ^ arr[N - 1];

        return res;
    }

    // Driver code
    public static void Main(String[] args)
    {
        int[] arr = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
        int N = arr.Length;

          // Function call
        Console.WriteLine(findRepeating(arr, N));
    }
}

// This code is contributed by Rajput-Ji

Javascript

// JavaScript program to find the only repeating
// element in an array where elements are
// from 1 to n-1.


function findRepeating(arr,n)
{

  // res is going to store value of
  // 1 ^ 2 ^ 3 .. ^ (n-1) ^ arr[0] ^
  // arr[1] ^ .... arr[n-1]
  let res = 0;
  for (let i=0; i<n-1; i++)
    res = res ^ (i+1) ^ arr[i];
  res = res ^ arr[n-1];
  return res;
}

// driver code

let arr = [ 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 ];
let n = arr.length;
document.write(findRepeating(arr, n));

PHP

<?php
// PHP program to find the only repeating
// element in an array where elements are
// from 1 to n-1.

function findRepeating($arr, $N)
{

    // res is going to store value of
    // 1 ^ 2 ^ 3 .. ^ (N-1) ^ arr[0] ^
    // arr[1] ^ .... arr[N-1]
    $res = 0;
    for($i = 0; $i < $N - 1; $i++)
        $res = $res ^ ($i + 1) ^ $arr[$i];
    $res = $res ^ $arr[$N - 1];

    return $res;
}

    // Driver Code
    $arr =array(9, 8, 2, 6, 1, 8, 5, 3, 4, 7);
    $N = sizeof($arr) ;

    // Function call
    echo findRepeating($arr, $N);

// This code is contributed by ajit
?>

Output

8

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

Find the only repetitive element using indexing:

As there are only positive numbers, so visit the index equal to the current element and make it negative. If an index value is already negative, then it means that current element is repeated

Follow the below steps to solve the problem:

• Iterate through the array.
• For every index visit arr[arr[i]], if it is positive change the sign of elements at that index, else print the element.

Below is the implementation of the above approach:

C++

// CPP program to find the only
// repeating element in an array
// where elements are from 1 to N-1.

#include <bits/stdc++.h>
using namespace std;

// Function to find repeated element
int findRepeating(int arr[], int N)
{
    int missingElement = 0;

    // indexing based
    for (int i = 0; i < N; i++) {

        int element = arr[abs(arr[i])];

        if (element < 0) {
            missingElement = arr[i];
            break;
        }

        arr[abs(arr[i])] = -arr[abs(arr[i])];
    }

    return abs(missingElement);
}

// Driver code
int main()
{
    int arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };

    int N = sizeof(arr) / sizeof(arr[0]);

    // Function call
    cout << findRepeating(arr, N);

    return 0;
}

Java

// Java program to find the only
// repeating element in an array
// where elements are from 1 to N-1.
import java.lang.Math.*;

class GFG {

    // Function to find repeated element
    static int findRepeating(int arr[], int N)
    {
        int missingElement = 0;

        // indexing based
        for (int i = 0; i < N; i++) {

            int absVal = Math.abs(arr[i]);
            int element = arr[absVal];

            if (element < 0) {
                missingElement = arr[i];
                break;
            }

            absVal = Math.abs(arr[i]);
            arr[absVal] = -arr[absVal];
        }

        return Math.abs(missingElement);
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };

        int N = arr.length;

          // Function call
        System.out.println(findRepeating(arr, N));
    }
}
// This code is contributed by
// Smitha Dinesh Semwal.

Python3

# Python3 program to find the only
# repeating element in an array
# where elements are from 1 to N-1.

# Function to find repeated element


def findRepeating(arr, N):

    missingElement = 0

    # indexing based
    for i in range(0, N):

        element = arr[abs(arr[i])]

        if(element < 0):
            missingElement = arr[i]
            break

        arr[abs(arr[i])] = -arr[abs(arr[i])]

    return abs(missingElement)


# Driver code
if __name__ == "__main__":
  arr = [9, 8, 2, 6, 1, 8, 5, 3, 4, 7]
  N = len(arr)

  # Function call
  print(findRepeating(arr, N))

# This code is contributed by Smitha Dinesh Semwal.

C#

// C# program to find the only
// repeating element in an array
// where elements are from 1 to N-1.
using System;

public class GFG {

    // Function to find repeated element
    static int findRepeating(int[] arr, int N)
    {
        int missingElement = 0;

        // indexing based
        for (int i = 0; i < N; i++) {

            int absVal = Math.Abs(arr[i]);
            int element = arr[absVal];

            if (element < 0) {
                missingElement = arr[i];
                break;
            }

            absVal = Math.Abs(arr[i]);
            arr[absVal] = -arr[absVal];
        }

        return Math.Abs(missingElement);
    }

    // Driver Code
    public static void Main(string[] args)
    {
        int[] arr = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };

        int N = arr.Length;

          // Function call
        Console.WriteLine(findRepeating(arr, N));
    }
}

// this code is contributed by phasing17

Javascript

     // JavaScript program for the above approach;

     // Function to find repeated element
     function findRepeating(arr, n) {
         let missingElement = 0;

         // indexing based
         for (let i = 0; i < n; i++) {

             let absVal = Math.abs(arr[i]);
             let element = arr[absVal];

             if (element < 0) {
                 missingElement = arr[i];
                 break;
             }

             let absVal = Math.abs(arr[i]);
             arr[absVal] = -arr[absVal];
         }

         return Math.abs(missingElement);

     }

     // driver code

     let arr = [5, 4, 3, 9, 8,
         9, 1, 6, 2, 5];

     let n = arr.length;

     document.write(findRepeating(arr, n));
// This code is contributed by Potta Lokesh

PHP

<?php
// PHP program to find the only
// repeating element in an array
// where elements are from 1 to n-1.

// Function to find repeated elements
function findRepeating($arr, $N)
{
    $missingElement = 0;

    // indexing based
    for ($i = 0; $i < $N; $i++)
    {

        $element = $arr[abs($arr[$i])];

        if($element < 0)
        {
            $missingElement = $arr[$i];
            break;
        }

    $arr[abs($arr[$i])] = -$arr[abs($arr[$i])];
}

return abs($missingElement);

}

// Driver Code
$arr = array (5, 4, 3, 9, 8,
              9, 1, 6, 2, 5);

$N = sizeof($arr);

// Function call
echo findRepeating($arr, $N);

// This code is contributed by ajit
?>

Output

8

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

Find the only repetitive element using Linked-List cycle method:

Use two pointers the fast and the slow. The fast one goes forward two steps each time, while the slow one goes only step each time. They must meet the same item when slow==fast. 

In fact, they meet in a circle, the duplicate number must be the entry point of the circle when visiting the array from array[0]. 

Next we just need to find the entry point. We use a point(we can use the fast one before) to visit from the beginning with one step each time, do the same job to slow. When fast==slow, they meet at the entry point of the circle. 

Follow the below steps to solve the problem:

• Declare two integer pointers as slow and fast
• Move the slow pointer one time and fast pointer two times, until slow is not equal to fast
• Once they are equal then again start the fast pointer from the start of the array
• Move both the pointers, one step at a time until both of them are equal
• Return slow or fast pointer as the answer

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;

int findDuplicate(vector<int>& nums)
{
    int slow = nums[0];//slow pointer
    int fast = nums[0];//fast pointer

    do {
        slow = nums[slow];//moves one step-->if speed is x
        fast = nums[nums[fast]];//it moves fast than slow pointer-->speed is 2x
    } while (slow != fast);

    fast = nums[0];//initialize fast to the starting point
  //loop untill  both pointer will meet at the same point
    while (slow != fast) {
        slow = nums[slow];
        fast = nums[fast];
    }
//return duplicate number
    return slow;
}

// Driver code
int main()
{
    vector<int> v{ 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };

      // Function call
    int ans = findDuplicate(v);
    cout << ans << endl;
    return 0;
}

Java

import java.io.*;
import java.util.*;

class GFG {

  public static int findDuplicate(ArrayList<Integer> nums)
  {
    int slow = nums.get(0);
    int fast = nums.get(0);

    do {
      slow = nums.get(slow);
      fast = nums.get(nums.get(fast));
    }
    while (slow != fast);

    fast = nums.get(0);
    while (slow != fast) {
      slow = nums.get(slow);
      fast = nums.get(fast);
    }

    return slow;
  }

  public static void main(String[] args)
  {
    ArrayList<Integer> arr = new ArrayList<Integer>( Arrays. asList( 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 ) );

    // Function call
    int ans = findDuplicate(arr);
    System.out.println(ans);
  }
}

// This code is contributed by adityapatil12

Python3

class GFG :
    @staticmethod
    def  findDuplicate( nums) :
        slow = nums[0]
        fast = nums[0]
        while True :
            slow = nums[slow]
            fast = nums[nums[fast]]
            if((slow != fast) == False) :
                    break
        fast = nums[0]
        while (slow != fast) :
            slow = nums[slow]
            fast = nums[fast]
        return slow
    @staticmethod
    def main( args) :
        arr = [9, 8, 2, 6, 1, 8, 5, 3, 4, 7]

        # Function call
        ans = GFG.findDuplicate(arr)
        print(ans)

if __name__=="__main__":
    GFG.main([])

    # This code is contributed by aadityaburujwale.

C#

// C# code for the above approach
using System;

public class GFG {

  static int findDuplicate(int[] nums)
  {
    int slow = nums[0];
    int fast = nums[0];

    do {
      slow = nums[slow];
      fast = nums[nums[fast]];
    } while (slow != fast);

    fast = nums[0];
    while (slow != fast) {
      slow = nums[slow];
      fast = nums[fast];
    }

    return slow;
  }

  static public void Main()
  {

    // Code
    int[] v = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };

    // Function call
    int ans = findDuplicate(v);
    Console.Write(ans);
  }
}

// This code is contributed by lokesh.

Javascript

function findDuplicate(nums){
    var slow = nums[0];
    var fast = nums[0];

    do {
      slow = nums[slow];
      fast = nums[nums[fast]];
    }
    while (slow != fast);

    fast = nums[0];
    while (slow != fast) {
      slow = nums[slow];
      fast = nums[fast];
    }

    return slow;
}

    var arr = [ 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 ];

    // Function call
    var ans = findDuplicate(arr);
    console.log(ans);

    // This code is contributed by aadityaburujwale.

Output

8

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

Find the only repetitive element Using Binary Search:

Initialize low to 1 (since the array has values in the range of 1 to n, where n is the size of the array) and high to nums.size() – 1. Enter a while loop that iteratively reduces the search range by half: Calculate mid as the average of low and high. Iterate over the nums array and count the number of elements that are less than or equal to mid. Based on the count of elements less than or equal to mid, perform a binary search on the left half of the array if the count is less than or equal to mid, or on the right half of the array otherwise. Return the low variable, which is the duplicate number in the array. The reason for this is that the code performs a binary search on the array, which has a time complexity of O(log n), and it iterates over the array to count the number of elements less than or equal to the midpoint in each iteration. The worst-case time complexity of this iteration is O(n log n) because the total number of iterations required is proportional to the height of the binary search tree, which is log n, and at each iteration, we are visiting all elements in the array, which has a time complexity of O(n). Therefore, the overall time complexity is O(n log n). implementation is given below:-

C++

#include <bits/stdc++.h>
using namespace std;

int findDuplicate(vector<int>& nums)
{
    int low = 1, high = nums.size() - 1, cnt;

        while(low <=  high)
        {
            int mid = low + (high - low) / 2;
            cnt = 0;
            // cnt number less than equal to mid
            for(int n : nums)
            {
                if(n <= mid)
                    ++cnt;
            }
            // binary search on left
            if(cnt <= mid)
                low = mid + 1;
            else
            // binary search on right
                high = mid - 1;

        }
        return low;
}

// Driver code
int main()
{
    vector<int> v{ 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };

      // Function call
    int ans = findDuplicate(v);
    cout << ans << endl;//8
    return 0;
}

//priyapramanick17

Java

import java.util.ArrayList;
import java.util.List;

public class FindDuplicate {
    public static void main(String[] args) {
        List<Integer> nums = new ArrayList<>();
        nums.add(9);
        nums.add(8);
        nums.add(2);
        nums.add(6);
        nums.add(1);
        nums.add(8);
        nums.add(5);
        nums.add(3);
        nums.add(4);
        nums.add(7);

        // Function call
        int ans = findDuplicate(nums);
        System.out.println(ans); // 8
    }

    public static int findDuplicate(List<Integer> nums) {
        int low = 1;
        int high = nums.size() - 1;
        int cnt;

        while (low <= high) {
            int mid = low + (high - low) / 2;
            cnt = 0;

            // Count the number of elements less than or equal to mid
            for (int n : nums) {
                if (n <= mid)
                    cnt++;
            }

            // Binary search on the left
            if (cnt <= mid)
                low = mid + 1;
            else
                // Binary search on the right
                high = mid - 1;
        }
        return low;
    }
}

Python

def find_duplicate(nums):
    # Initialize the low and high values for binary search
    low = 1
    high = len(nums) - 1

    while low <= high:
        mid = low + (high - low) // 2  # Calculate the middle index
        cnt = 0

        # Count the numbers less than or equal to mid
        for n in nums:
            if n <= mid:
                cnt += 1

        # Perform binary search on the left half
        if cnt <= mid:
            low = mid + 1
        else:
            # Perform binary search on the right half
            high = mid - 1

    return low

# Driver code
if __name__ == "__main__":
    v = [9, 8, 2, 6, 1, 8, 5, 3, 4, 7]

    # Function call
    ans = find_duplicate(v)
    print(ans)  # Output should be 8

C#

using System;
using System.Collections.Generic;

class Program
{
    // Function to find the duplicate element in a list of integers
    public static int FindDuplicate(List<int> nums)
    {
        int low = 1;
        int high = nums.Count - 1;
        int cnt;

        while (low <= high)
        {
            int mid = low + (high - low) / 2;
            cnt = 0;

            // Count numbers less than or equal to mid in the list
            foreach (int n in nums)
            {
                if (n <= mid)
                {
                    cnt++;
                }
            }

            // If the count of numbers less than or equal to mid is less than or equal to mid,
            // it means the duplicate element is on the right side of mid.
            if (cnt <= mid)
            {
                low = mid + 1; // Binary search on the right side
            }
            else
            {
                high = mid - 1; // Binary search on the left side
            }
        }

        return low; // The duplicate element is found at 'low'
    }

    static void Main(string[] args)
    {
        List<int> nums = new List<int> { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };

        // Function call
        int ans = FindDuplicate(nums);
        Console.WriteLine(ans); // Print the duplicate element
    }
}

Javascript

function findDuplicate(nums) {
    let low = 1;
    let high = nums.length - 1;
    let cnt;

    while (low <= high) {
        let mid = low + Math.floor((high - low) / 2);
        cnt = 0;
        // cnt number less than equal to mid
        for (let n of nums) {
            if (n <= mid)
                ++cnt;
        }
        // binary search on left
        if (cnt <= mid)
            low = mid + 1;
        else
            // binary search on right
            high = mid - 1;

    }
    return low;
}

// Driver code
let v = [9, 8, 2, 6, 1, 8, 5, 3, 4, 7];

// Function call
let ans = findDuplicate(v);
console.log(ans); //8