Find the only repeating element in a sorted array of size n

Given a sorted array of n elements containing elements in range from 1 to n-1 i.e. one element occurs twice, the task is to find the repeating element in an array.

Examples :

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

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

Brute Force:

1. Traverse the input array using a for a loop.
2. For each element in the array, traverse the remaining part of the array using another for loop.
3. For each subsequent element, check if it is equal to the current element.
4. If a match is found, return the current element.
5. If no match is found, continue with the next element in the outer loop.
6. If the outer loop completes without finding a match, return -1 to indicate that there is no repeating element in the array.

Below is the implementation of the above approach:

C++

// C++ program to find the only repeating element in an
// array of size n and elements from range 1 to n-1.
#include <bits/stdc++.h>
using namespace std;


// Returns index of second appearance of a repeating element
// The function assumes that array elements are in range from
// 1 to n-1.
int FindRepeatingElement(int arr[], int size){
    for(int i=0; i<size; i++){
        for(int j=i+1; j<size; j++){
            if(arr[i] == arr[j])
                return i;
        }
    }
    return -1;
}


// Driver code
int main()
{
    int arr[] = {1, 2 , 3 , 4 , 4};
    int n = sizeof(arr) / sizeof(arr[0]);
    int index = FindRepeatingElement(arr, n);
    if (index != -1)
        cout << arr[index];
    return 0;
}

Java

import java.util.*;

public class Main {

    // Returns index of second appearance of a repeating element
    // The function assumes that array elements are in range from
    // 1 to n-1.
    public static int findRepeatingElement(int arr[], int size){
        for(int i=0; i<size; i++){
            for(int j=i+1; j<size; j++){
                if(arr[i] == arr[j])
                    return i;
            }
        }
        return -1;
    }

    // Driver code
    public static void main(String[] args) {
        int arr[] = {1, 2 , 3 , 4 , 4};
        int n = arr.length;
        int index = findRepeatingElement(arr, n);
        if (index != -1)
            System.out.println(arr[index]);
    }
}

Python3

# Python3 program to find the only repeating element in an
# array of size n and elements from range 1 to n-1.

# Returns second appearance of a repeating element
# The function assumes that array elements are in range from
# 1 to n-1.
def FindRepeatingElement(arr, size):
    for i in range(size):
        for j in range(i+1, size):
            if arr[i] == arr[j]:
                return arr[i]
    return -1

# Driver code
if __name__ == '__main__':
    arr = [1, 2, 3, 4, 4]
    n = len(arr)
    element = FindRepeatingElement(arr, n)
    if element != -1:
        print(element)

C#

using System;

public class Program
{

  // Returns index of second appearance of a repeating element
  // The function assumes that array elements are in range from
  // 1 to n-1.
  public static int FindRepeatingElement(int[] arr, int size)
  {
    for (int i = 0; i < size; i++)
    {
      for (int j = i + 1; j < size; j++)
      {
        if (arr[i] == arr[j])
        {
          return i;
        }
      }
    }
    return -1;
  }

  // Driver code
  public static void Main()
  {
    int[] arr = { 1, 2, 3, 4, 4 };
    int n = arr.Length;
    int index = FindRepeatingElement(arr, n);
    if (index != -1)
    {
      Console.WriteLine(arr[index]);
    }
  }
}

Javascript

// JavaScript program to find the only repeating element in an
// array of size n and elements from range 1 to n-1.

// Returns index of second appearance of a repeating element
// The function assumes that array elements are in range from
// 1 to n-1.
function FindRepeatingElement(arr, size)
{
    for(let i=0; i<size; i++)
    {
        for(let j=i+1; j<size; j++)
        {
            if(arr[i] == arr[j])
                return i;
        }
    }
    return -1;
}

// Driver code
let arr = [1, 2 , 3 , 4 , 4];
let n = arr.length;
let index = FindRepeatingElement(arr, n);
if (index != -1)
console.log(arr[index]);

// This code is contributed by akashish__

Output

4

Time Complexity: O(N^2)
Auxiliary Space: O(1)

An efficient method is to use Binary Search.

Observation: If an element ‘X’ is repeating, then it must be at index ‘X’ in the array. So the problem reduces to find any element whose value is same as its index.

C++

// C++ program to find the only repeating element in an
// array of size n and elements from range 1 to n-1.
#include <bits/stdc++.h>
using namespace std;


// Returns index of second appearance of a repeating element
// The function assumes that array elements are in range from
// 1 to n-1.
int FindRepeatingElement(int arr[], int size){
    int lo = 0;
    int hi = size - 1;
    int mid;

    while(lo <= hi){
        mid = (lo+hi)/2;

        if(arr[mid] <= mid){
            hi = mid-1;
        }
        else{
            lo = mid + 1;
        }
    }

    return lo;
}

// Driver code
int main()
{
    int arr[] = {1, 2, 3, 3, 4, 5};
    int n = sizeof(arr) / sizeof(arr[0]);
    int index = FindRepeatingElement(arr, n);
    if (index != -1)
        cout << arr[index];
    return 0;
}

Java

// Java program to find the only repeating element in an
// array of size n and elements from range 1 to n-1.

class Test
{
    // Returns index of second appearance of a repeating element
    // The function assumes that array elements are in range from
    // 1 to n-1.
    static int findRepeatingElement(int arr[], int low, int high)
    {
        // low = 0 , high = n-1;
        if (low > high)
            return -1;

        int mid = (low + high) / 2;

        // Check if the mid element is the repeating one
        if (arr[mid] != mid + 1)
        {
            if (mid > 0 && arr[mid]==arr[mid-1])
                return mid;

            // If mid element is not at its position that means
            // the repeated element is in left
            return  findRepeatingElement(arr, low, mid-1);
        }

        // If mid is at proper position then repeated one is in
        // right.
        return findRepeatingElement(arr, mid+1, high);
    }

    // Driver method
    public static void main(String[] args)
    {
        int  arr[] = {1, 2, 3, 3, 4, 5};
        int index = findRepeatingElement(arr, 0, arr.length-1);
        if (index != -1)
            System.out.println(arr[index]);
    }
}

Python3

# Python program to find the only repeating element in an
# array of size n and elements from range 1 to n-1

# Returns index of second appearance of a repeating element
# The function assumes that array elements are in range from
# 1 to n-1.
def findRepeatingElement(arr, low, high):

    # low = 0 , high = n-1
    if low > high:
        return -1

    mid = (low + high) // 2

    # Check if the mid element is the repeating one
    if (arr[mid] != mid + 1):

        if (mid > 0 and arr[mid]==arr[mid-1]):
            return mid

        # If mid element is not at its position that means
        # the repeated element is in left
        return  findRepeatingElement(arr, low, mid-1)

    # If mid is at proper position then repeated one is in
    # right.
    return findRepeatingElement(arr, mid+1, high)

# Driver code
arr = [1, 2, 3, 3, 4, 5]
n = len(arr)
index = findRepeatingElement(arr, 0, n-1)
if (index is not -1):
    print (arr[index])

#This code is contributed by Afzal Ansari

Javascript

<script>
      // JavaScript program to find the only repeating element in an
      // array of size n and elements from range 1 to n-1.

      // Returns index of second appearance of a repeating element
      // The function assumes that array elements are in range from
      // 1 to n-1.
      function findRepeatingElement(arr, low, high)
      {

        // low = 0 , high = n-1;
        if (low > high) return -1;

        var mid = parseInt((low + high) / 2);

        // Check if the mid element is the repeating one
        if (arr[mid] != mid + 1)
        {
          if (mid > 0 && arr[mid] == arr[mid - 1]) return mid;

          // If mid element is not at its position that means
          // the repeated element is in left
          return findRepeatingElement(arr, low, mid - 1);
        }

        // If mid is at proper position then repeated one is in
        // right.
        return findRepeatingElement(arr, mid + 1, high);
      }

      // Driver code
      var arr = [1, 2, 3, 3, 4, 5];
      var n = arr.length;
      var index = findRepeatingElement(arr, 0, n - 1);
      if (index != -1) document.write(arr[index]);

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

C#

// C# program to find the only repeating
// element in an array of size n and
// elements from range 1 to n-1.
using System;

class Test
{
    // Returns index of second appearance of a
    // repeating element. The function assumes that
    // array elements are in range from 1 to n-1.
    static int findRepeatingElement(int []arr, int low,
                                              int high)
    {
        // low = 0 , high = n-1;
        if (low > high)
            return -1;

        int mid = (low + high) / 2;

        // Check if the mid element
        // is the repeating one
        if (arr[mid] != mid + 1)
        {
            if (mid > 0 && arr[mid]==arr[mid-1])
                return mid;

            // If mid element is not at its position
            // that means the repeated element is in left
            return findRepeatingElement(arr, low, mid-1);
        }

        // If mid is at proper position
        // then repeated one is in right.
        return findRepeatingElement(arr, mid+1, high);
    }

    // Driver method
    public static void Main()
    {
        int []arr = {1, 2, 3, 3, 4, 5};
        int index = findRepeatingElement(arr, 0, arr.Length-1);
        if (index != -1)
        Console.Write(arr[index]);
    }
}

// This code is contributed by Nitin Mittal.

PHP

<?php
// PHP program to find the only
// repeating element in an array
// of size n and elements from
// range 1 to n-1.

// Returns index of second
// appearance of a repeating
// element. The function assumes
// that array elements are in
// range from 1 to n-1.
function findRepeatingElement($arr,
                              $low,
                              $high)
{
    // low = 0 , high = n-1;
    if ($low > $high)
        return -1;

    $mid = floor(($low + $high) / 2);

    // Check if the mid element
    // is the repeating one
    if ($arr[$mid] != $mid + 1)
    {
        if ($mid > 0 && $arr[$mid] ==
                        $arr[$mid - 1])
            return $mid;

        // If mid element is not at
        // its position that means
        // the repeated element is in left
        return findRepeatingElement($arr, $low,
                                    $mid - 1);
    }

    // If mid is at proper position
    // then repeated one is in right.
    return findRepeatingElement($arr, $mid + 1,
                                        $high);
}

// Driver code
$arr = array(1, 2, 3, 3, 4, 5);
$n = sizeof($arr);
$index = findRepeatingElement($arr, 0,
                              $n - 1);
if ($index != -1)
echo $arr[$index];

// This code is contributed
// by nitin mittal.
?>

Output

3

Time Complexity : O(log n)

Space Complexity: O(1)

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