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Find the only repetitive element between 1 to n-1

  • Difficulty Level : Easy
  • Last Updated : 17 Aug, 2021

We are given an array arr[] of size n. Numbers are from 1 to (n-1) in random order. The array has only one repetitive element. We need to find the repetitive element.

Examples:

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

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

Method 1 (Simple) We use two nested loops. The outer loop traverses through all elements and the inner loop check if the element picked by outer loop appears anywhere else.

Time Complexity: O(n*n)

Method 2 (Using Sum Formula): 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.



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 code
int main()
{  
    int arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
    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 code
    public static void main(String args[])
    {
        int[] arr = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
        int n = arr.length;
        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 Code
arr = [9, 8, 2, 6, 1, 8, 5, 3, 4, 7]
n = len(arr)
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 code
    public static void Main(String []args)
    {
        int[] arr = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
        int n = arr.Length;
        Console.WriteLine(findRepeating(arr, n));
    }
}
 
/* This code contributed by PrinciRaj1992 */
Output :
8
 
Time Complexity: O(n)
Auxiliary Space: O(1)
Causes overflow for large arrays.
Method 3 (Use Hashing): Use a hash table to store elements visited. If a seen element appears again, we return it.
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)
{
   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 code
int main()
{  
    int arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
    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 code
public static void main(String[] args)
{
    int arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
    int n = arr.length;
    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
    rteurn -1
 
# Driver code
arr = [9, 8, 2, 6, 1, 8, 5, 3]
n = len(arr)
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;
    Console.WriteLine(findRepeating(arr, n));;
}
}
 
// This code has been contributed by 29AjayKumar
Output :
8
 
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 4(Use XOR): The idea is based on the fact that x ^ x = 0 and x ^ y = y ^ x.
1) Compute XOR of elements from 1 to n-1.
2) Compute XOR of array elements.
3) XOR of above two would be our result.
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]);
  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.
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;
        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
arr = [9, 8, 2, 6, 1, 8, 5, 3, 4, 7]
n = len(arr)
print(findRepeating(arr, n))
 
# This code is contributed by Smitha Dinesh Semwal.
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) ;
    echo findRepeating($arr, $n);
     
// This code is contributed by ajit
?>
Output:
8
 
Time Complexity: O(n)
Auxiliary Space: O(1)
Method 5: Using indexing.
1. Iterate through the array.
2. For every index visit a[index], if it is positive change the sign of element at a[index] index, else print the element.
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[] = { 5, 4, 3, 9, 8,
                  9, 1, 6, 2, 5};
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    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 element = arr[Math.abs(arr[i])];
     
            if(element < 0){
                missingElement = arr[i];
                break;
            }
         
        arr[Math.abs(arr[i])] = -arr[Math.abs(arr[i])];
    }
     
    return Math.abs(missingElement);
     
    }
     
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 5, 4, 3, 9, 8,
                    9, 1, 6, 2, 5};
     
        int n = arr.length;
     
        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
arr = [5, 4, 3, 9, 8, 9, 1, 6, 2, 5]
n = len(arr)
print(findRepeating(arr, n))
 
# This code is contributed by Smitha Dinesh Semwal.
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);
 
echo findRepeating($arr, $n);
 
// This code is contributed by ajit
?>
Javascript



<script>
 
       // 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 element = arr[Math.abs(arr[i])];
 
               if (element < 0) {
                   missingElement = arr[i];
                   break;
               }
 
               arr[Math.abs(arr[i])] = -arr[Math.abs(arr[i])];
           }
 
           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
   </script>
Output :
9
Time Complexity: O(n)
Auxiliary Space: O(1)
 
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Article Contributed By :
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Shashank_Pathak
@Shashank_Pathak
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