Skip to content
Related Articles

Related Articles

Improve Article

Find the repeating element in an Array of size N consisting of first M natural numbers

  • Last Updated : 23 Jul, 2021

Given an array arr[] of size N, which contains a permutation of numbers from 1 to M, as well as an element that is repeated(one or more times), the task is to find the repeating element.

Examples:

Input: arr[]={2, 6, 4, 3, 1, 5, 2}, N=7
Output:
2
Explanation: In arr[], all elements from 0 to 6 occurs once, except 2 which is repeated once.

Input: arr[]={2, 1, 3, 1, 1, 1}, N=6
Output:
1

Naive Approach: The naive approach would be to sort the array and check for adjacent elements that are equal.



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

Approach: Follow the steps below to solve the problem:

  1. Initialize two variables M and sum to store the maximum element and the sum of the array respectively.
  2. Traverse array arr and do the following:
    1. Add the current element to sum
    2. Compare the current element to M to calculate the maximum element.
  3. Store the sum of permutation from 1 to M in a variable say, sum1, using the formula:
Sum of elements from 1 to X= X*(X+1)/2
  1. Calculate the answer as the difference between sum and sum1 divided by the number of extra characters i.e. (sum-sum1)/(N-M).

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate the repeating character in a given
// permutation
int repeatingElement(int arr[], int N)
{
    // variables to store maximum element and sum of the
    // array respectively.
    int M = 0, sum = 0;
    for (int i = 0; i < N; i++) {
 
        // calculate sum of array
        sum += arr[i];
 
        // calculate maximum element in the array
        M = max(M, arr[i]);
    }
 
    // calculating sum of permutation
    int sum1 = M * (M + 1) / 2;
 
    // calculate required answer
    int ans = (sum - sum1) / (N - M);
    return ans;
}
// Driver code
int main()
{
    // Input
    int arr[] = { 2, 6, 4, 3, 1, 5, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    cout << repeatingElement(arr, N) << endl;
    return 0;
}

Java




// Java Program for the above approach
import java.io.*;
 
class GFG
{
   
  // Function to calculate the repeating character in a given
  // permutation
  public static int repeatingElement(int arr[], int N)
  {
     
    // variables to store maximum element and sum of the
    // array respectively.
    int M = 0, sum = 0;
    for (int i = 0; i < N; i++) {
 
      // calculate sum of array
      sum += arr[i];
 
      // calculate maximum element in the array
      M = Math.max(M, arr[i]);
    }
 
    // calculating sum of permutation
    int sum1 = M * (M + 1) / 2;
 
    // calculate required answer
    int ans = (sum - sum1) / (N - M);
    return ans;
  }
 
  // Driver code
  public static void main (String[] args)
  {
     
    // Input
    int arr[] = { 2, 6, 4, 3, 1, 5, 2 };
    int N = arr.length;
 
    // Function call
    System.out.println(repeatingElement(arr, N));
  }
}
 
// This code is contributed by lokeshpotta20

Python3




# Python 3 program for the above approach
 
# Function to calculate the repeating character in a given
# permutation
def repeatingElement(arr, N):
   
    # variables to store maximum element and sum of the
    # array respectively.
    M = 0
    sum = 0
    for i in range(N):
       
        # calculate sum of array
        sum += arr[i]
 
        # calculate maximum element in the array
        M = max(M, arr[i])
 
    # calculating sum of permutation
    sum1 = M * (M + 1) // 2
 
    # calculate required answer
    ans = (sum - sum1) // (N - M)
    return ans
 
# Driver code
if __name__ == '__main__':
   
    # Input
    arr = [2, 6, 4, 3, 1, 5, 2]
    N = len(arr)
 
    # Function call
    print(repeatingElement(arr, N))
     
    # This code is contributed by SURENDRA_GANGWAR.

C#




// C++ program for the above approach
using System;
 
// Function to calculate the repeating character in a given
// permutation
public class GFG
{
    public static int repeatingElement(int[] arr, int N)
    {
       
        // variables to store maximum element and sum of the
        // array respectively.
        int M = 0, sum = 0;
        for (int i = 0; i < N; i++) {
 
            // calculate sum of array
            sum += arr[i];
 
            // calculate maximum element in the array
            M = Math.Max(M, arr[i]);
        }
 
        // calculating sum of permutation
        int sum1 = M * (M + 1) / 2;
 
        // calculate required answer
        int ans = (sum - sum1) / (N - M);
        return ans;
    }
   
    // Driver code
    public static void Main()
    {
        // Input
        int[] arr = { 2, 6, 4, 3, 1, 5, 2 };
        int N = 7;
 
        // Function call
        Console.WriteLine(repeatingElement(arr, N));
    }
}
 
// This code is contributed by Sohom Das

Javascript




// JavaScript program for the above approach
 
        // Function to calculate the repeating character in a given
        // permutation
        function repeatingElement(arr, N)
        {
         
            // variables to store maximum element and sum of the
            // array respectively.
            let M = 0, sum = 0;
            for (let i = 0; i < N; i++) {
 
                // calculate sum of array
                sum += arr[i];
 
                // calculate maximum element in the array
                M = Math.max(M, arr[i]);
            }
 
            // calculating sum of permutation
            let sum1 = parseInt(M * (M + 1) / 2);
 
            // calculate required answer
            let ans = parseInt((sum - sum1) / (N - M));
            return ans;
        }
        // Driver code
 
        // Input
        let arr = [2, 6, 4, 3, 1, 5, 2];
        let N = arr.length;
 
        // Function call
        document.write(repeatingElement(arr, N));
 
  // This code is contributed by Potta Lokesh
    </script>

 
 

Output
2

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :