Find the only element that appears b times

Given an array where every element occurs a times, except one element which occurs b (a>b) times. Find the element that occurs b times.

Examples:

Input : arr[]  = [1, 1, 2, 2, 2, 3, 3, 3]
         a = 3, b = 2
Output : 1

Add each number once and multiply the sum by a, we will get a times the sum of each element of the array. Store it as a_sum. Subtract the sum of the whole array from the a_sum and divide the result by (a-b). The number we get is the required number (which appears b times in the array).



C++

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// CPP program to find the only element that 
// appears b times
#include <bits/stdc++.h>
using namespace std;
  
int appearsbTimes(int arr[], int n, int a, int b)
{
    unordered_set<int> s;
  
    int a_sum = 0, sum = 0;
  
    for (int i = 0; i < n; i++) {
        if (s.find(arr[i]) == s.end()) {
            s.insert(arr[i]);
            a_sum += arr[i];
        }
  
        sum += arr[i];
    }
  
    a_sum = a * a_sum;
  
    return ((a_sum - sum) / (a - b));
}
  
int main()
{
    int arr[] = { 1, 1, 2, 2, 2, 3, 3, 3 };
    int a = 3, b = 2;
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << appearsbTimes(arr, n, a, b);
    return 0;
}

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Java














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// Java program to find the only element that  


// appears b times 


import java.util.*; 


  


class GFG  


{ 


static int appearsbTimes(int arr[], int n,  


                         int a, int b) 


{ 


    HashSet<Integer> s = new HashSet<Integer>(); 


  


    int a_sum = 0, sum = 0; 


  


    for (int i = 0; i < n; i++) 


    { 


        if (!s.contains(arr[i])) 


        { 


            s.add(arr[i]); 


            a_sum += arr[i]; 


        } 


  


        sum += arr[i]; 


    } 


    a_sum = a * a_sum; 


  


    return ((a_sum - sum) / (a - b)); 


} 


  


// Driver Code 


public static void main(String[] args)  


{ 


    int arr[] = { 1, 1, 2, 2, 2, 3, 3, 3 }; 


    int a = 3, b = 2; 


    int n = arr.length; 


    System.out.println(appearsbTimes(arr, n, a, b)); 


} 


} 


  


// This code is contributed by 29AjayKumar 



















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Python3














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# Python3 program to find the only  


# element that appears b times 


def appearsbTimes(arr, n, a, b): 


  


    s = dict() 


  


    a_Sum = 0


    Sum = 0


  


    for i in range(n): 


        if (arr[i] not in s.keys()): 


            s[arr[i]] = 1


            a_Sum += arr[i] 


  


        Sum += arr[i] 


          


    a_Sum = a * a_Sum 


  


    return ((a_Sum - Sum) // (a - b)) 


  


# Driver code 


arr = [1, 1, 2, 2, 2, 3, 3, 3] 


a, b = 3, 2


n = len(arr) 


print(appearsbTimes(arr, n, a, b)) 


  


# This code is contributed by mohit kumar 



















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C#














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// C# program to find the only element that  


// appears b times 


using System; 


using System.Collections.Generic; 


  


class GFG  


{ 


      


static int appearsbTimes(int []arr, int n,  


                        int a, int b) 


{ 


    HashSet<int> s = new HashSet<int>(); 


  


    int a_sum = 0, sum = 0; 


  


    for (int i = 0; i < n; i++) 


    { 


        if (!s.Contains(arr[i])) 


        { 


            s.Add(arr[i]); 


            a_sum += arr[i]; 


        } 


  


        sum += arr[i]; 


    } 


    a_sum = a * a_sum; 


  


    return ((a_sum - sum) / (a - b)); 


} 


  


// Driver Code 


public static void Main(String[] args)  


{ 


    int []arr = { 1, 1, 2, 2, 2, 3, 3, 3 }; 


    int a = 3, b = 2; 


    int n = arr.Length; 


    Console.WriteLine(appearsbTimes(arr, n, a, b)); 


} 


} 


  


// This code is contributed by Princi Singh 



















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Output:


1





Please refer below article for more approaches.


Find the only element that appears k times.


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