# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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).

 // CPP program to find the only element that  // appears b times #include using namespace std;    int appearsbTimes(int arr[], int n, int a, int b) {     unordered_set 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; }

 // 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 s = new HashSet();        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

 # 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

 // 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 s = new HashSet();        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

Output:
1

Please refer below article for more approaches.

Find the only element that appears k times.

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