# 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<``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); ` `    ``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<``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 `

Output:
```1
```

Please refer below article for more approaches.

Find the only element that appears k times.

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