# 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++

`// 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;` `}` |

## Java

`// 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` |

## Python3

`# 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#

`// 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` |

## Javascript

`<script>` `// Javascript program to find the only element that` `// appears b times` `function` `appearsbTimes(arr, n, a, b)` `{` ` ` `var` `s = ` `new` `Set();` ` ` `var` `a_sum = 0, sum = 0;` ` ` `for` `(` `var` `i = 0; i < n; i++) {` ` ` `if` `(!s.has(arr[i])) {` ` ` `s.add(arr[i]);` ` ` `a_sum += arr[i];` ` ` `}` ` ` `sum += arr[i];` ` ` `}` ` ` `a_sum = a * a_sum;` ` ` `return` `((a_sum - sum) / (a - b));` `}` `var` `arr = [1, 1, 2, 2, 2, 3, 3, 3];` `var` `a = 3, b = 2;` `var` `n = arr.length;` `document.write( appearsbTimes(arr, n, a, b));` `</script>` |

**Output:**

1

Please refer to the article below for more approaches.

Find the only element that appears k times.

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