# 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; ` `} ` |

*chevron_right*

*filter_none*

## 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

**Output:**

1

Please refer below article for more approaches.

Find the only element that appears k times.

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