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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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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Improved By : mohit kumar 29