# Unique element in an array where all elements occur K times except one | Set 2

Given an array arr[] where every element occurs K times except one element which occurs only once, the task is to find that unique element.

Examples:

Input: arr[] = {12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7}, k = 3
Output: 7
Explanation:
7 is the only element which occurs once while others occurs k times.

Input: arr[] = {12, 12, 2, 2, 3}, k = 2
Output: 3
Explanation:
3 is the only element which occurs once while others occurs k times.

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

Approach: Suppose we have every element K times then the difference between the sum of all elements in the given array and the K*sum of all unique elements is (K-1) times the unique element.
For Example:

arr[] = {a, a, a, b, b, b, c, c, c, d}, k = 3
unique elements = {a, b, c, d}
Difference = 3*(a + b + c + d) – (a + a + a + b + b + b + c + c + c + d) = 2*d

Therefore, Generalizing the equation:
The unique element can be given by: Below are the steps:

1. Store all the element of the given array in the set to get the unique elements.
2. Find the sum of all element in the array (say sum_array) and sum of all the element in the set(say sum_set).
3. The unique element is given by (K*sum_set – sum_array)/(K – 1).

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach     #include  using namespace std;     // Function that find the unique element  // in the array arr[]  int findUniqueElements(int arr[], int N,                         int K)  {      // Store all unique element in set      unordered_set<int> s(arr, arr + N);         // Sum of all element of the array      int arr_sum = accumulate(arr, arr + N, 0);         // Sum of element in the set      int set_sum = accumulate(s.begin(),                               s.end(),                               0);         // Print the unique element using formula      cout << (K * set_sum - arr_sum) / (K - 1);  }     // Driver Code  int main()  {         int arr[] = { 12, 1, 12, 3, 12, 1,                    1, 2, 3, 2, 2, 3, 7 };      int N = sizeof(arr) / sizeof(arr);      int K = 3;         // Function call      findUniqueElements(arr, N, K);         return 0;  }

## Python 3

 # Python 3 program for the above approach     # Function that find the unique element  # in the array arr[]  def findUniqueElements(arr, N, K):             # Store all unique element in set      s = set()      for x in arr:          s.add(x)         # Sum of all element of the array      arr_sum = sum(arr)         # Sum of element in the set      set_sum = 0     for x in s:          set_sum += x         # Print the unique element using formula      print((K * set_sum - arr_sum) // (K - 1))     # Driver Code  if __name__ == '__main__':             arr = [ 12, 1, 12, 3, 12, 1,              1, 2, 3, 2, 2, 3, 7 ]      N = len(arr)      K = 3        # Function call      findUniqueElements(arr, N, K)     # This code is contributed by Samarth

Output:

7


Time Complexity: O(N), where N is the number of elements in the array
Auxiliary Space Complexity: O(N/K), where K is the frequency.

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