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

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

Naive 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: 

\frac{(K*(\text{sum of unique element}) - (\text{sum of all element}))}{(K - 1)}

Below are the steps:  

  1. Store all the elements of the given array in the set to get the unique elements.
  2. Find the sum of all elements in the array (say sum_array) and the sum of all the elements 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 <bits/stdc++.h>
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[0]);
    int K = 3;
 
    // Function call
    findUniqueElements(arr, N, K);
 
    return 0;
}

                    

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function that find the unique element
// in the array arr[]
static void findUniqueElements(int arr[],
                               int N, int K)
{
     
    // Store all unique element in set
    Set<Integer> s = new HashSet<>();
    for(int i = 0; i < N; i++)
        s.add(arr[i]);
 
    // Sum of all element of the array
    int arr_sum = 0;
    for(int i = 0; i < N; i++)
        arr_sum += arr[i];
 
    // Sum of element in the set
    int set_sum = 0;
    Iterator it = s.iterator();
 
    while (it.hasNext())
    {
        set_sum += (int)it.next();
    }
 
    // Print the unique element using formula
    System.out.println((K * set_sum - arr_sum) /
                       (K - 1));
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 12, 1, 12, 3, 12, 1,
                  1, 2, 3, 2, 2, 3, 7 };
    int N = arr.length;
    int K = 3;
 
    // Function call
    findUniqueElements(arr, N, K);
}
}
 
// This code is contributed by offbeat

                    

Python3

# Python3 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

                    

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
     
// Function that find the unique element
// in the array []arr
static void findUniqueElements(int []arr,
                               int N, int K)
{
     
    // Store all unique element in set
    HashSet<int> s = new HashSet<int>();
    for(int i = 0; i < N; i++)
        s.Add(arr[i]);
 
    // Sum of all element of the array
    int arr_sum = 0;
    for(int i = 0; i < N; i++)
        arr_sum += arr[i];
 
    // Sum of element in the set
    int set_sum = 0;
    foreach(int i in s)
        set_sum += i;
 
 
    // Print the unique element using formula
    Console.WriteLine((K * set_sum - arr_sum) /
                      (K - 1));
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 12, 1, 12, 3, 12, 1,
                  1, 2, 3, 2, 2, 3, 7 };
    int N = arr.Length;
    int K = 3;
 
    // Function call
    findUniqueElements(arr, N, K);
}
}
 
// This code is contributed by gauravrajput1

                    

Javascript

<script>
 
// JavaScript implementation of the above approach
 
// Function that find the unique element
// in the array arr[]
function findUniqueElements(arr, N, K)
{
    // Store all unique element in set
    var s = new Set(arr);
 
    // Sum of all element of the array
    var arr_sum = arr.reduce((a, b) => a + b, 0);
 
    // Sum of element in the set
    var set_sum = 0;
    s.forEach(function(value) {
        set_sum += value;
    })
 
    // Print the unique element using formula
document.write(Math.floor((K * set_sum - arr_sum) / (K - 1)));
}
 
// Driver Code
var arr = [ 12, 1, 12, 3, 12, 1,
                  1, 2, 3, 2, 2, 3, 7 ];
var N = arr.length;
var K = 3;
 
findUniqueElements(arr, N,K);
 
</script>

                    

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.

Another Approach:  The idea is to use hashing but it will take O(n) time and requires extra space. We can also do it by sorting but it will take O(N log N) time.

Efficient Approach: The above problem can be optimized in terms of constant space complexity. Use bit manipulation approach for this problem.

  • Let’s assume we have a case where all the elements appear k times except 1 element.
  • So, count the bit of every element for 32 bits.
  • Count 0th bit of every element and take modulo by k will eliminate the bit of elements which present k times and we remain with the only bit of element which presents one time.
  • Apply this process for all the 32 bits and by taking modulo by k we will eliminate repeated elements bits.
  • At every step, generate the results from these remaining bits.
  • To handle any negative number, check if the result is present in the array or not if present then print it. Else print negative of the result.

For Example arr[] = {2, 2, 4, 2, 2, 2, 1, 1, 1, 1, 1} , k = 5

IndexElements2nd bit1st bit0th bit
02= 010010
12= 010010
24 =100100
32= 010010
42= 010010
52= 010010
61 = 001001
71 = 001001
81 = 001001
91 = 001001
101 = 001001
Sum%k 1%5=15%5=05%5=0
Result 100

Our result is (100) in binary = 4 in decimal. So the final answer will be 4, because it presents 1 time.

Below is the implementation of the above approach:

C++

// CPP program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find single occurrence element
int findunique(vector<int>& a, int k)
{
    int res = 0;
 
    for (int i = 0; i < 32; i++) {
        int p = 0;
 
        for (int j = 0; j < a.size(); j++) {
            // By shifting 1 to left ith
            // time and taking and with 1
            // will give us that ith
            // bit of a[j] is 1 or 0
            p += (abs(a[j]) & (
                  1 << i)) != 0 ? 1 : 0;
        }
 
        // Taking modulo of p with k
        p %= k;
 
        // Generate result
        res += pow(2, i) * p;
    }
 
    int c = 0;
 
    // Loop for negative numbers
    for (auto x : a)
 
        if (x == res) {
            c = 1;
            break;
        }
   
    // Check if the calculated value res
    // is present in array, then mark c=1
    // and if c = 1 return res
    // else res must be -ve
    return c == 1 ? res : -res;
}
 
// Driver code
int main()
{
 
    vector<int> a = { 12, 12, 2, 2, 3 };
    int k = 2;
 
    // Function call
    cout << findunique(a, k) << "\n";
}
// This article is contributed by ajaykr00kj

                    

Java

// Java program for the above approach
import java.util.Arrays;
class Main{
     
// Function to find single
// occurrence element
public static int findunique(int a[],
                             int k)
{
  int res = 0;
 
  for (int i = 0; i < 32; i++)
  {
    int p = 0;
 
    for (int j = 0; j < a.length; j++)
    {
      // By shifting 1 to left ith
      // time and taking and with 1
      // will give us that ith
      // bit of a[j] is 1 or 0
      p += (Math.abs(a[j]) &
           (1 << i)) != 0 ? 1 : 0;
    }
 
    // Taking modulo of p with k
    p %= k;
 
    // Generate result
    res += Math.pow(2, i) * p;
  }
 
  int c = 0;
 
  // Loop for negative numbers
  for (int x = 0; x < a.length; x++)
 
    if (a[x] == res)
    {
      c = 1;
      break;
    }
 
  // Check if the calculated value res
  // is present in array, then mark c=1
  // and if c = 1 return res
  // else res must be -ve
  return c == 1 ? res : -res;
}
 
// Driver code
public static void main(String[] args)
{
  int a[] = {12, 12, 2, 2, 3};
  int k = 2;
 
  // Function call
  System.out.println(findunique(a, k));
}
}
 
// This code is contributed by divyeshrabadiya07

                    

Python3

# Python3 program for the above approach
 
# Function to find single occurrence element
def findunique(a, k):
 
    res = 0
 
    for i in range(32):
        p = 0
 
        for j in range(len(a)):
           
            # By shifting 1 to left ith
            # time and taking and with 1
            # will give us that ith
            # bit of a[j] is 1 or 0
            if (abs(a[j]) & (1 << i)) != 0 :
                p += 1
 
        # Taking modulo of p with k
        p %= k
 
        # Generate result
        res += pow(2, i) * p
 
    c = 0
 
    # Loop for negative numbers
    for x in a:
 
        if (x == res) :
            c = 1
            break
 
    # Check if the calculated value res
    # is present in array, then mark c=1
    # and if c = 1 return res
    # else res must be -ve
    if c == 1 :
        return res
    else :
        return -res
 
# Driver code
a = [ 12, 12, 2, 2, 3 ]
k = 2
 
# Function call
print(findunique(a, k))
 
# This code is contributed by divyesh072019

                    

C#

// C# program for the
// above approach
using System;
class GFG{
      
// Function to find single
// occurrence element
public static int findunique(int []a,
                             int k)
{
  int res = 0;
  
  for (int i = 0; i < 32; i++)
  {
    int p = 0;
  
    for (int j = 0; j < a.Length; j++)
    {
      // By shifting 1 to left ith
      // time and taking and with 1
      // will give us that ith
      // bit of a[j] is 1 or 0
      p += (Math.Abs(a[j]) &
           (1 << i)) != 0 ? 1 : 0;
    }
  
    // Taking modulo of p with k
    p %= k;
  
    // Generate result
    res += (int)Math.Pow(2, i) * p;
  }
  
  int c = 0;
  
  // Loop for negative numbers
  for (int x = 0; x < a.Length; x++)
  
    if (a[x] == res)
    {
      c = 1;
      break;
    }
  
  // Check if the calculated value res
  // is present in array, then mark c=1
  // and if c = 1 return res
  // else res must be -ve
  return c == 1 ? res : -res;
}
  
// Driver code
public static void Main(string []args)
{
  int []a = {12, 12, 2, 2, 3};
  int k = 2;
  
  // Function call
  Console.Write(findunique(a, k));
}
}
 
// This code is contributed by Rutvik_56

                    

Javascript

<script>
 
// Javascript program for the above approach
 
// Function to find single occurrence element
function findunique(a, k)
{
    var res = 0;
 
    for (var i = 0; i < 32; i++) {
        var p = 0;
 
        for (var j = 0; j < a.length; j++) {
            // By shifting 1 to left ith
            // time and taking and with 1
            // will give us that ith
            // bit of a[j] is 1 or 0
            p += (Math.abs(a[j]) & (
                  1 << i)) != 0 ? 1 : 0;
        }
 
        // Taking modulo of p with k
        p %= k;
 
        // Generate result
        res += Math.pow(2, i) * p;
    }
 
    var c = 0;
 
    // Loop for negative numbers
    for (var i =0;i<a.length;i++)
 
        if (a[i] == res) {
            c = 1;
            break;
        }
   
    // Check if the calculated value res
    // is present in array, then mark c=1
    // and if c = 1 return res
    // else res must be -ve
    return c == 1 ? res : -res;
}
 
// Driver code
 
var a = [ 12, 12, 2, 2, 3 ];
var k = 2;
 
// Function call
document.write( findunique(a, k) + "<br>");
 
</script>

                    

 
 


Output
3


 

Time Complexity: O(32 * n) = O(n)
Auxiliary Space: O(1)
 


 Using Counter and List Comprehension in python:

Approach:

We can use Python’s Counter module to count the occurrences of each element in the given array. Then, we can filter out the elements that occur k times using list comprehension and find the unique element using set difference operation.

Import the Counter class from the collections module. This will allow us to count the occurrences of each element in the input array.
Define the find_unique_element function that takes an input array arr and an integer k.
Use the Counter class to count the number of occurrences of each element in the input array.
Create a set k_elements that contains all elements in the input array that occur exactly k times.
Create a set unique_elements that contains all elements in the input array that occur less than k times.
Compute the set difference between unique_elements and k_elements to obtain a set containing only the unique element in the input array that occurs less than k times.
Convert the resulting set to a list and return the first element (which is the only element in the set, since it is unique).

Python3

from collections import Counter
 
def find_unique_element(arr, k):
    count = Counter(arr)
    k_elements = set([key for key in count if count[key] == k])
    unique_element = list(set(arr) - k_elements)[0]
    return unique_element
 
arr1 = [12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7]
k1 = 3
print(find_unique_element(arr1, k1))  # Output: 7
 
arr2 = [12, 12, 2, 2, 3]
k2 = 2
print(find_unique_element(arr2, k2))  # Output: 3

                    

Output
7
3

Time Complexity: O(n)
Space Complexity: O(n)



Last Updated : 19 Apr, 2023
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