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.
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:
Below are the steps:
- Store all the elements of the given array in the set to get the unique elements.
- 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).
- 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 Codeint 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 approachimport 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 codepublic 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 Codeif __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 approachusing System;using System.Collections.Generic;class GFG{ // Function that find the unique element// in the array []arrstatic 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 codepublic 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 formuladocument.write(Math.floor((K * set_sum - arr_sum) / (K - 1)));}// Driver Codevar 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> |
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
Index Elements 2nd bit 1st bit 0th bit 0 2= 010 0 1 0 1 2= 010 0 1 0 2 4 =100 1 0 0 3 2= 010 0 1 0 4 2= 010 0 1 0 5 2= 010 0 1 0 6 1 = 001 0 0 1 7 1 = 001 0 0 1 8 1 = 001 0 0 1 9 1 = 001 0 0 1 10 1 = 001 0 0 1 Sum%k 1%5=1 5%5=0 5%5=0 Result 1 0 0 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 elementint 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 codeint 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 approachimport java.util.Arrays;class Main{ // Function to find single // occurrence elementpublic 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 codepublic 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 elementdef 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 codea = [ 12, 12, 2, 2, 3 ]k = 2# Function callprint(findunique(a, k))# This code is contributed by divyesh072019 |
C#
// C# program for the// above approachusing System;class GFG{ // Function to find single // occurrence elementpublic 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 elementfunction 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 codevar a = [ 12, 12, 2, 2, 3 ];var k = 2;// Function calldocument.write( findunique(a, k) + "<br>");</script> |
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 Counterdef 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_elementarr1 = [12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7]k1 = 3print(find_unique_element(arr1, k1)) # Output: 7arr2 = [12, 12, 2, 2, 3]k2 = 2print(find_unique_element(arr2, k2)) # Output: 3 |
7 3
Time Complexity: O(n)
Space Complexity: O(n)
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