Given an array which contains all elements occurring k times, but one occurs only once. Find that unique element.
Examples:
Input : arr[] = {6, 2, 5, 2, 2, 6, 6}
k = 3
Output : 5
Every element appears 3 times accept 5.
Input : arr[] = {2, 2, 2, 10, 2}
k = 4
Output : 10
Every element appears 4 times accept 10.
A Simple Solution is to use two nested loops. The outer loop picks an element one by one starting from the leftmost element. The inner loop checks if the element is present k times or not. If present, then ignores the element, else prints the element.
Time Complexity of above solution is O(n2). We can Use Sorting to solve the problem in O(nLogn) time. The idea is simple, first sort the array so that all occurrences of every element become consecutive. Once the occurrences become consecutive, we can traverse the sorted array and print the unique element in O(n) time.
We can Use Hashing to solve this in O(n) time on average. The idea is to traverse the given array from left to right and keep track of visited elements in a hash table. Finally print the element with count 1.
The hashing based solution requires O(n) extra space. We can use bitwise AND to find the unique element in O(n) time and constant extra space.
- Create an array count[] of size equal to number of bits in binary representations of numbers.
- Fill count array such that count[i] stores count of array elements with i-th bit set.
- Form result using count array. We put 1 at a position i in result if count[i] is not multiple of k. Else we put 0.
C++
// CPP program to find unique element where // every element appears k times except one #include <bits/stdc++.h> using namespace std; int findUnique(unsigned int a[], int n, int k) { // Create a count array to store count of // numbers that have a particular bit set. // count[i] stores count of array elements // with i-th bit set. int INT_SIZE = 8 * sizeof(unsigned int); int count[INT_SIZE]; memset(count, 0, sizeof(count)); // AND(bitwise) each element of the array // with each set digit (one at a time) // to get the count of set bits at each // position for (int i = 0; i < INT_SIZE; i++) for (int j = 0; j < n; j++) if ((a[j] & (1 << i)) != 0) count[i] += 1; // Now consider all bits whose count is // not multiple of k to form the required // number. unsigned res = 0; for (int i = 0; i < INT_SIZE; i++) res += (count[i] % k) * (1 << i); return res; } // Driver Code int main() { unsigned int a[] = { 6, 2, 5, 2, 2, 6, 6 }; int n = sizeof(a) / sizeof(a[0]); int k = 3; cout << findUnique(a, n, k); return 0; } |
Java
// Java program to find unique element where // every element appears k times except one class GFG { static int findUnique(int a[], int n, int k) { // Create a count array to store count of // numbers that have a particular bit set. // count[i] stores count of array elements // with i-th bit set. byte sizeof_int = 4; int INT_SIZE = 8 * sizeof_int; int count[] = new int[INT_SIZE]; // AND(bitwise) each element of the array // with each set digit (one at a time) // to get the count of set bits at each // position for (int i = 0; i < INT_SIZE; i++) for (int j = 0; j < n; j++) if ((a[j] & (1 << i)) != 0) count[i] += 1; // Now consider all bits whose count is // not multiple of k to form the required // number. int res = 0; for (int i = 0; i < INT_SIZE; i++) res += (count[i] % k) * (1 << i); return res; } // Driver Code public static void main(String[] args) { int a[] = { 6, 2, 5, 2, 2, 6, 6 }; int n = a.length; int k = 3; System.out.println(findUnique(a, n, k)); } } // This code is contributed by 29AjayKumar |
Python3
# Python 3 program to find unique element where # every element appears k times except one import sys def findUnique(a, n, k): # Create a count array to store count of # numbers that have a particular bit set. # count[i] stores count of array elements # with i-th bit set. INT_SIZE = 8 * sys.getsizeof(int) count = [0] * INT_SIZE # AND(bitwise) each element of the array # with each set digit (one at a time) # to get the count of set bits at each # position for i in range(INT_SIZE): for j in range(n): if ((a[j] & (1 << i)) != 0): count[i] += 1 # Now consider all bits whose count is # not multiple of k to form the required # number. res = 0 for i in range(INT_SIZE): res += (count[i] % k) * (1 << i) return res # Driver Code if __name__ == '__main__': a = [6, 2, 5, 2, 2, 6, 6] n = len(a) k = 3 print(findUnique(a, n, k)); # This code is contributed by # Surendra_Gangwar |
C#
// C# program to find unique element where // every element appears k times except one using System; class GFG { static int findUnique(int []a, int n, int k) { // Create a count array to store count of // numbers that have a particular bit set. // count[i] stores count of array elements // with i-th bit set. byte sizeof_int = 4; int INT_SIZE = 8 * sizeof_int; int []count = new int[INT_SIZE]; // AND(bitwise) each element of the array // with each set digit (one at a time) // to get the count of set bits at each // position for (int i = 0; i < INT_SIZE; i++) for (int j = 0; j < n; j++) if ((a[j] & (1 << i)) != 0) count[i] += 1; // Now consider all bits whose count is // not multiple of k to form the required // number. int res = 0; for (int i = 0; i < INT_SIZE; i++) res += (count[i] % k) * (1 << i); return res; } // Driver Code public static void Main(String[] args) { int []a = { 6, 2, 5, 2, 2, 6, 6 }; int n = a.Length; int k = 3; Console.WriteLine(findUnique(a, n, k)); } } // This code is contributed by PrinciRaj1992 |
PHP
<?php //PHP program to find unique element where // every element appears k times except one function findUnique($a, $n, $k) { // Create a count array to store count of // numbers that have a particular bit set. // count[i] stores count of array elements // with i-th bit set. $INT_SIZE = 8 * PHP_INT_SIZE; $count = array(); for($i=0; $i< $INT_SIZE; $i++) $count[$i] = 0; // AND(bitwise) each element of the array // with each set digit (one at a time) // to get the count of set bits at each // position for ( $i = 0; $i < $INT_SIZE; $i++) for ( $j = 0; $j < $n; $j++) if (($a[$j] & (1 << $i)) != 0) $count[$i] += 1; // Now consider all bits whose count is // not multiple of k to form the required // number. $res = 0; for ($i = 0; $i < $INT_SIZE; $i++) $res += ($count[$i] % $k) * (1 << $i); return $res; } // Driver Code $a = array( 6, 2, 5, 2, 2, 6, 6 ); $n = count($a); $k = 3; echo findUnique($a, $n, $k); // This code is contributed by Rajput-Ji ?> |
Output:
5
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