# Unique element in an array where all elements occur k times except one

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.
```

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

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.

1. Create an array count[] of size equal to number of bits in binary representations of numbers.
2. Fill count array such that count[i] stores count of array elements with i-th bit set.
3. 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 ` `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); ` `    ``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

 ` `

Output:

```5
```

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