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

Last Updated : 07 Nov, 2023

Given an array that 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
Explanation: Every element appears 3 times accept 5.

Input  : arr[] = {2, 2, 2, 10, 2}
k = 4
Output: 10
Explanation: Every element appears 4 times accept 10.

Recommended Practice

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.

The Time Complexity of the above solution is O(n2). We can Use Sorting to solve the problem in O(nLogn) time. The idea is simple, the 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.
1. 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.

Below is the implementation of the above approach:

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);`   `    ``// Before returning the res we need` `    ``// to check the occurrence  of that` `    ``// unique element and divide it` `    ``res = res / (n % k);` `    ``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);`   `        ``// Before returning the res we need` `        ``// to check the occurrence  of that` `        ``// unique element and divide it` `        ``res = res / (n % k);`   `        ``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)`   `    ``# Before returning the res we need` `    ``# to check the occurrence  of that` `    ``# unique element and divide it` `    ``res ``=` `res ``/` `(n ``%` `k)` `    ``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);`   `        ``// Before returning the res we need` `        ``// to check the occurrence  of that` `        ``// unique element and divide it` `        ``res = res / (n % k);` `        ``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`

Javascript

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Output

```5

```

Time Complexity: O(n)
Auxiliary Space: O(1)

Using Set:

Approach:

In this approach, we create two sets: one to store the unique elements in the array, and the other to store the elements that appear k times. Then we return the difference between these two sets.

• Create two empty sets: unique_set and k_set.
• Iterate through each element in the array arr.
• If the element is already in the unique_set, it means it has appeared before. Add it to k_set.
• If the element is not in the unique_set, add it to unique_set.
• Take the set difference between unique_set and k_set using the – operator. This gives us a set with only the unique element in it.
• Use the pop() method to get the element from the set (since there is only one element in it).

C++

 `#include ` `#include ` `#include `   `using` `namespace` `std;`   `// Function to find a unique element with a specified frequency (k)` `int` `findUniqueElement(``int` `arr[], ``int` `n, ``int` `k) {` `    ``unordered_map<``int``, ``int``> freqMap; ``// Create an unordered_map to store frequency`   `    ``// Populate the frequency map` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``int` `num = arr[i];` `        ``freqMap[num]++; ``// Increment the frequency of the current element` `    ``}`   `    ``// Iterate through the frequency map` `    ``for` `(``const` `auto``& entry : freqMap) {` `        ``if` `(entry.second != k) { ``// If the frequency is not equal to k` `            ``return` `entry.first; ``// Return the unique element` `        ``}` `    ``}`   `    ``return` `-1; ``// Return -1 if no unique element is found` `}`   `int` `main() {` `    ``int` `arr[] = {6, 2, 5, 2, 2, 6, 6};` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``int` `k = 3;`   `    ``int` `uniqueElement = findUniqueElement(arr, n, k);`   `    ``if` `(uniqueElement != -1) {` `        ``cout << ``"Unique element: "` `<< uniqueElement << endl;` `    ``} ``else` `{` `        ``cout << ``"No unique element found."` `<< endl;` `    ``}`   `    ``return` `0;` `}`

Java

 `import` `java.util.HashMap;` `import` `java.util.Map;`   `public` `class` `Main {` `    ``public` `static` `int` `findUniqueElement(``int``[] arr, ``int` `k) {` `        ``Map freqMap = ``new` `HashMap<>();` `        ``for` `(``int` `i = ``0``; i < arr.length; i++) {` `            ``int` `num = arr[i];` `            ``freqMap.put(num, freqMap.getOrDefault(num, ``0``) + ``1``);` `        ``}` `        ``for` `(Map.Entry entry : freqMap.entrySet()) {` `            ``if` `(entry.getValue() != k) {` `                ``return` `entry.getKey();` `            ``}` `        ``}` `        ``return` `-``1``; ``// or any suitable default value to indicate no unique element found` `    ``}`   `    ``public` `static` `void` `main(String[] args) {` `        ``int``[] arr = {``6``, ``2``, ``5``, ``2``, ``2``, ``6``, ``6``};` `        ``int` `k = ``3``;` `        ``int` `uniqueElement = findUniqueElement(arr, k);` `        ``if` `(uniqueElement != -``1``) {` `            ``System.out.println(``"Unique element: "` `+ uniqueElement);` `        ``} ``else` `{` `            ``System.out.println(``"No unique element found."``);` `        ``}` `    ``}` `}`

Python3

 `from` `collections ``import` `Counter`   `def` `find_unique_element(arr, k):` `    ``freq_dict ``=` `Counter(arr)` `    ``for` `key, val ``in` `freq_dict.items():` `        ``if` `val !``=` `k:` `            ``return` `key`   `# example usage` `arr ``=` `[``6``, ``2``, ``5``, ``2``, ``2``, ``6``, ``6``]` `k ``=` `3` `print``(find_unique_element(arr, k))`

C#

 `using` `System;` `using` `System.Collections.Generic;`   `class` `Program` `{` `    ``// Function to find a unique element with a specified frequency (k)` `    ``static` `int` `FindUniqueElement(``int``[] arr, ``int` `k)` `    ``{` `        ``Dictionary<``int``, ``int``> freqMap = ``new` `Dictionary<``int``, ``int``>(); ``// Create a Dictionary to store frequency`   `        ``// Populate the frequency map` `        ``foreach` `(``int` `num ``in` `arr)` `        ``{` `            ``if` `(freqMap.ContainsKey(num))` `            ``{` `                ``freqMap[num]++; ``// Increment the frequency of the current element` `            ``}` `            ``else` `            ``{` `                ``freqMap[num] = 1; ``// Initialize the frequency of the current element` `            ``}` `        ``}`   `        ``// Iterate through the frequency map` `        ``foreach` `(``var` `entry ``in` `freqMap)` `        ``{` `            ``if` `(entry.Value != k) ``// If the frequency is not equal to k` `            ``{` `                ``return` `entry.Key; ``// Return the unique element` `            ``}` `        ``}`   `        ``return` `-1; ``// Return -1 if no unique element is found` `    ``}`   `    ``static` `void` `Main()` `    ``{` `        ``int``[] arr = { 6, 2, 5, 2, 2, 6, 6 };` `        ``int` `k = 3;`   `        ``int` `uniqueElement = FindUniqueElement(arr, k);`   `        ``if` `(uniqueElement != -1)` `        ``{` `            ``Console.WriteLine(``"Unique element: "` `+ uniqueElement);` `        ``}` `        ``else` `        ``{` `            ``Console.WriteLine(``"No unique element found."``);` `        ``}` `    ``}` `}`

Javascript

 `// Function to find a unique element with a specified frequency (k)` `function` `findUniqueElement(arr, k) {` `    ``let freqMap = ``new` `Map(); ``// Create a Map to store frequency`   `    ``// Populate the frequency map` `    ``for` `(let i = 0; i < arr.length; i++) {` `        ``let num = arr[i];` `        ``if` `(freqMap.has(num)) {` `            ``freqMap.set(num, freqMap.get(num) + 1); ``// Increment the frequency of the current element` `        ``} ``else` `{` `            ``freqMap.set(num, 1);` `        ``}` `    ``}`   `    ``// Iterate through the frequency map` `    ``for` `(const [key, value] of freqMap) {` `        ``if` `(value !== k) { ``// If the frequency is not equal to k` `            ``return` `key; ``// Return the unique element` `        ``}` `    ``}`   `    ``return` `-1; ``// Return -1 if no unique element is found` `}`   `const arr = [6, 2, 5, 2, 2, 6, 6];` `const k = 3;`   `const uniqueElement = findUniqueElement(arr, k);`   `if` `(uniqueElement !== -1) {` `    ``console.log(``"Unique element:"``, uniqueElement);` `} ``else` `{` `    ``console.log(``"No unique element found."``);` `}`

Output

```5

```

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