Elements that occurred only once in the array

Given an array arr that has numbers appearing twice or once. The task is to identify numbers that occur only once in the array.

Note: Duplicates appear side by side every time. There might be a few numbers that can occur at one time and just assume this is a right rotating array (just say an array can rotate k times towards right). The order of the elements in the output doesn’t matter.

Examples:

```Input: arr[] = { 7, 7, 8, 8, 9, 1, 1, 4, 2, 2 }
Output: 9 4

Input: arr[] = {-9, -8, 4, 4, 5, 5, -1}
Output: -9 -8 -1```

Method-1: Using Sorting.

• Sort the array.
• Check for each element at index i (except the first and last element), if
`arr[i] != arr[i-1] && arr [i] != arr[i+1]`
• For the first element, check if arr[0] != arr[1].
• For the last element, check if arr[n-1] != arr[n-2].

Algorithm:

1. Sort the given array in non-decreasing order using any sorting algorithm.
2. Traverse the sorted array and compare each element with its adjacent element.
3. If an element is not equal to its adjacent elements, then print it.
4. For the first element, check if it is different from the second element. If yes, print it.
5. For the last element, check if it is different from the second last element. If yes, print it.

Pseudocode:

```occurredOnce(arr[], n):
sort(arr, arr + n)
for i = 0 to n-1 do
if i == 0 and arr[i] != arr[i+1] then
print arr[i]
else if i == n-1 and arr[i] != arr[i-1] then
print arr[i]
else if arr[i] != arr[i-1] and arr[i] != arr[i+1] then
print arr[i]```

Below is the implementation of the above approach:

C++

 `// C++ implementation of above approach` `#include ` `using` `namespace` `std;`   `// Function to find the elements that` `// appeared only once in the array` `void` `occurredOnce(``int` `arr[], ``int` `n)` `{` `    ``// Sort the array` `    ``sort(arr, arr + n);`   `    ``// Check for first element` `    ``if` `(arr[0] != arr[1])` `        ``cout << arr[0] << ``" "``;`   `    ``// Check for all the elements if it is different` `    ``// its adjacent elements` `    ``for` `(``int` `i = 1; i < n - 1; i++)` `        ``if` `(arr[i] != arr[i + 1] && arr[i] != arr[i - 1])` `            ``cout << arr[i] << ``" "``;`   `    ``// Check for the last element` `    ``if` `(arr[n - 2] != arr[n - 1])` `        ``cout << arr[n - 1] << ``" "``;` `}`   `// Driver code` `int` `main()` `{`   `    ``int` `arr[] = { 7, 7, 8, 8, 9, 1, 1, 4, 2, 2 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``occurredOnce(arr, n);`   `    ``return` `0;` `}`

Java

 `// Java implementation` `// of above approach` `import` `java.util.*;`   `class` `GFG` `{`   `// Function to find the elements that ` `// appeared only once in the array` `static` `void` `occurredOnce(``int` `arr[], ``int` `n)` `{` `    ``// Sort the array` `    ``Arrays.sort(arr);`   `    ``// Check for first element` `    ``if` `(arr[``0``] != arr[``1``])` `        ``System.out.println(arr[``0``] + ``" "``);`   `    ``// Check for all the elements ` `    ``// if it is different` `    ``// its adjacent elements` `    ``for` `(``int` `i = ``1``; i < n - ``1``; i++)` `        ``if` `(arr[i] != arr[i + ``1``] && ` `            ``arr[i] != arr[i - ``1``])` `            ``System.out.print(arr[i] + ``" "``);`   `    ``// Check for the last element` `    ``if` `(arr[n - ``2``] != arr[n - ``1``])` `        ``System.out.print(arr[n - ``1``] + ``" "``);` `}`   `// Driver code` `public` `static` `void` `main(String args[])` `{` `    ``int` `arr[] = {``7``, ``7``, ``8``, ``8``, ``9``, ` `                 ``1``, ``1``, ``4``, ``2``, ``2``};` `    ``int` `n = arr.length;`   `    ``occurredOnce(arr, n);` `}` `}`   `// This code is contributed` `// by Arnab Kundu`

Python3

 `# Python 3 implementation ` `# of above approach`   `# Function to find the elements ` `# that appeared only once in ` `# the array` `def` `occurredOnce(arr, n):` `    `  `    ``# Sort the array` `    ``arr.sort()`   `    ``# Check for first element` `    ``if` `arr[``0``] !``=` `arr[``1``]:` `        ``print``(arr[``0``], end ``=` `" "``)`   `    ``# Check for all the elements` `    ``# if it is different its ` `    ``# adjacent elements` `    ``for` `i ``in` `range``(``1``, n ``-` `1``):` `        ``if` `(arr[i] !``=` `arr[i ``+` `1``] ``and` `            ``arr[i] !``=` `arr[i ``-` `1``]):` `            ``print``( arr[i], end ``=` `" "``)`   `    ``# Check for the last element` `    ``if` `arr[n ``-` `2``] !``=` `arr[n ``-` `1``]:` `        ``print``(arr[n ``-` `1``], end ``=` `" "``)`   `# Driver code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``arr ``=` `[ ``7``, ``7``, ``8``, ``8``, ``9``, ` `            ``1``, ``1``, ``4``, ``2``, ``2` `]` `    ``n ``=` `len``(arr)` `    ``occurredOnce(arr, n)`   `# This code is contributed ` `# by ChitraNayal`

Javascript

 ``

C#

 `// C# implementation` `// of above approach` `using` `System;`   `class` `GFG` `{`   `// Function to find the elements that ` `// appeared only once in the array` `static` `void` `occurredOnce(``int``[] arr, ``int` `n)` `{` `    ``// Sort the array` `    ``Array.Sort(arr);`   `    ``// Check for first element` `    ``if` `(arr[0] != arr[1])` `        ``Console.Write(arr[0] + ``" "``);`   `    ``// Check for all the elements ` `    ``// if it is different` `    ``// its adjacent elements` `    ``for` `(``int` `i = 1; i < n - 1; i++)` `        ``if` `(arr[i] != arr[i + 1] && ` `            ``arr[i] != arr[i - 1])` `            ``Console.Write(arr[i] + ``" "``);`   `    ``// Check for the last element` `    ``if` `(arr[n - 2] != arr[n - 1])` `        ``Console.Write(arr[n - 1] + ``" "``);` `}`   `// Driver code` `public` `static` `void` `Main()` `{` `    ``int``[] arr = {7, 7, 8, 8, 9, ` `                ``1, 1, 4, 2, 2};` `    ``int` `n = arr.Length;`   `    ``occurredOnce(arr, n);` `}` `}`   `// This code is contributed ` `// by ChitraNayal`

PHP

 ``

Output

`4 9 `

Complexity Analysis:

• Time Complexity: O(Nlogn)
• Auxiliary Space: O(1)

Method-2: (Using Hashing): In C++, unordered_map can be used for hashing.

• Traverse the array.
• Store each element with its occurrence in the unordered_map.
• Traverse the unordered_map and print all the elements with occurrence 1.

Below is the implementation of the above approach:

C++

 `// C++ implementation to find elements` `// that appeared only once` `#include ` `using` `namespace` `std;`   `// Function to find the elements that` `// appeared only once in the array` `void` `occurredOnce(``int` `arr[], ``int` `n)` `{` `    ``unordered_map<``int``, ``int``> mp;`   `    ``// Store all the elements in the map with` `    ``// their occurrence` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``mp[arr[i]]++;`   `    ``// Traverse the map and print all the` `    ``// elements with occurrence 1` `    ``for` `(``auto` `it = mp.begin(); it != mp.end(); it++)` `        ``if` `(it->second == 1)` `            ``cout << it->first << ``" "``;` `}`   `// Driver code` `int` `main()` `{`   `    ``int` `arr[] = { 7, 7, 8, 8, 9, 1, 1, 4, 2, 2 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``occurredOnce(arr, n);`   `    ``return` `0;` `}`

Java

 `// Java implementation to find elements ` `// that appeared only once ` `import` `java.util.*;` `import` `java.io.*;`   `class` `GFG ` `{` `    ``// Function to find the elements that ` `    ``// appeared only once in the array ` `    ``static` `void` `occurredOnce(``int``[] arr, ``int` `n)` `    ``{` `            ``HashMap mp = ``new` `HashMap<>();` `            `  `            ``// Store all the elements in the map with ` `            ``// their occurrence` `            ``for` `(``int` `i = ``0``; i < n; i++) ` `            ``{` `                ``if` `(mp.containsKey(arr[i]))` `                    ``mp.put(arr[i], ``1` `+ mp.get(arr[i]));` `                ``else` `                    ``mp.put(arr[i], ``1``);` `            ``}`   `            ``// Traverse the map and print all the ` `            ``// elements with occurrence 1` `            ``for` `(Map.Entry entry : mp.entrySet()) ` `            ``{` `                ``if` `(Integer.parseInt(String.valueOf(entry.getValue())) == ``1``)` `                    ``System.out.print(entry.getKey() + ``" "``);` `            ``}` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String args[]) ` `    ``{` `            ``int``[] arr = { ``7``, ``7``, ``8``, ``8``, ``9``, ``1``, ``1``, ``4``, ``2``, ``2` `};` `            ``int` `n = arr.length;`   `            ``occurredOnce(arr, n); ` `    ``}` `}`   `// This code is contributed by rachana soma`

Python3

 `# Python3 implementation to find elements` `# that appeared only once` `import` `math as mt`   `# Function to find the elements that` `# appeared only once in the array` `def` `occurredOnce(arr, n):`   `    ``mp ``=` `dict``()`   `    ``# Store all the elements in the ` `    ``# map with their occurrence` `    ``for` `i ``in` `range``(n):` `        ``if` `arr[i] ``in` `mp.keys():` `            ``mp[arr[i]] ``+``=` `1` `        ``else``:` `            ``mp[arr[i]] ``=` `1`   `    ``# Traverse the map and print all ` `    ``# the elements with occurrence 1` `    ``for` `it ``in` `mp:` `        ``if` `mp[it] ``=``=` `1``:` `            ``print``(it, end ``=` `" "``)`   `# Driver code` `arr ``=` `[``7``, ``7``, ``8``, ``8``, ``9``, ``1``, ``1``, ``4``, ``2``, ``2``]` `n ``=` `len``(arr)`   `occurredOnce(arr, n)`   `# This code is contributed by ` `# Mohit Kumar 29`

Javascript

 ``

C#

 `// C# implementation to find elements ` `// that appeared only once ` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG ` `{` `    ``// Function to find the elements that ` `    ``// appeared only once in the array ` `    ``static` `void` `occurredOnce(``int``[] arr, ``int` `n)` `    ``{` `      ``Dictionary<``int``, ``int``> mp = ``new` `Dictionary<``int``, ``int``>();`   `      ``// Store all the elements in the map with ` `      ``// their occurrence` `      ``for` `(``int` `i = 0; i < n; i++) ` `      ``{` `        ``if` `(mp.ContainsKey(arr[i]))` `          ``mp[arr[i]] = 1 + mp[arr[i]];` `        ``else` `          ``mp.Add(arr[i], 1);` `      ``}`   `      ``// Traverse the map and print all the ` `      ``// elements with occurrence 1` `      ``foreach``(KeyValuePair<``int``, ``int``> entry ``in` `mp)` `      ``{` `        ``if` `(Int32.Parse(String.Join(``""``, entry.Value)) == 1)` `          ``Console.Write(entry.Key + ``" "``);` `      ``}` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main(String []args) ` `    ``{` `      ``int``[] arr = { 7, 7, 8, 8, 9, 1, 1, 4, 2, 2 };` `      ``int` `n = arr.Length;` `      ``occurredOnce(arr, n); ` `    ``}` `}`   `// This code is contributed by shikhasingrajput`

Output

`4 9 `

Complexity Analysis:

• Time Complexity: O(N)
• Auxiliary Space: O(N)

Method-3: Using given assumptions.

It is given that an array can be rotated any time and duplicates will appear side by side every time. So, after rotating, the first and last elements will appear side by side.

• Check if the first and last elements are equal. If yes, then start traversing the elements between them.
• Check if the current element is equal to the element in the immediate previous index. If yes, check the same for the next element.
• If not, print the current element.

Implementation:

C++

 `// C++ implementation to find elements` `// that appeared only once` `#include ` `using` `namespace` `std;`   `// Function to find the elements that` `// appeared only once in the array` `void` `occurredOnce(``int` `arr[], ``int` `n)` `{` `    ``int` `i = 1, len = n;`   `    ``// Check if the first and last element is equal` `    ``// If yes, remove those elements` `    ``if` `(arr[0] == arr[len - 1]) {` `        ``i = 2;` `        ``len--;` `    ``}`   `    ``// Start traversing the remaining elements` `    ``for` `(; i < n; i++)`   `        ``// Check if current element is equal to` `        ``// the element at immediate previous index` `        ``// If yes, check the same for next element` `        ``if` `(arr[i] == arr[i - 1])` `            ``i++;`   `        ``// Else print the current element` `        ``else` `            ``cout << arr[i - 1] << ``" "``;`   `    ``// Check for the last element` `    ``if` `(arr[n - 1] != arr[0] && arr[n - 1] != arr[n - 2])` `        ``cout << arr[n - 1];` `}`   `// Driver code` `int` `main()` `{`   `    ``int` `arr[] = { 7, 7, 8, 8, 9, 1, 1, 4, 2, 2 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``occurredOnce(arr, n);`   `    ``return` `0;` `}`

Java

 `// Java implementation to find ` `// elements that appeared only once` `class` `GFG` `{` `// Function to find the elements that` `// appeared only once in the array` `static` `void` `occurredOnce(``int` `arr[], ``int` `n)` `{` `    ``int` `i = ``1``, len = n;`   `    ``// Check if the first and last ` `    ``// element is equal. If yes, ` `    ``// remove those elements` `    ``if` `(arr[``0``] == arr[len - ``1``]) ` `    ``{` `        ``i = ``2``;` `        ``len--;` `    ``}`   `    ``// Start traversing the` `    ``// remaining elements` `    ``for` `(; i < n; i++)`   `        ``// Check if current element is ` `        ``// equal to the element at ` `        ``// immediate previous index` `        ``// If yes, check the same` `        ``// for next element` `        ``if` `(arr[i] == arr[i - ``1``])` `            ``i++;`   `        ``// Else print the current element` `        ``else` `            ``System.out.print(arr[i - ``1``] + ``" "``);`   `    ``// Check for the last element` `    ``if` `(arr[n - ``1``] != arr[``0``] && ` `        ``arr[n - ``1``] != arr[n - ``2``])` `        ``System.out.print(arr[n - ``1``]);` `}`   `// Driver code` `public` `static` `void` `main(String args[])` `{` `    ``int` `arr[] = {``7``, ``7``, ``8``, ``8``, ``9``,` `                 ``1``, ``1``, ``4``, ``2``, ``2``};` `    ``int` `n = arr.length;`   `    ``occurredOnce(arr, n);` `}` `}`   `// This code is contributed ` `// by Arnab Kundu`

Python3

 `# Python 3 implementation to find ` `# elements that appeared only once`   `# Function to find the elements that` `# appeared only once in the array` `def` `occurredOnce(arr, n):` `    ``i ``=` `1` `    ``len` `=` `n`   `    ``# Check if the first and ` `    ``# last element is equal` `    ``# If yes, remove those elements` `    ``if` `arr[``0``] ``=``=` `arr[``len` `-` `1``]: ` `        ``i ``=` `2` `        ``len` `-``=` `1`   `    ``# Start traversing the` `    ``# remaining elements` `    ``while` `i < n:`   `        ``# Check if current element is ` `        ``# equal to the element at` `        ``# immediate previous index` `        ``# If yes, check the same for` `        ``# next element` `        ``if` `arr[i] ``=``=` `arr[i ``-` `1``]:` `            ``i ``+``=` `1`   `        ``# Else print the current element` `        ``else``:` `            ``print``(arr[i ``-` `1``], end ``=` `" "``)` `            `  `        ``i ``+``=` `1`   `    ``# Check for the last element` `    ``if` `(arr[n ``-` `1``] !``=` `arr[``0``] ``and` `        ``arr[n ``-` `1``] !``=` `arr[n ``-` `2``]):` `        ``print``(arr[n ``-` `1``])`   `# Driver code` `if` `__name__ ``=``=` `"__main__"``: ` `    ``arr ``=` `[ ``7``, ``7``, ``8``, ``8``, ``9``, ``1``, ``1``, ``4``, ``2``, ``2` `]` `    ``n ``=` `len``(arr)`   `    ``occurredOnce(arr, n)`   `# This code is contributed ` `# by ChitraNayal`

Javascript

 ``

C#

 `// C# implementation to find ` `// elements that appeared only once` `using` `System;`   `class` `GFG` `{` `// Function to find the elements that` `// appeared only once in the array` `static` `void` `occurredOnce(``int``[] arr, ``int` `n)` `{` `    ``int` `i = 1, len = n;`   `    ``// Check if the first and last ` `    ``// element is equal. If yes, ` `    ``// remove those elements` `    ``if` `(arr[0] == arr[len - 1]) ` `    ``{` `        ``i = 2;` `        ``len--;` `    ``}`   `    ``// Start traversing the` `    ``// remaining elements` `    ``for` `(; i < n; i++)`   `        ``// Check if current element is ` `        ``// equal to the element at ` `        ``// immediate previous index` `        ``// If yes, check the same` `        ``// for next element` `        ``if` `(arr[i] == arr[i - 1])` `            ``i++;`   `        ``// Else print the current element` `        ``else` `            ``Console.Write(arr[i - 1] + ``" "``);`   `    ``// Check for the last element` `    ``if` `(arr[n - 1] != arr[0] && ` `        ``arr[n - 1] != arr[n - 2])` `        ``Console.Write(arr[n - 1]);` `}`   `// Driver code` `public` `static` `void` `Main()` `{` `    ``int``[] arr = {7, 7, 8, 8, 9,` `                ``1, 1, 4, 2, 2};` `    ``int` `n = arr.Length;`   `    ``occurredOnce(arr, n);` `}` `}`   `// This code is contributed ` `// by ChitraNayal`

PHP

 ``

Output

`9 4 `

Complexity Analysis:

• Time Complexity: O(N)
• Auxiliary Space: O(1)

Method #4:Using built-in Python functions:

• Count the frequencies of every element using the Counter function
• Traverse the frequency array and print all the elements with occurrence 1.

Below is the implementation

C++

 `// C++ program for the above approach` `#include` `using` `namespace` `std;`   `// Function to find the elements that` `// appeared only once in the array` `void` `OccurredOnce(``int` `arr[], ``int` `n) {` `    ``// counting frequency of every element` `    ``// using unordered_map` `    ``unordered_map<``int``, ``int``> mp;` `    ``for``(``int` `i=0;i

Java

 `import` `java.util.HashMap;`   `public` `class` `Main {` `    ``public` `static` `void` `OccurredOnce(``int``[] arr) {` `        ``// Counting frequency of every element using HashMap` `        ``HashMap mp = ``new` `HashMap<>();` `        ``for` `(``int` `i = ``0``; i < arr.length; i++) {` `            ``if` `(mp.containsKey(arr[i])) {` `                ``mp.put(arr[i], mp.get(arr[i]) + ``1``);` `            ``} ``else` `{` `                ``mp.put(arr[i], ``1``);` `            ``}` `        ``}` `        ``// Traverse the map and print all the elements with occurrence 1` `        ``for` `(``int` `it : mp.keySet()) {` `            ``if` `(mp.get(it) == ``1``) {` `                ``System.out.print(it + ``" "``);` `            ``}` `        ``}` `    ``}`   `    ``public` `static` `void` `main(String[] args) {` `        ``int``[] arr = {``7``, ``7``, ``8``, ``8``, ``9``, ``1``, ``1``, ``4``, ``2``, ``2``};` `        ``OccurredOnce(arr);` `    ``}` `}`

Python3

 `# Python3 implementation to find elements` `# that appeared only once` `from` `collections ``import` `Counter`   `# Function to find the elements that` `# appeared only once in the array` `def` `occurredOnce(arr, n):`   `    ``#counting frequency of every element using Counter` `    ``mp``=``Counter(arr)` `    ``# Traverse the map and print all ` `    ``# the elements with occurrence 1` `    ``for` `it ``in` `mp:` `        ``if` `mp[it] ``=``=` `1``:` `            ``print``(it, end ``=` `" "``)`   `# Driver code` `arr ``=` `[``7``, ``7``, ``8``, ``8``, ``9``, ``1``, ``1``, ``4``, ``2``, ``2``]` `n ``=` `len``(arr)`   `occurredOnce(arr, n)`   `# This code is contributed by vikkycirus`

Javascript

 `function` `occurredOnce(arr, n) {` `// counting frequency of every element` `// using Map` `let mp = ``new` `Map();` `for``(let i=0; i

C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;` `using` `System.Linq;`   `class` `Program {` `    `  `    ``// Function to find the elements that` `    ``// appeared only once in the array` `    ``static` `void` `OccurredOnce(``int``[] arr, ``int` `n) {` `        ``// counting frequency of every element using .Count() method` `        ``Dictionary<``int``, ``int``> mp = arr.GroupBy(x => x).ToDictionary(g => g.Key, g => g.Count());` `        `  `        ``// Traverse the map and print all ` `        ``// the elements with occurrence 1` `        ``foreach` `(``var` `item ``in` `mp) {` `            ``if` `(item.Value == 1) {` `                ``Console.Write(item.Key + ``" "``);` `            ``}` `        ``}` `    ``}` `    `  `    ``// Driver code ` `    ``static` `void` `Main(``string``[] args) {` `        ``int``[] arr = {7, 7, 8, 8, 9, 1, 1, 4, 2, 2};` `        ``int` `n = arr.Length;` `        ``OccurredOnce(arr, n);` `    ``}` `}`   `// This code is contributed by Prince Kumar`

Output

`9 4 `

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

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