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Code Array

# Find element that appears only once in array

Finding an element that appears only once in an array is a common programming problem. The goal is to identify the element in the array that doesn't have any duplicate occurrences.

## Problem Statement

Given an array of integers, find and display all the elements that appear only once in the array.

## Example

Consider the following array:

``arr = [8, 3, 2, 6, 4, -5, 1, 6, 8, 9, 7]``

The elements that appear only once in this array are: `3`, `2`, `4`, `-5`, `1`, `9`, and `7`.

## Idea to Solve

Iterate through the array and for each element, check if it has any duplicates by comparing it with all other elements in the array.

## Pseudocode

``````function unique_element(arr, size):
for i from 0 to size - 1:
status = 1
for j from 0 to size - 1:
if i ≠ j and arr[i] = arr[j]:
status = 0
break
if status = 1:
print(arr[i])

// Example usage
arr = [8, 3, 2, 6, 4, -5, 1, 6, 8, 9, 7]
size = size(arr)
unique_element(arr, size)``````

## Algorithm Explanation

1. The `unique_element` function takes the array and its size as parameters.
2. For each element in the array, it initializes `status` to `1`, indicating that the element is unique until proven otherwise.
3. It compares the current element with all other elements in the array using a nested loop.
4. If it finds a duplicate, it sets `status` to `0` and breaks out of the inner loop.
5. After checking all other elements, if `status` is still `1`, it means the current element is unique and it prints the element.

## Code Solution

``````//C Program
//Find element that appears only once in array
#include <stdio.h>

//Find and display all non repeated element
void unique_element(int arr[],int size)
{

if(size<=0)
{
//When invalid size of array
return;
}
else
{
int status = 1;

for (int i = 0; i < size; ++i)
{
//initial set element is a unique
status = 1;

for (int j = 0; j < size && status == 1; ++j)
{
if(i!=j && arr[i]==arr[j])
{
//When find duplicate elements
status=0;
}
}
if(status==1)
{
//When get a unique
printf("  %d",arr[i] );
}
}

}
}

int main()
{
//Define array elements
int arr[] ={8,3,2,6,4,-5,1,6,8,9,7};

//Get the size of array elements
int size=sizeof(arr) / sizeof(arr[0]);

unique_element(arr,size);

return 0;

}```
```

#### Output

``  3  2  4  -5  1  9  7``
``````/*
C++ Program
Find element that appears only once in array
*/
#include<iostream>
using namespace std;

class MyArray {
public:

//Find and display all non repeated element
void unique_element(int arr[], int size) {
if (size <= 0) {
return;
} else {
int status = 1;
for (int i = 0; i < size; ++i) {
//initial set element is a unique
status = 1;
for (int j = 0; j < size && status == 1; ++j) {
if (i != j && arr[i] == arr[j]) {
//When find duplicate elements
status = 0;
}
}
if (status == 1) {
//When get a unique

cout << " " << arr[i];
}
}
}
}
};
int main() {
MyArray obj ;
int arr[] = {
8,
3,
2,
6,
4,
-5,
1,
6,
8,
9,
7
};
//Count size of array
int size = sizeof(arr) / sizeof(arr[0]);
obj.unique_element(arr, size);
return 0;
}```
```

#### Output

`` 3 2 4 -5 1 9 7``
``````/*
Java Program
Find element that appears only once in array
*/
public class MyArray {

//Find and display all non repeated element
public void unique_element(int []arr,int size)
{

if(size<=0)
{
//When invalid size of array
return;
}
else
{
int status = 1;

for (int i = 0; i < size; ++i)
{
//initial set element is a unique
status = 1;

for (int j = 0; j < size && status == 1; ++j)
{
if(i!=j && arr[i]==arr[j])
{
//When find duplicate elements
status=0;
}
}
if(status==1)
{
//When get a unique
System.out.print("  "+arr[i] );
}
}

}
}

public static void main(String[] args)
{

MyArray obj = new MyArray();
//Define array elements
int []arr =  {8,3,2,6,4,-5,1,6,8,9,7};
//Count size of array
int size=arr.length;
obj.unique_element(arr,size);

}
}```
```

#### Output

`` 3 2 4 -5 1 9 7``
``````using System;

/*
C# Program
Find element that appears only once in array
*/

public class MyArray {
//Find and display all non repeated element
public void unique_element(int[] arr, int size) {
if (size <= 0) {
return;
} else {
int status = 1;
for (int i = 0; i < size; ++i) {
//initial set element is a unique
status = 1;
for (int j = 0; j < size && status == 1; ++j) {
if (i != j && arr[i] == arr[j]) {
//When find duplicate elements
status = 0;
}
}
if (status == 1) {
Console.Write(" " + arr[i]);
}
}
}
}
public static void Main(String[] args) {
MyArray obj = new MyArray();
//Define array elements
int[] arr = {
8,
3,
2,
6,
4,
-5,
1,
6,
8,
9,
7
};
//Count size of array
int size = arr.Length;
obj.unique_element(arr, size);
}
}```
```

#### Output

`` 3 2 4 -5 1 9 7``
``````<?php
/*
Php Program
Find element that appears only once in array
*/
class MyArray {
//Find and display all non repeated element

public 	function unique_element(\$arr, \$size) {
if (\$size <= 0) {
return;
} else {
\$status = 1;
for (\$i = 0; \$i < \$size; ++\$i) {
//initial set element is a unique
\$status = 1;
for (\$j = 0; \$j < \$size && \$status == 1; ++\$j) {
if (\$i != \$j && \$arr[\$i] == \$arr[\$j]) {
//When find duplicate elements
\$status = 0;
}
}
if (\$status == 1) {
//When get a unique

echo(" ". \$arr[\$i]);
}
}
}
}
}

function main() {
\$obj = new MyArray();
//Define array elements
\$arr = array(8, 3, 2, 6, 4, -5, 1, 6, 8, 9, 7);
//Count size of array
\$size = count(\$arr);
\$obj->unique_element(\$arr, \$size);

}
main();```
```

#### Output

`` 3 2 4 -5 1 9 7``
``````/*
Node Js Program
Find element that appears only once in array
*/
class MyArray {
//Find and display all non repeated element
unique_element(arr, size) {
if (size <= 0) {
return;
} else {
var status = 1;
for (var i = 0; i < size; ++i) {
//initial set element is a unique
status = 1;
for (var j = 0; j < size && status == 1; ++j) {
if (i != j && arr[i] == arr[j]) {
//When find duplicate elements
status = 0;
}
}

if (status == 1) {
//When get a unique

process.stdout.write(" " + arr[i]);
}
}
}
}
}

function main(args) {
var obj = new MyArray();
//Define array elements
var arr = [8, 3, 2, 6, 4, -5, 1, 6, 8, 9, 7];
//Count size of array
var size = arr.length;
obj.unique_element(arr, size);
}

main();```
```

#### Output

`` 3 2 4 -5 1 9 7``
``````# Python 3 Program
# Find element that appears only once in array
class MyArray :
# Find and display all non repeated element
def unique_element(self, arr, size) :
if (size <= 0) :
return
else :
status = 1
i = 0
while (i < size) :
# initial set element is a unique
status = 1
j = 0
while (j < size and status == 1) :
if (i != j and arr[i] == arr[j]) :
# When find duplicate elements
status = 0

j += 1

if (status == 1) :
print(" ", arr[i], end = "")

i += 1

def main() :
obj = MyArray()
arr = [8, 3, 2, 6, 4, -5, 1, 6, 8, 9, 7]
size = len(arr)
obj.unique_element(arr, size)

if __name__ == "__main__":
main()```
```

#### Output

``  3  2  4  -5  1  9  7``
``````# Ruby Program
# Find element that appears only once in array
class MyArray
# Find and display all non repeated element
def unique_element(arr, size)
if (size <= 0)
return
else
status = 1
i = 0
while (i < size)
# initial set element is a unique
status = 1
j = 0
while (j < size && status == 1)
if (i != j && arr[i] == arr[j])
# When find duplicate elements
status = 0
end
j += 1
end
if (status == 1)
print(" ", arr[i])
end
i += 1
end
end
end
end
def main()
obj = MyArray.new()
arr = [8, 3, 2, 6, 4, -5, 1, 6, 8, 9, 7]
size = arr.length
obj.unique_element(arr, size)
end
main()```
```

#### Output

`` 3 2 4 -5 1 9 7``
``````/*
Scala Program
Find element that appears only once in array
*/
class MyArray {
//Find and display all non repeated element
def unique_element(arr: Array[Int], size: Int): Unit = {
if (size <= 0) {
return;
} else {
var status: Int = 1;
var i: Int = 0;
while (i < size) {
//initial set element is a unique
status = 1;
var j: Int = 0;
while (j < size && status == 1) {
if (i != j && arr(i) == arr(j)) {
//When find duplicate elements
status = 0;
}
j += 1;
}
if (status == 1) {
print(" " + arr(i));
}
i += 1;
}
}
}
}
object Main {
def main(args: Array[String]): Unit = {
val obj: MyArray = new MyArray();
val arr: Array[Int] = Array(8, 3, 2, 6, 4, -5, 1, 6, 8, 9, 7);
val size: Int = arr.length;
obj.unique_element(arr, size);
}
}```
```

#### Output

`` 3 2 4 -5 1 9 7``
``````/*
Swift Program
Find element that appears only once in array
*/
class MyArray {
//Find and display all non repeated element
func unique_element(_ arr: [Int], _ size: Int) {
if (size <= 0) {
return;
} else {
var status: Int = 1;
var i: Int = 0;
while (i < size) {
//initial set element is a unique
status = 1;
var j: Int = 0;
while (j < size && status == 1) {
if (i != j && arr[i] == arr[j]) {
//When find duplicate elements
status = 0;
}
j += 1;
}
if (status == 1) {
print(" ", arr[i], terminator: "");
}
i += 1;
}
}
}
}
func main() {
let obj: MyArray = MyArray();
let arr: [Int] = [8, 3, 2, 6, 4, -5, 1, 6, 8, 9, 7];
let size: Int = arr.count;
obj.unique_element(arr, size);
}
main();```
```

#### Output

``  3  2  4  -5  1  9  7``

## Time Complexity

The algorithm compares each element with all other elements, resulting in a nested loop structure. Therefore, the time complexity of the algorithm is `O(n^2)`, where `n` is the size of the array.

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