How to print a unique elements of an Array?

private static IEnumerable<int> FindUniqueNumbersQuick(int[] numbers, out IEnumerable<int> duplicates)
{
var unique = new List<int>();
var dups = new List<int>();
foreach (var n in numbers)
{
if (dups.Contains(n))
continue;
if (unique.Contains(n))
{
unique.Remove(n);
dups.Add(n);
continue;
}
unique.Add(n);
}
duplicates = dups;
return unique;
}
// look full application here:
// https://github.com/sergeklokov/FindUniqueNumbers/blob/master/FindUniqueNumbers/FindUniqueNumbers.cs

I have seen many answeres here but all the answers are in old style but C# has grown now. By using LINQ you can achieve it in a minimal code. Try to find unique elements in array using below simple approach.

class Program
{
static void Main(string[] args)
{
int[] items = { 2, 5, 7, 5, 8, 10, 8 };
IEnumerable<int> uniqueItems = items.Distinct<int>();
Console.WriteLine("Unique array items: " + string.Join(",", uniqueItems));
Console.ReadLine();
}
}
/*
Output:
Unique array items: 2, 5, 7, 8, 10
*/

/*
* C program to print all unique elements of an unsorted array
*/
#include
#include
int main() {
int array[100], size, i, j;
printf("Enter number of elements in array\n");
scanf("%d", &size);
printf("Enter %d numbers\n", size);
for(i = 0; i < size; i++){
scanf("%d", &array[i]);
}
printf("Unique Elements\n");
for(i = 0; i < size; i++) {
for (j=0; j
if (array[i] == array[j])
break;
}
if (i == j){
/* No duplicate element found between index 0 to i */
printf("%d ", array[i]);
}
}
getch();
return 0;
}

Refer the below code to print unique elements in array:

using System;
public class Exercise6
{
public static void Main()
{
int n,ctr=0;
int[] arr1 = new int[100];
int i, j, k;
Console.Write("\n\nPrint all unique elements of an array:\n");
Console.Write("------------------------------------------\n");
Console.Write("Input the number of elements to be stored in the array :");
n = Convert.ToInt32(Console.ReadLine());
Console.Write("Input {0} elements in the array :\n",n);
for(i=0;i
{
Console.Write("element - {0} : ",i);
arr1[i] = Convert.ToInt32(Console.ReadLine());
}
/*Checking duplicate elements in the array */
Console.Write("\nThe unique elements found in the array are : \n");
for(i=0; i
{
ctr=0;
/*Check duplicate bifore the current position and
increase counter by 1 if found.*/
for(j=0; j
{
/*Increment the counter when the seaarch value is duplicate.*/
if(arr1[i]==arr1[j])
{
ctr++;
}
}
/*Check duplicate after the current position and
increase counter by 1 if found.*/
for(k=i+1; k
{
/*Increment the counter when the seaarch value is duplicate.*/
if(arr1[i]==arr1[k])
{
ctr++;
}
}
/*Print the value of the current position of the array as unique value
when counter remain contains its initial value.*/
if(ctr==0)
{
Console.Write("{0} ",arr1[i]);
}
}
Console.Write("\n\n");
}
}

can you explain with C# Program using foreach statement to print an unique array.Thank You...

#include

using namespace std;

void printDistinct(int arr[], int n)

{

// Pick all elements one by one

for (int i=0; i

{

// Check if the picked element is already printed

int j;

for (j=0; j

if (arr[i] == arr[j])

break;

// If not printed earlier, then print it

if (i == j)

cout << arr[i] << " ";

}

// Driver program to test above function

int main()

{

int arr[] = {6, 10, 5, 4, 9, 120, 4, 6, 10};

int n = sizeof(arr)/sizeof(arr[0]);

printDistinct(arr, n);

return 0;

}

Run on IDE

6 Replies

Know the answer? Post it — somebody with the same question will find it here.

Serge Klokov6y ago
1. private static IEnumerable<int> FindUniqueNumbersQuick(int[] numbers, out IEnumerable<int> duplicates)
2. {
3. var unique = new List<int>();
4. var dups = new List<int>();
5. foreach (var n in numbers)
6. {
7. if (dups.Contains(n))
8. continue;
9. if (unique.Contains(n))
10. {
11. unique.Remove(n);
12. dups.Add(n);
13. continue;
14. }
15. unique.Add(n);
16. }
17. duplicates = dups;
18. return unique;
19. }
20. // look full application here:
21. // https://github.com/sergeklokov/FindUniqueNumbers/blob/master/FindUniqueNumbers/FindUniqueNumbers.cs
0