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import java.util.*;
public class UniqueElement {
static int getU(int[] list){
int unique =0;
for(int i=0;i<list.length;i++){
unique = list[i];
int count=0;
for(int j=0;j<list.length;j++){
if(list[j]==unique){
count++;
}
}
if(count==1){
return unique;
}
}
return -1;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] list = new int[n];
for(int i=0;i<n;i++){
list[i]= sc.nextInt();
}
int unique= getU(list);
System.out.println(unique);
}
}
list_no=[]
n= int(input(“Enter the number of values in list to be added”))
for i in range(0,n):
value= int(input(“Enter the value”))
list_no.append(value)
for i in range(len(list_no)):
count = 0;
count=list_no.count(list_no[i])
if count == 1:
print(list_no[i])
break
count = 0
#Python
n=int(input())
l=list(map(int,input().split(” “)))
a1=2*sum(list(set(l)))
a2=sum(l)
print(a1-a2)
IN PYHTON CODE WOULD BE LIKE :
def lonelyinteger(nos):
for i in nos:
if nos.count(i)==1:
break
return i
n = int(input())
nos = list(map(int,input().split()))
print(lonelyinteger(nos))
The C++ code for the same
#include
using namespace std;
int main() {
int count[100];int arr[100]; int n;
cout<>n; cout<<"\n";
for(int i=0;i>arr[i];
}
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
if(arr[j]==arr[i]&&i!=j)
{
count[i]++;
}
}
}
for(int i=0;i<n;i++)
{
if(count[i]==0)
{
cout<<"\n"<<arr[i];
}
}
}
import java.util.*;
public class char_count_String {
public static void main(String[] args)
{ Scanner sc = new Scanner (System.in);
int L = sc.nextInt();
int arr [] = new int [L];
for(int i =0;i<L;i++) {arr[i]=sc.nextInt();}
System.out.println(fuction(arr));
}
static int fuction(int[] arr) {
Map hm = new HashMap();
for(int x : arr) {hm.put(x, hm.getOrDefault(x,0)+1 );}
for(int x : arr) {if(hm.get(x )==1 )return x;}
return -1;
}
}
#include
using namespace std;
int getLonely(int *arr,int n)
{
int res=arr[0];
for(int i=0;i>n;
int arr[100];
for(int i=0;i>arr[i];
cout<<getLonely(arr,n)<<endl;
}
in c++
def lonely_integer(a):
result = 0
for i in a:
result ^= i ”’XOR OPERATION
A^A = 0
0^B = B So, for each pair of numbers the xor operation results in 0.
Since there is only one number that does not have a pair, the result is its lonly number,
since 0^b =b (or b^0 = b).”’
return result
n=int(input())
b = map(int,input().strip().split(” “))
print(lonely_integer(b))
#include
int main()
{
int i,a[100],n,j;
scanf(“%d”,&n);
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
if(a[i]==a[j])
{
a[i]=0;
a[j]=0;
}
}
}
for(i=0;i<n;i++){
if(a[i]!=0)
printf("%d",a[i]);
}
}
in c
Thanks for contributing the code Srimukhi
can i get the python code for this program??
can i get the c code for this program??
Here is a C code for this problem
#include
#define MAX_SIZE 100
int main()
{
int arr[MAX_SIZE], freq[MAX_SIZE];
int size, i, j, count;
/* Input size of array and elements in array */
printf(“Enter size of array: “);
scanf(“%d”, &size);
printf(“Enter elements in array: “);
for(i=0; i
sir,How to write this code in python
You can solve this using list and compare function in python
n = int(input(‘Enter the number of integers in the array’))
x = input().split()
def countUnique(x, n):
a = []
for i in range(n):
if x[i] in a:
continue
else:
a.append(x[i])
return len(a)
b = countUnique(x, n)
print(‘unique elements = {}’.format(b))
lst=[1,2,3,4,3,2,1]
for i in range(len(lst)):
count=0
for j in range(len(lst)):
if (lst[i]==lst[j]) and i!=j:
count+=1
if count == 0 :
print(lst[i])
n=int(input(‘Enter a length of list’))
lst=[]
for i in range(n):
lst.append(int(input(” “)))
for i in range(len(lst)):
count=0
for j in range(len(lst)):
if (lst[i]==lst[j]) and i!=j:
count+=1
if count == 0 :
print(lst[i])