Find all the unique pairs from the unsorted array.
Answers
class prog {
int check(int n)
{String a=""+n;int y=0;
int l=a.length();int v[]=new int[l];
while(n>0)
{v[y]=n%10;
n=n/10;y++;
}
y=1;
for(int i=0;i<l-1;i++)
{
for(int j=i+1;j<l;j++)
{
if (v[i]==v[j])
{y=0;
break;}
}
}
return y;
}
void main(int s[])
{
int u=s.length();
for (int i=0;i<=u;i++)
{
int r=check(s[i]);
if (r==1)
{
System.out.println(s[i]+"is a unique number");
}
}
}}
Hope this helps you.
I have put in quite an effort typing this on phone
please mark it as the brainliest answer.
/* A simple program to count pairs with difference k*/
using namespace std;
int countPairsWithDiffK(int arr[], int n, int k)
{
int count = 0;
// Pick all elements one by one
for (int i = 0; i < n; i++)
{
// See if there is a pair of this picked element
for (int j = i+1; j < n; j++)
if (arr[i] - arr[j] == k || arr[j] - arr[i] == k )
count++;
}
return count;
}
// Driver program to test above function
int main()
{
int arr[] = {1, 5, 3, 4, 2};
int n = sizeof(arr)/sizeof(arr[0]);
int k = 3;
cout << "Count of pairs with given diff is "
<< countPairsWithDiffK(arr, n, k);
return 0;
}