Priya and AND Codechef Solution: Priya loves bitwise AND, but she hates programming. Help her solve this problem.
Given an array AA of size NN, let BijBij denote the bitwise AND of A[i]A[i] and A[j]A[j]. You have to find the number of pairs (i,j)(i,j), such that i<ji<j and Bij=A[i]Bij=A[i].
Input:
- The first line of the input contains a single integer TT denoting the number of test cases.
- The first line of each test case consists of a single integer NN, denoting the Size of Array AA.
- The second line of each test case contains NN space-separated integers A1,A2,A3…ANA1,A2,A3…AN.
Output:
For each test case, output a single line, count of such pairs.
Constraints
- 1≤T≤1001≤T≤100
- 1≤N≤1001≤N≤100
- 1≤A[i]≤1001≤A[i]≤100
Sample Input:
2 5 1 1 1 1 1 1 10
Sample Output:
10
Explanation
Example case 1: Number of valid pairs are -(1,2),(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,4),(3,5)(1,2),(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,4),(3,5) and (4,5)(4,5). Therefore, total valid pairs =10=10.
Example case 2: Since N=1N=1, therefore there are no valid pairs
Priya and AND – CodeChef Solution in JAVA
import java.util.*;
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner s=new Scanner(System.in);
int t=s.nextInt();
while(t-->0)
{
int n=s.nextInt();
int[] a=new int[n];
for(int i=0;i<n;i++)
{
a[i]=s.nextInt();
}
if(n==1)
{
System.out.println(0);
continue;
}
System.out.println(find(a));
}
}
public static int find(int[] a)
{
int count=0;
for(int i=0;i<a.length;i++)
{
for(int j=i+1;j<a.length;j++)
{
if((a[i]&a[j])==a[i])
{
count++;
}
}
}
return count;
}
}Priya and AND – CodeChef Solution in CPP
#include<iostream>
#define ll long long
using namespace std;
int main()
{
ll t;
cin>>t;
while(t--)
{
ll n;
cin>>n;
ll arr[n];
for(int i=0;i<n;i++)
{
cin>>arr[i];
}
ll count=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<i;j++)
{
if((arr[i]&arr[j])==arr[j])
{
count++;
}
}
}
cout<<count<<endl;
}
return 0;
}Priya and AND -CodeChef Solution in Python
for _ in range(int(input())):
k=int(input())
p=list(map(int,input().split()))
cnt=0
for i in range(k-1):
for j in range(i+1,k):
if((p[i] & p[j]) == p[i]):
cnt+=1
print(cnt)Disclaimer: The above Problem(Priya and AND) is generated by CodeChef but the solution is provided by Chase2learn.This tutorial is only for Educational and Learning purpose.
