Given an array arr[] of size n contains distinct elements, the task is to find the number of ordered pairs (A, B) that can be made where A and B are subsets of the given array arr[] and A ∩ B = Φ (i.e, Both A and B subset should be disjoint or no common elements between them).
Example:
Input: arr = {1, 2}
Output: 9
Explanation: The subsets of the array are {}, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, and {1, 2, 3}. The ordered pairs (A, B) where A and B disjoint subsets are: ({}, {}), ({}, {1}), ({}, {2}), ({1}, {}), ({2}, {}), ({1, 2}, {}), ({}, {1, 2}, ({1}, {2}, ({2}, {1})Input: arr = {1, 2, 3}
Output: 27
Approach:
Every element has three options, It will either be included in A, B, or not included in either A or B. Hence, there would be 3^n possible combinations for given array of size n.
Steps-by-step approach:
- Set the modulo constant MOD to 1e9 + 7.
-
Define a function binpow() that calculates the power using binary exponentiation.
- Take the modulo of the base to keep it within the specified range.
- Initialize result to 1.
- Iterate until the exponent is greater than 0.
- If the current bit of the exponent is 1, multiply the result by the base.
- Square the base and divide the exponent by 2 in each iteration.
- Return the final result.
-
Solve Function:
- Define a function solve that takes an integer n and a vector arr.
- Call binpow(3, n) to calculate (3^n) mod (1e9 + 7).
- Return the result.
Below is the implementation of the above approach:
#include <bits/stdc++.h> using namespace std;
// Define the modulo constant const int MOD = 1e9 + 7;
// Function to calculate power using binary exponentiation long long binpow( long long base, long long exp )
{ base %= MOD; // Take modulo of base to keep it within
// range
long long result = 1; // Initialize result
while ( exp
> 0) { // Loop until exponent is greater than 0
if ( exp & 1) // If exponent is odd
result = (result * base)
% MOD; // Multiply result with base
base = (base * base) % MOD; // Square the base
exp >>= 1; // Divide the exponent by 2
}
return result; // Return the final result
} int solve( int n, vector< int >& arr)
{ // Use binary exponentiation instead of pow
// This calculates (3^n) mod (1e9 + 7)
return binpow(3, n);
} int main()
{ int n = 3; // The number of elements
vector< int > arr = { 1, 2, 3 };
// Call the solve function and print the result
cout << solve(n, arr);
return 0;
} |
import java.util.*;
public class Main {
// Define the modulo constant
static final int MOD = 1000000007 ;
// Function to calculate power using binary exponentiation
static long binpow( long base, long exp) {
base %= MOD; // Take modulo of base to keep it within range
long result = 1 ; // Initialize result
while (exp > 0 ) { // Loop until exponent is greater than 0
if ((exp & 1 ) == 1 ) // If exponent is odd
result = (result * base) % MOD; // Multiply result with base
base = (base * base) % MOD; // Square the base
exp >>= 1 ; // Divide the exponent by 2
}
return result; // Return the final result
}
static int solve( int n, List<Integer> arr) {
// Use binary exponentiation instead of pow
// This calculates (3^n) mod (1e9 + 7)
return ( int ) binpow( 3 , n);
}
public static void main(String[] args) {
int n = 3 ; // The number of elements
List<Integer> arr = Arrays.asList( 1 , 2 , 3 );
// Call the solve function and print the result
System.out.println(solve(n, arr));
}
} |
using System;
using System.Collections.Generic;
public class Program
{ // Define the modulo constant
const int MOD = 1000000007;
// Function to calculate power using binary exponentiation
static long Binpow( long baseVal, long exp)
{
baseVal %= MOD; // Take modulo of base to keep it within range
long result = 1; // Initialize result
while (exp > 0) // Loop until exponent is greater than 0
{
if ((exp & 1) == 1) // If exponent is odd
result = (result * baseVal) % MOD; // Multiply result with base
baseVal = (baseVal * baseVal) % MOD; // Square the base
exp >>= 1; // Divide the exponent by 2
}
return result; // Return the final result
}
static int Solve( int n, List< int > arr)
{
// Use binary exponentiation instead of pow
// This calculates (3^n) mod (1e9 + 7)
return ( int )Binpow(3, n);
}
public static void Main( string [] args)
{
int n = 3; // The number of elements
List< int > arr = new List< int > { 1, 2, 3 };
// Call the Solve function and print the result
Console.WriteLine(Solve(n, arr));
}
} |
// Define the modulo constant const MOD = BigInt(1e9 + 7); // Function to calculate power using binary exponentiation function binpow(base, exp) {
base %= MOD; // Take modulo of base to keep it within range
let result = BigInt(1); // Initialize result
while (exp > 0n) { // Loop until exponent is greater than 0
if (exp & 1n) // If exponent is odd
result = (result * base) % MOD; // Multiply result with base
base = (base * base) % MOD; // Square the base
exp >>= 1n; // Divide the exponent by 2
}
return result; // Return the final result
} // Function to solve the problem function solve(n, arr) {
// Use binary exponentiation instead of pow
// This calculates (3^n) mod (1e9 + 7)
return binpow(3n, BigInt(n));
} // Sample usage const n = 3; // The number of elements
const arr = [1, 2, 3]; // Call the solve function and print the result console.log(solve(n, arr).toString()); //This code is contributed by Monu. |
# Python Implementation # Function to calculate power using binary exponentiation def binpow(base, exp, mod):
base % = mod # Take modulo of base to keep it within range
result = 1 # Initialize result
while exp > 0 : # Loop until exponent is greater than 0
if exp & 1 : # If exponent is odd
result = (result * base) % mod # Multiply result with base
base = (base * base) % mod # Square the base
exp >> = 1 # Divide the exponent by 2
return result # Return the final result
def solve(n, arr):
MOD = 10 * * 9 + 7 # Define the modulo constant
# Use binary exponentiation instead of pow
# This calculates (3^n) mod (1e9 + 7)
return binpow( 3 , n, MOD)
if __name__ = = "__main__" :
n = 3 # The number of elements
arr = [ 1 , 2 , 3 ]
# Call the solve function and print the result
print (solve(n, arr))
# This code is contributed by Sakshi |
27
Time Complexity: O(log(n))
Auxiliary Space: O(log(n))