Find 2^(2^A) % B

Given two integers A and B, the task is to calculate 2^2A % B.

Examples:

Input: A = 3, B = 5 
Output: 1 
2²³ % 5 = 2⁸ % 5 = 256 % 5 = 1.

Input: A = 10, B = 13 
Output: 3 

Approach: The problem can be efficiently solved by breaking it into sub-problems without overflow of integer by using recursion. 

Let F(A, B) = 2^2A % B. 
Now, F(A, B) = 2^2A % B 
= 2^{2 * 2A – 1} % B 
= (2^{2A – 1 + 2A – 1}) % B 
= (2^{2A – 1} * 2^{2A – 1}) % B 
= (F(A – 1, B) * F(A – 1, B)) % B 
Therefore, F(A, B) = (F(A – 1, B) * F(A – 1, B)) % B. 
The base case is F(1, B) = 2²¹ % B = 4 % B. 

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long

// Function to return 2^(2^A) % B
ll F(ll A, ll B)
{

    // Base case, 2^(2^1) % B = 4 % B
    if (A == 1)
        return (4 % B);
    else
    {
        ll temp =  F(A - 1, B);
        return (temp * temp) % B;
    }
}

// Driver code
int main()
{
    ll A = 25, B = 50;

    // Print 2^(2^A) % B
    cout << F(A, B);

    return 0;
}

Java

// Java implementation of the above approach
class GFG
{
    // Function to return 2^(2^A) % B
    static long F(long A, long B)
    {

        // Base case, 2^(2^1) % B = 4 % B
        if (A == 1)
            return (4 % B);
        else
        {
            long temp = F(A - 1, B);
            return (temp * temp) % B;
        }
    }

    // Driver code
    public static void main(String args[])
    {
        long A = 25, B = 50;

        // Print 2^(2^A) % B
        System.out.println(F(A, B));
    }
}

// This code is contributed by Ryuga

Python3

# Python3 implementation of the approach

# Function to return 2^(2^A) % B
def F(A, B):

    # Base case, 2^(2^1) % B = 4 % B
    if (A == 1):
        return (4 % B);
    else:
        temp = F(A - 1, B);
        return (temp * temp) % B;

# Driver code
A = 25;
B = 50;

# Print 2^(2^A) % B
print(F(A, B));

# This code is contributed by mits

C#

// C# implementation of the approach
class GFG
{

// Function to return 2^(2^A) % B
static long F(long A, long B)
{

    // Base case, 2^(2^1) % B = 4 % B
    if (A == 1)
        return (4 % B);
    else
    {
        long temp = F(A - 1, B);
        return (temp * temp) % B;
    }
}

// Driver code
static void Main()
{
    long A = 25, B = 50;

    // Print 2^(2^A) % B
    System.Console.WriteLine(F(A, B));
}
}

// This code is contributed by mits

PHP

<?php
// PHP implementation of the approach

// Function to return 2^(2^A) % B
function F($A, $B)
{

    // $Base case, 2^(2^1) % B = 4 % B
    if ($A == 1)
        return (4 % $B);
    else
    {
        $temp =  F($A - 1, $B);
        return ($temp * $temp) % $B;
    }
}

// Driver code
$A = 25;
$B = 50;

 // Print 2^(2^$A) % $B
echo F($A, $B);

// This code contributed by Rajput-Ji

Javascript

<script>

// Javascript implementation of the approach

// Function to return 2^(2^A) % B
function F(A, B)
{

    // Base case, 2^(2^1) % B = 4 % B
    if (A == 1)
        return (4 % B);
    else
    {
        var temp =  F(A - 1, B);
        return (temp * temp) % B;
    }
}

// Driver code
var A = 25, B = 50;

// Print 2^(2^A) % B
document.write( F(A, B));

// This code is contributed by noob2000

</script>

Output:

46

Time Complexity: O(A)
Auxiliary Space: O(A), due to recursive call stack