Find value of y mod (2 raised to power x)

Given two positive integer x and y. we have to find the value of y mod 2^x. That is remainder when y is divided by 2^x. 

Examples:

Input : x = 3, y = 14
Output : 6
Explanation : 14 % 23 =  14 % 8 = 6.

Input : x = 4, y = 14
Output : 14
Explanation : 14 % 24 =  14 % 16 = 14.

To solve this question we can use pow() and modulo operator and can easily find the remainder. 
But there are some points we should care about:  

• For higher value of x such that 2^x is greater than long long int range, we can not obtain proper result.
• Whenever y < 2^x the remainder will always be y. So, in that case we can restrict ourselves to calculate value of 2^x as:

y < 2x
log y < x
// means if log y is less than x, then 
// we can print y as remainder.

• The maximum value of 2^x for which we can store 2^x in a variable is 2⁶³. This means for x > 63, y is always going to be smaller than 2^x and in that case remainder of y mod 2^x is y itself.

keeping in mind the above points we can approach this problem as :  

if (log y < x)
    return y;
else if (x  < 63)
    return y;
else 
    return (y % (pow(2, x)))

Note: As python is limit free we can directly use mod and pow() function 

C++

// C++ Program to find the
// value of y mod 2^x
#include <bits/stdc++.h>
using namespace std;

// function to calculate y mod 2^x
long long int yMod(long long int y,
                    long long int x)
{
    // case 1 when y < 2^x
    if (log2(y) < x)
        return y;

    // case 2 when 2^x is out of
    // range means again y < 2^x
    if (x > 63)
        return y;

    // if y > 2^x
    return (y % (1 << x));
}

// driver program
int main()
{
    long long int y = 12345;
    long long int x = 11;
    cout << yMod(y, x);
    return 0;
}

Java

// Java Program to find
// the value of y mod 2^x
import java.io.*;

class GFG
{
    // function to calculate
    // y mod 2^x
    static long yMod(long y,
                     long x)
    {
        // case 1 when y < 2^x
        if ((Math.log(y) /
             Math.log(2)) < x)
            return y;

        // case 2 when 2^x is
        // out of range means
        // again y < 2^x
        if (x > 63)
            return y;

        // if y > 2^x
        return (y % (1 << (int)x));
    }

    // Driver Code
    public static void main(String args[])
    {
        long y = 12345;
        long x = 11;
        System.out.print(yMod(y, x));
    }
}

// This code is contributed by
// Manish Shaw(manishshaw1)

Python3

# Program to find the value
# of y mod 2 ^ x function to
# return y mod 2 ^ x
def yMod(y, x) :
    return (y % pow(2, x))

# Driver code
y = 12345
x = 11
print(yMod(y, x))

C#

// C# Program to find the
// value of y mod 2^x
using System;

class GFG
{
    // function to calculate
    // y mod 2^x
    static long yMod(long y,
                     long x)
    {
        // case 1 when y < 2^x
        if (Math.Log(y, 2) < x)
            return y;

        // case 2 when 2^x is
        // out of range means
        // again y < 2^x
        if (x > 63)
            return y;

        // if y > 2^x
        return (y % (1 << (int)x));
    }

    // Driver Code
    static void Main()
    {
        long y = 12345;
        long x = 11;
        Console.Write(yMod(y, x));
    }
}

// This code is contributed by
// Manish Shaw(manishshaw1)

PHP

<?php
// PHP Program to find the
// value of y mod 2^x

// function to
// calculate y mod 2^x
function yMod($y, $x)
{
    // case 1 when y < 2^x
    if ((log($y) / log(2)) < $x)
        return $y;

    // case 2 when 2^x is
    // out of range means
    // again y < 2^x
    if ($x > 63)
        return $y;

    // if y > 2^x
    return ($y % (1 << $x));
}

// Driver Code
$y = 12345;
$x = 11;
echo (yMod($y, $x));

// This code is contributed by
// Manish Shaw(manishshaw1)
?>

Javascript

<script>

// Javascript program to find the
// value of y mod 2^x

// Function to calculate y mod 2^x
function yMod(y, x)
{

    // Case 1 when y < 2^x
    if ((Math.log(y) / Math.log(2)) < x)
        return y;

    // Case 2 when 2^x is
    // out of range means
    // again y < 2^x
    if (x > 63)
        return y;

    // If y > 2^x
    return (y % (1 << x));
}

// Driver Code
let y = 12345;
let x = 11;

document.write(yMod(y, x));

// This code is contributed by _saurabh_jaiswal

</script>

Output

57

Time Complexity: O(x)
Auxiliary Space: O(1)

Approach:  Bitwise manipulation

This approach is called bitwise manipulation. Here are the steps:

1. Calculate 2^x using the left shift operator (<<). For example, 1 << 3 will give us 8.
2. Subtract 1 from the result of step 1 to get a binary number with x number of 1’s. For example, 2^3 – 1 = 7, which in binary is 111.
3. Perform a bitwise AND operation between y and the result of step 2. The result will be the remainder when y is divided by 2^x.

C++

#include <iostream>

int find_modulo(int x, int y)
{
    int result = y & ((1 << x) - 1);
    std::cout << "The remainder of " << y << " when divided by 2^" << x << " is " << result << "." << std::endl;
    return result;
}

int main()
{
    find_modulo(3, 14);
    find_modulo(4, 14);
    return 0;
}

Java

public class Main {
    public static int find_modulo(int x, int y) {
        int result = y & ((1 << x) - 1);
        System.out.println("The remainder of " + y + " when divided by 2^" + x + " is " + result + ".");
        return result;
    }

    public static void main(String[] args) {
        find_modulo(3, 14);
        find_modulo(4, 14);
    }
}

Python3

def find_modulo(x, y):
    result = y & ((1 << x) - 1)
    print(f"The remainder of {y} when divided by 2^{x} is {result}.")
    return result

find_modulo(3, 14)
find_modulo(4, 14)

C#

using System;

class Program {
    static int FindModulo(int x, int y) {
        int result = y & ((1 << x) - 1);
        Console.WriteLine($"The remainder of {y} when divided by 2^{x} is {result}.");
        return result;
    }
    // Drive code
    static void Main(string[] args) {
        FindModulo(3, 14);
        FindModulo(4, 14);
    }
}
// This code is contributed by Shiv1o43g

Javascript

// Javascript program to find remainder of y when divided by 2^x
function findModulo(x, y) {
let result = y & ((1 << x) - 1);
console.log(The remainder of ${y} when divided by 2^${x} is ${result}.);
return result;
}

// Driver code
findModulo(3, 14);
findModulo(4, 14);

Output

The remainder of 14 when divided by 2^3 is 6.
The remainder of 14 when divided by 2^4 is 14.

The time complexity:  O(1) 
The auxiliary space: O(1)

Compute modulus division by a power-of-2-number