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Fast Exponention using Bit Manipulation

  • Difficulty Level : Medium
  • Last Updated : 17 Jul, 2021

Given two integers A and N, the task is to calculate A raised to power N (i.e. AN).

Examples: 

Input: A = 3, N = 5 
Output: 243 
Explanation: 
3 raised to power 5 = (3*3*3*3*3) = 243

Input: A = 21, N = 4
Output: 194481 
Explanation: 
21 raised to power 4 = (21*21*21*21) = 194481 
 

Naive Approach: 
The simplest approach to solve this problem is to repetitively multiply A, N times and print the product. 

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

Efficient Approach: 
To optimize the above approach, the idea is to use Bit Manipulation. Convert the integer N to its binary form and follow the steps below:  

  • Initialize ans to store the final answer of AN.
  • Traverse until N > 0 and in each iteration, perform Right Shift operation on it.
  • Also, in each iteration, multiply A with itself and update it.
  • If current LSB is set, then multiply current value of A to ans.
  • Finally, after completing the above steps, print ans.

Below is the implementation of the above approach:

C++




// C++ Program to implement
// the above approach
#include <iostream>
using namespace std;
 
// Function to return a^n
int powerOptimised(int a, int n)
{
 
    // Stores final answer
    int ans = 1;
 
    while (n > 0) {
 
        int last_bit = (n & 1);
 
        // Check if current LSB
        // is set
        if (last_bit) {
            ans = ans * a;
        }
 
        a = a * a;
 
        // Right shift
        n = n >> 1;
    }
 
    return ans;
}
 
// Driver Code
int main()
{
    int a = 3, n = 5;
 
    cout << powerOptimised(a, n);
 
    return 0;
}

Java




// Java program to implement
// the above approach
class GFG{
 
// Function to return a^n
static int powerOptimised(int a, int n)
{
 
    // Stores final answer
    int ans = 1;
 
    while (n > 0)
    {
        int last_bit = (n & 1);
 
        // Check if current LSB
        // is set
        if (last_bit > 0)
        {
            ans = ans * a;
        }
         
        a = a * a;
 
        // Right shift
        n = n >> 1;
    }
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    int a = 3, n = 5;
 
    System.out.print(powerOptimised(a, n));
}
}
 
// This code is contributed by PrinciRaj1992

Python3




# Python3 program to implement
# the above approach
 
# Function to return a^n
def powerOptimised(a, n):
     
    # Stores final answer
    ans = 1
     
    while (n > 0):
        last_bit = (n & 1)
         
        # Check if current LSB
        # is set
        if (last_bit):
            ans = ans * a
        a = a * a
         
        # Right shift
        n = n >> 1
         
    return ans
 
# Driver code
if __name__ == '__main__':
     
    a = 3
    n = 5
     
    print(powerOptimised(a,n))
 
# This code is contributed by virusbuddah_

C#




// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to return a^n
static int powerOptimised(int a, int n)
{
     
    // Stores readonly answer
    int ans = 1;
 
    while (n > 0)
    {
        int last_bit = (n & 1);
 
        // Check if current LSB
        // is set
        if (last_bit > 0)
        {
            ans = ans * a;
        }
        a = a * a;
 
        // Right shift
        n = n >> 1;
    }
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
    int a = 3, n = 5;
 
    Console.Write(powerOptimised(a, n));
}
}
 
// This code is contributed by Princi Singh

Javascript




<script>
 
// Javascript program to implement
// the above approach
 
// Function to return a^n
function powerOptimised(a, n)
{
   
    // Stores final answer
    let ans = 1;
   
    while (n > 0)
    {
        let last_bit = (n & 1);
   
        // Check if current LSB
        // is set
        if (last_bit > 0)
        {
            ans = ans * a;
        }
           
        a = a * a;
   
        // Right shift
        n = n >> 1;
    }
    return ans;
 
    // Driver Code
         
    let a = 3, n = 5;
   
    document.write(powerOptimised(a, n));
       
</script>
Output: 
243

 

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


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