Fast Exponention using Bit Manipulation

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++

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// C++ Program to implement
// the above appraoch
#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;
}

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Java

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// Java program to implement 
// the above appraoch 
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

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Python3

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# 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_

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C#

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// C# program to implement 
// the above appraoch 
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

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Output: 

243

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

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