Calculate square of a number without using *, / and pow()

Given an integer n, calculate square of a number without using *, / and pow().

Examples :

Input: n = 5
Output: 25

Input: 7
Output: 49

Input: n = 12
Output: 144

A Simple Solution is to repeatedly add n to result. Below is the implementation of this idea.

C++

// Simple solution to calculate square without 
// using * and pow() 
#include<iostream> 
using namespace std; 
  
int square(int n) 
{ 
   // handle negative input 
   if (n<0) n = -n; 
  
   // Initialize result 
   int res = n; 
  
   // Add n to res n-1 times 
   for (int i=1; i<n; i++) 
       res += n; 
  
   return res; 
} 
  
// driver program 
int main() 
{ 
    for (int n = 1; n<=5; n++) 
        cout << "n = " << n << ", n^2 = "
             << square(n) << endl; 
    return 0; 
} 

Java

// Java Simple solution to calculate  
// square without using * and pow() 
import java.io.*; 
  
class GFG { 
      
    public static int square(int n) { 
  
        // handle negative input 
        if (n < 0) 
            n = -n; 
      
        // Initialize result 
        int res = n; 
      
        // Add n to res n-1 times 
        for (int i = 1; i < n; i++) 
            res += n; 
      
        return res; 
    } 
      
    // Driver code 
    public static void main(String[] args) { 
  
        for (int n = 1; n <= 5; n++) 
            System.out.println("n = " + n +  
                   ", n^2 = " + square(n)); 
    } 
} 
  
// This code is contributed by sunnysingh 

Python3

# Simple solution to 
# calculate square without 
# using * and pow() 
  
def square(n): 
  
    # handle negative input 
    if (n < 0): 
        n = -n 
      
    # Initialize result 
    res = n 
      
    # Add n to res n-1 times 
    for i in range(1, n): 
        res += n 
      
    return res 
  
  
# Driver Code 
for n in range(1, 6): 
    print("n =", n , end=", ") 
    print("n^2 =", square(n)) 
  
# This code is contributed by 
# Smitha Dinesh Semwal 

C#

// C# Simple solution to calculate  
// square without using * and pow() 
using System; 
  
class GFG 
{ 
    public static int square(int n) { 
  
        // handle negative input 
        if (n < 0) 
            n = -n; 
      
        // Initialize result 
        int res = n; 
      
        // Add n to res n-1 times 
        for (int i = 1; i < n; i++) 
            res += n; 
      
        return res; 
    } 
      
    // Driver code 
    public static void Main() { 
  
        for (int n = 1; n <= 5; n++) 
        Console.WriteLine("n = " + n +  
                ", n^2 = " + square(n)); 
    } 
} 
  
// This code is contributed by Sam007 

PHP

<?php 
// PHP implementation to  
// calculate square  
// without using * and pow() 
  
function square($n) 
{ 
  
// handle negative input 
    if ($n < 0) $n = -$n; 
  
// Initialize result 
        $res = $n; 
  
// Add n to res n-1 times 
for ($i = 1; $i < $n; $i++) 
    $res += $n; 
  
return $res; 
} 
  
// Driver Code 
for ($n = 1; $n<=5; $n++) 
    echo "n = ", $n, ", ", "n^2 = ",  
                  square($n), "\n "; 
      
// This code is contributed by Ajit  
?> 

Output :

n = 1, n^2 = 1
n = 2, n^2 = 4
n = 3, n^2 = 9
n = 4, n^2 = 16
n = 5, n^2 = 25

Time complexity of above solution is O(n). We can do it in O(Logn) time using bitwise operators. The idea is based on the following fact.

  square(n) = 0 if n == 0
  if n is even 
     square(n) = 4*square(n/2) 
  if n is odd
     square(n) = 4*square(floor(n/2)) + 4*floor(n/2) + 1 

Examples
  square(6) = 4*square(3)
  square(3) = 4*(square(1)) + 4*1 + 1 = 9
  square(7) = 4*square(3) + 4*3 + 1 = 4*9 + 4*3 + 1 = 49

How does this work?

If n is even, it can be written as
  n = 2*x 
  n² = (2*x)² = 4*x²
If n is odd, it can be written as 
  n = 2*x + 1
  n² = (2*x + 1)² = 4*x² + 4*x + 1

floor(n/2) can be calculated using bitwise right shift operator. 2*x and 4*x can be calculated

Below is the implementation based on above idea.

C++

// Square of a number using bitwise operators 
#include<iostream> 
using namespace std; 
  
int square(int n) 
{ 
    // Base case 
    if (n==0) return 0; 
  
    // Handle negative number 
    if (n < 0) n = -n; 
  
    // Get floor(n/2) using right shift 
    int x = n>>1; 
  
    // If n is odd 
    if (n&1) 
        return ((square(x)<<2) + (x<<2) + 1); 
    else // If n is even 
        return (square(x)<<2); 
} 
  
// Driver Code 
int main() 
{ 
    for (int n = 1; n<=5; n++) 
        cout << "n = " << n << ", n^2 = " << square(n) << endl; 
    return 0; 
} 

Java

// Square of a number using  
// bitwise operators 
class GFG  
{ 
    static int square(int n) 
    { 
          
        // Base case 
        if (n == 0)  
            return 0; 
      
        // Handle negative number 
        if (n < 0)  
            n = -n; 
      
        // Get floor(n/2) using 
        // right shift 
        int x = n >> 1; 
      
        // If n is odd 
        ; 
        if (n % 2 != 0) 
            return ((square(x) << 2)  
                    + (x << 2) + 1); 
        else // If n is even 
            return (square(x) << 2); 
    } 
  
      
      
    public static void main(String args[])  
    { 
        for (int n = 1; n <= 5; n++) 
        System.out.println("n = " + n + 
                            " n^2 = " +  
                            square(n));  
    } 
} 
  
// This code is contributed by Sam007 

Python3

# Square of a number using bitwise 
# operators 
  
def square(n): 
      
    # Base case 
    if (n == 0): 
        return 0
  
    # Handle negative number 
    if (n < 0): 
        n = -n 
  
    # Get floor(n/2) using 
    # right shift 
    x = n>>1
  
    # If n is odd 
    if (n & 1): 
        return ((square(x) << 2)  
                 + (x << 2) + 1); 
                   
    # If n is even 
    else: 
        return (square(x) << 2); 
  
# Driver Code 
for n in range (1, 6): 
    print("n = " , n ," n^2 = ", 
                       square(n)) 
# This code is contributed by Sam007 

C#

// Square of a number using bitwise 
// operators 
using System; 
  
class GFG { 
      
    static int square(int n) 
    { 
          
        // Base case 
        if (n == 0)  
            return 0; 
      
        // Handle negative number 
        if (n < 0)  
            n = -n; 
      
        // Get floor(n/2) using 
        // right shift 
        int x = n>>1; 
      
        // If n is odd 
        ; 
        if (n % 2 != 0) 
            return ((square(x) << 2)  
                     + (x << 2) + 1); 
        else // If n is even 
            return (square(x)<<2); 
    } 
  
    static void Main() 
    { 
        for (int n = 1; n <= 5; n++) 
        Console.WriteLine( "n = " + n 
            + " n^2 = " + square(n));  
    } 
} 
  
// This code is contributed by Sam0007. 

PHP

<?php 
// Square of a number using 
// bitwise operators 
  
function square($n) 
{ 
      
    // Base case 
    if ($n==0) return 0; 
  
    // Handle negative number 
    if ($n < 0) $n = -$n; 
  
    // Get floor(n/2) 
    // using right shift 
    $x = $n >> 1; 
  
    // If n is odd 
    if ($n & 1) 
        return ((square($x) << 2) +  
                    ($x << 2) + 1); 
    else // If n is even 
        return (square($x) << 2); 
} 
  
    // Driver Code 
    for ($n = 1; $n <= 5; $n++) 
        echo "n = ", $n, ", n^2 = ", square($n),"\n"; 
      
// This code is contributed by ajit 
?> 

Output :

n = 1, n^2 = 1
n = 2, n^2 = 4
n = 3, n^2 = 9
n = 4, n^2 = 16
n = 5, n^2 = 25

Time complexity of the above solution is O(Logn).

This article is contributed by Ujjwal Jain. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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