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

Last Updated : 29 Mar, 2023

Given an integer n, calculate the 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 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 code 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

n=−n = -n=−n; // Initialize result res=res = res=n; // Add n to res n-1 times for ($i = 1; i<i < i<n; $i++) res+=res += res+=n; return $res; } // Driver Code for ($n = 1; n<=5;n<=5; n<=5;n++) echo "n = ", $n, ", ", "n^2 = ", square($n), "\n "; // This code is contributed by Ajit ?>

JavaScript

`

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: O(n)

Auxiliary Space: O(1)

Approach 2:

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

Examples square(6) = 4square(3) square(3) = 4(square(1)) + 41 + 1 = 9 square(7) = 4square(3) + 43 + 1 = 49 + 4*3 + 1 = 49

How does this work?

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

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

Below is the implementation based on the above idea.

C++ `

// Square of a number using bitwise operators #include <bits/stdc++.h> 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() { // Function calls 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);
}

// Driver code
public static void main(String args[])
{
    // Function calls
    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);
}

// Driver code
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

n=−n = -n=−n; // Get floor(n/2) // using right shift x=x = 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 <= 5; n<=5;n++) echo "n = ", n,",n2=",square(n, ", n^2 = ", square(n,",n2=",square(n),"\n"; // This code is contributed by ajit ?>

JavaScript

`

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: O(log n)
Auxiliary Space: O(log n) as well, as the number of function calls stored in the call stack will be logarithmic to the size of the input

Approach 3:

For a given number num we get square of it by multiplying number as num * num. Now write one of num in square num * num in terms of power of 2. Check below examples.

Eg: num = 10, square(num) = 10 * 10 = 10 * (8 + 2) = (10 * 8) + (10 * 2) num = 15, square(num) = 15 * 15 = 15 * (8 + 4 + 2 + 1) = (15 * 8) + (15 * 4) + (15 * 2) + (15 * 1)

Multiplication with power of 2's can be done by left shift bitwise operator.

Below is the implementation based on the above idea.

C++ `

// Simple solution to calculate square without // using * and pow() #include using namespace std;

int square(int num) { // handle negative input if (num < 0) num = -num;

// Initialize result
int result = 0, times = num;

while (times > 0) 
{
    int possibleShifts = 0, currTimes = 1;

    while ((currTimes << 1) <= times) 
    {
        currTimes = currTimes << 1;
        ++possibleShifts;
    }

    result = result + (num << possibleShifts);
    times = times - currTimes;
}

return result;

}

// Driver code int main() { // Function calls for (int n = 10; n <= 15; ++n) cout << "n = " << n << ", n^2 = " << square(n) << endl; return 0; }

// This code is contributed by sanjay235

Java

// Simple solution to calculate square // without using * and pow() import java.io.*;

class GFG{

public static int square(int num) {

// Handle negative input
if (num < 0)
    num = -num;

// Initialize result
int result = 0, times = num;

while (times > 0)
{
    int possibleShifts = 0, 
             currTimes = 1;

    while ((currTimes << 1) <= times)
    {
        currTimes = currTimes << 1;
        ++possibleShifts;
    }

    result = result + (num << possibleShifts);
    times = times - currTimes;
}
return result;

}

// Driver code public static void main(String[] args) { for(int n = 10; n <= 15; ++n) { System.out.println("n = " + n + ", n^2 = " + square(n)); } } }

// This code is contributed by RohitOberoi

Python3

Simple solution to calculate square without

using * and pow()

def square(num):

# Handle negative input
if (num < 0):
    num = -num

# Initialize result
result, times = 0, num

while (times > 0):
    possibleShifts, currTimes = 0, 1

    while ((currTimes << 1) <= times):
        currTimes = currTimes << 1
        possibleShifts += 1

    result = result + (num << possibleShifts)
    times = times - currTimes

return result

Driver Code

Function calls

for n in range(10, 16): print("n =", n, ", n^2 =", square(n))

This code is contributed by divyesh072019

C#

// Simple solution to calculate square // without using * and pow() using System; class GFG {

static int square(int num)
{
     
    // Handle negative input
    if (num < 0)
        num = -num;
 
    // Initialize result
    int result = 0, times = num;
 
    while (times > 0)
    {
        int possibleShifts = 0, 
                 currTimes = 1;
 
        while ((currTimes << 1) <= times)
        {
            currTimes = currTimes << 1;
            ++possibleShifts;
        }
 
        result = result + (num << possibleShifts);
        times = times - currTimes;
    }
    return result;
}

static void Main() { for(int n = 10; n <= 15; ++n) { Console.WriteLine("n = " + n + ", n^2 = " + square(n)); } } }

// This code is contributed by divyeshrabadiy07

JavaScript

`

Output

n = 10, n^2 = 100 n = 11, n^2 = 121 n = 12, n^2 = 144 n = 13, n^2 = 169 n = 14, n^2 = 196 n = 15, n^2 = 225

Time Complexity: O(logn)

Auxiliary Space: O(1)

Thanks to Sanjay for approach 3 solution.

C++ `

// Simple solution to calculate square without // using * and pow() #include using namespace std;

int square(int num) { // handle negative input if (num < 0) num = -num;

// Initialize power of 2 and result
int power = 0, result = 0;
int temp = num;

while (temp) {
    if (temp & 1) {
        // result=result+(num*(2^power))
        result += (num << power);
    }
    power++;

    // temp=temp/2
    temp = temp >> 1;
}

return result;

}

// Driver code int main() { // Function calls for (int n = 10; n <= 15; ++n) cout << "n = " << n << ", n^2 = " << square(n) << endl; return 0; }

// This code is contributed by Aditya Verma

Java

/*package whatever //do not write package name here */

import java.io.; import java.util.; // java program for Simple solution to calculate square without // using * and pow()

// This code is contributed by Aditya Verma public class Main {

public static int square(int num)
{
    // handle negative input
    if (num < 0)
        num = -num;

    // Initialize power of 2 and result
    int power = 0, result = 0;
    int temp = num;

    while (temp > 0) {
        if ((temp & 1) > 0) {
            // result=result+(num*(2^power))
            result += (num << power);
        }
        power++;

        // temp=temp/2
        temp = temp >> 1;
    }

    return result;
}

public static void main(String[] args) {
    // Function calls
    for (int n = 10; n <= 15; ++n)
        System.out.println("n = " + n + ", n^2 = " + square(n));
}

}

// The code is contributed by Nidhi goel.

Python3

def square(num): # handle negative input if num < 0: num = -num

# Initialize power of 2 and result
power, result = 0, 0
temp = num

while temp:
    if temp & 1:
        # result=result+(num*(2^power))
        result += (num << power)
    power += 1

    # temp=temp/2
    temp = temp >> 1

return result

Driver code

for n in range(10, 16): print(f"n = {n}, n^2 = {square(n)}")

` JavaScript ``

// Simple solution to calculate square without // using * and pow()

function square(num) {

// handle negative input
if (num < 0)
num = -num;

// Initialize power of 2 and result
let power = 0, result = 0;
let temp = num;

while (temp) {
    if (temp & 1) {
        // result=result+(num*(2^power))
        result += (num << power);
    }
    power++;

    // temp=temp/2
    temp = temp >> 1;
}

return result;

}

// Driver code

// Function calls for (let n = 10; n <= 15; ++n) console.log(n = <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo separator="true">,</mo><msup><mi>n</mi><mn>2</mn></msup><mo>=</mo></mrow><annotation encoding="application/x-tex">{n}, n^2 = </annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.0085em;vertical-align:-0.1944em;"></span><span class="mord"><span class="mord mathnormal">n</span></span><span class="mpunct">,</span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mord"><span class="mord mathnormal">n</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">=</span></span></span></span>{square(n)});

// This code is contributed by phasing17

`` C# `

// Simple solution to calculate square without // using * and pow() using System;

public class Program { public static int Square(int num) { // handle negative input if (num < 0) num = -num;

    // Initialize power of 2 and result
    int power = 0, result = 0;
    int temp = num;

    while (temp > 0)
    {
        if ((temp & 1) > 0)
        {
            // result=result+(num*(2^power))
            result += (num << power);
        }
        power++;

        // temp=temp/2
        temp = temp >> 1;
    }

    return result;
}

public static void Main()
{
    // Function calls
    for (int n = 10; n <= 15; ++n)
        Console.WriteLine("n = " + n + ", n^2 = " + Square(n));
}

}

// Contributed by adityasha4x71

`

Output

n = 10, n^2 = 100 n = 11, n^2 = 121 n = 12, n^2 = 144 n = 13, n^2 = 169 n = 14, n^2 = 196 n = 15, n^2 = 225

Time Complexity: O(logn)

Auxiliary Space: O(1)