Multiply a given Integer with 3.5

Given an integer x, write a function that multiplies x with 3.5 and returns the integer result. You are not allowed to use %, /, *.

Examples :
Input: 2
Output: 7

Input: 5
Output: 17 (Ignore the digits after decimal point)

Solution:
1. We can get x*3.5 by adding 2*x, x and x/2. To calculate 2*x, left shift x by 1 and to calculate x/2, right shift x by 1.
Below is the implementation of the above approach:

C++

// C++ program to multiply
// a number with 3.5
#include <iostream>
using namespace std;

int multiplyWith3Point5(int x)
{
    return (x<<1) + x + (x>>1);
} 

/* Driver program to test above functions*/
int main()
{
    int x = 4; 
    cout << " "<< multiplyWith3Point5(x);
    getchar();
    return 0;
}


// This code is contributed by shivanisinghss2110.

// C program to multiply
// a number with 3.5
#include <stdio.h>

int multiplyWith3Point5(int x)
{
    return (x<<1) + x + (x>>1);
} 

/* Driver program to test above functions*/
int main()
{
    int x = 4; 
    printf("%d", multiplyWith3Point5(x));
    getchar();
    return 0;
}

Java

// Java Program to multiply
// a number with 3.5

class GFG {
        
    static int multiplyWith3Point5(int x)
    {
        return (x<<1) + x + (x>>1);
    } 
    
    /* Driver program to test above functions*/
    public static void main(String[] args)
    {
        int x = 2; 
        System.out.println(multiplyWith3Point5(x));
    }
}

// This code is contributed by prerna saini.

Python3

# Python 3 program to multiply
# a number with 3.5

def multiplyWith3Point5(x):

    return (x<<1) + x + (x>>1)
 

# Driver program to 
# test above functions
x = 4 
print(multiplyWith3Point5(x))

# This code is contributed by
# Smitha Dinesh Semwal

// C# Program to multiply
// a number with 3.5
using System;

class GFG
{
        
    static int multiplyWith3Point5(int x)
    {
        return (x<<1) + x + (x>>1);
    } 
    
    /* Driver program to test above functions*/
    public static void Main()
    {
        int x = 2; 
        Console.Write(multiplyWith3Point5(x));
    }
    
}

// This code is contributed by Sam007

JavaScript

<script>
// javascript Program to multiply
// a number with 3.5  

function multiplyWith3Point5(x)
{
    return (x<<1) + x + (x>>1);
} 

/* Driver program to test above functions*/
var x = 4; 
document.write(multiplyWith3Point5(x));

// This code is contributed by Amit Katiyar 
</script>

PHP

<?php
// PHP program to multiply
// a number with 3.5


function multiplyWith3Point5( $x)
{
    return ($x << 1) + $x + ($x >> 1);
} 

// Driver Code
$x = 4; 
echo multiplyWith3Point5($x);
    
// This code is contributed by vt_m.
?>

Output

Time Complexity : O(1)

Space Complexity : O(1)

2. Another way of doing this could be (8*x - x)/2 (See below code). Thanks to Ajaym for suggesting this.

C++

// C++ program approach

#include <iostream>
using namespace std;

int multiplyWith3Point5(int x)
{
  return ((x<<3) - x)>>1;
}    

// This code is contributed by shivanisinghss2110

#include <stdio.h>
int multiplyWith3Point5(int x)
{
  return ((x<<3) - x)>>1;
}

Java

// Java program for above approach
import java.io.*;

class GFG 
{

  // Function 
  static int multiplyWith3Point5(int x)
  {
    return ((x<<3) - x)>>1;
  }
} 

// This code is contributed by shivanisinghss2110

Python

# Python program for above approach
# Function 
def multiplyWith3Point5(x):
  
    return ((x<<3) - x)>>1
  
# This code is contributed by shivanisinghss2110

// C# program for above approach
using System;

class GFG{

// Function to multiple number
// with 3.5
static int multiplyWith3Point5(int x)
 {
    return ((x<<3) - x)>>1;
  }
} 

// This code is contributed by shivanisinghss2110.

JavaScript

// JavaScript program for above approach
// Function 
function multiplyWith3Point5(x)
  {
    return ((x<<3) - x)>>1;
  }

// This code is contributed by shivanisinghss2110

Time Complexity : O(1)

Space Complexity : O(1)

Another Approach:

Another way of doing this could be by doing a binary multiplication by 7 then divide by 2 using only <<, ^, &, and >>.

But here we have to mention that only positive numbers can be passed to this method.

Below is the implementation of the above approach:

C++

// C++ program for above approach
#include <iostream>
using namespace std;

// Function to multiple number
// with 3.5
int multiplyWith3Point5(int x)
{
    int r = 0;

    // The 3.5 is 7/2, so multiply  by 7 (x * 7) then divide
    // the result by 2  (result/2) x * 7 -> 7 is 0111 so by
    // doing multiply by 7 it means we do 2 shifting for
    // the number but since we doing multiply we need to
    // take care of carry one.
    int x1Shift = x << 1;
    int x2Shifts = x << 2;

    r = (x ^ x1Shift) ^ x2Shifts;
    int c = (x & x1Shift) | (x & x2Shifts) | (x1Shift & x2Shifts);
    while (c > 0) {
        c <<= 1;
        int t = r;
        r ^= c;
        c &= t;
    }

    // Then divide by 2
    // r / 2
    r = r >> 1;
    return r;
}

// Driver Code
int main()
{
    cout << (multiplyWith3Point5(5));
    return 0;
}

// This code is contributed by Sania Kumari Gupta (kriSania804)

// C program for above approach
#include <stdio.h>

// Function to multiple number
// with 3.5
int multiplyWith3Point5(int x)
{
    int r = 0;

    // The 3.5 is 7/2, so multiply  by 7 (x * 7) then divide
    // the result by 2  (result/2) x * 7 -> 7 is 0111 so by
    // doing multiply by 7 it means we do 2 shifting for
    // the number but since we doing multiply we need to
    // take care of carry one.
    int x1Shift = x << 1;
    int x2Shifts = x << 2;

    r = (x ^ x1Shift) ^ x2Shifts;
    int c = (x & x1Shift) | (x & x2Shifts) | (x1Shift & x2Shifts);
    while (c > 0) {
        c <<= 1;
        int t = r;
        r ^= c;
        c &= t;
    }

    // Then divide by 2
    // r / 2
    r = r >> 1;
    return r;
}

// Driver Code
int main()
{
    printf("%d",(multiplyWith3Point5(5)));
    return 0;
}

// This code is contributed by Sania Kumari Gupta (kriSania804)

Java

// Java program for above approach
import java.io.*;

class GFG 
{
   
    // Function to multiple number
    // with 3.5
    static int multiplyWith3Point5(int x)
    {
        int r = 0;

        // The 3.5 is 7/2, so multiply 
        // by 7 (x * 7) then
        // divide the result by 2 
        // (result/2) x * 7 -> 7 is
        // 0111 so by doing multiply 
        // by 7 it means we do 2
        // shifting for the number 
        // but since we doing
        // multiply we need to take 
        // care of carry one.
        int x1Shift = x << 1;
        int x2Shifts = x << 2;

        r = (x ^ x1Shift) ^ x2Shifts;
        int c = (x & x1Shift) | (x & x2Shifts)
                | (x1Shift & x2Shifts);
        while (c > 0) {
            c <<= 1;
            int t = r;
            r ^= c;
            c &= t;
        }

        // Then divide by 2 
        // r / 2
        r = r >> 1;
        return r;
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        System.out.println(multiplyWith3Point5(5));
    }
}

Python

# Python3 program for the above approach

# Function to multiple number
# with 3.5
def multiplyWith3Point5(x):   
    r = 0

    # The 3.5 is 7/2, so multiply
    # by 7 (x * 7) then
    # divide the result by 2
    # (result/2) x * 7 -> 7 is
    # 0111 so by doing multiply
    # by 7 it means we do 2
    # shifting for the number
    # but since we doing
    # multiply we need to take
    # care of carry one.    
    x1Shift = x << 1
    x2Shifts = x << 2

    r = (x ^ x1Shift) ^ x2Shifts
    c = (x & x1Shift) | (x & x2Shifts) | (x1Shift & x2Shifts)
    while (c > 0):
        c <<= 1
        t = r
        r ^= c
        c &= t

    # Then divide by 2
    # r / 2
    r = r >> 1
    return r

 # Driver Code
if __name__ == '__main__':
    print(multiplyWith3Point5(5))

# This code is contributed by nirajgusain5

// C# program for above approach
using System;

class GFG{

// Function to multiple number
// with 3.5
static int multiplyWith3Point5(int x)
{
    int r = 0;

    // The 3.5 is 7/2, so multiply
    // by 7 (x * 7) then
    // divide the result by 2
    // (result/2) x * 7 -> 7 is
    // 0111 so by doing multiply
    // by 7 it means we do 2
    // shifting for the number
    // but since we doing
    // multiply we need to take
    // care of carry one.
    int x1Shift = x << 1;
    int x2Shifts = x << 2;

    r = (x ^ x1Shift) ^ x2Shifts;
    int c = (x & x1Shift) | (x & x2Shifts) | 
      (x1Shift & x2Shifts);
      
    while (c > 0) 
    {
        c <<= 1;
        int t = r;
        r ^= c;
        c &= t;
    }

    // Then divide by 2
    // r / 2
    r = r >> 1;
    return r;
}

// Driver Code
public static void Main(string[] args)
{
    Console.WriteLine(multiplyWith3Point5(5));
}
}

// This code is contributed by ukasp

JavaScript

<script>

// Javascript program for the above approach

// Function to multiple number
// with 3.5
function multiplyWith3Polet5(x)
{
    let r = 0;

    // The 3.5 is 7/2, so multiply 
    // by 7 (x * 7) then
    // divide the result by 2 
    // (result/2) x * 7 -> 7 is
    // 0111 so by doing multiply 
    // by 7 it means we do 2
    // shifting for the number 
    // but since we doing
    // multiply we need to take 
    // care of carry one.
    let x1Shift = x << 1;
    let x2Shifts = x << 2;

    r = (x ^ x1Shift) ^ x2Shifts;
    let c = (x & x1Shift) | (x & x2Shifts) | 
      (x1Shift & x2Shifts);
      
    while (c > 0)
    {
        c <<= 1;
        let t = r;
        r ^= c;
        c &= t;
    }

    // Then divide by 2 
    // r / 2
    r = r >> 1;
    return r;
}
    
// Driver code
document.write(multiplyWith3Polet5(5));    

// This code is contributed by avijitmondal1998

</script>

Output

Time Complexity : O(log n)

Space Complexity : O(1)

Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem

Multiply a given Integer with 3.5

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