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Product of 2 numbers using recursion | Set 2

  • Difficulty Level : Easy
  • Last Updated : 18 Aug, 2021

Given two numbers N and M. The task is to find the product of the 2 numbers using recursion.
Note: The numbers can be both positive or negative.

Examples

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Input : N = 5 ,  M = 3
Output : 15

Input : N = 5  ,  M = -3
Output : -15

Input : N = -5  ,  M = 3
Output : -15

Input : N = -5  ,  M = -3
Output:15

A recursive solution to the above problem for only positive numbers is already discussed in the previous article. In this post, a recursive solution for finding the product for both positive and negative numbers is discussed. 



Below is the step by step approach:  

  1. Check if one or both of the numbers are negative.
  2. If the number passed in the second parameter is negative swap the parameters and call the function again.
  3. If both of the parameters are negative call the function again and pass the absolute values of the numbers as parameters.
  4. If n>m call the function with swapped parameters for less execution time of the function.
  5. As long as m is not 0 keep on calling the function with subcase n, m-1 and return n + multrecur(n, m-1).

Below is the implementation of the above approach:  

C++




// C++ program to find product of two numbers
// using recursion
#include <iostream>
using namespace std;
 
// Recursive function to calculate the product
// of 2 integers
int multrecur(int n, int m)
{
    // case 1 : n<0 and m>0
    // swap the position of n and m to keep second
    // parameter positive
    if (n > 0 && m < 0) {
        return multrecur(m, n);
    }
    // case 2 : both n and m are less than 0
    // return the product of their absolute values
    else if (n < 0 && m < 0) {
        return multrecur((-1 * n), (-1 * m));
    }
     
    // if n>m , swap n and m so that recursion
    // takes less time
    if (n > m) {
        return multrecur(m, n);
    }
     
    // as long as m is not 0 recursively call multrecur for 
    // n and m-1 return sum of n and the product of n times m-1
    else if (m != 0) {
        return n + multrecur(n, m - 1);
    }
     
    // m=0 then return 0
    else {
        return 0;
    }
}
// Driver code
int main()
{
    cout << "5 * 3 = " << multrecur(5, 3) << endl;
    cout << "5 * (-3) = " << multrecur(5, -3) << endl;
    cout << "(-5) * 3 = " << multrecur(-5, 3) << endl;
    cout << "(-5) * (-3) = " << multrecur(-5, -3) << endl;
     
    return 0;
}

Java




//Java program to find product of two numbers
//using recursion
public class GFG {
 
    //Recursive function to calculate the product
    //of 2 integers
    static int multrecur(int n, int m)
    {
    // case 1 : n<0 and m>0
    // swap the position of n and m to keep second
    // parameter positive
    if (n > 0 && m < 0) {
        return multrecur(m, n);
    }
    // case 2 : both n and m are less than 0
    // return the product of their absolute values
    else if (n < 0 && m < 0) {
        return multrecur((-1 * n), (-1 * m));
    }
     
    // if n>m , swap n and m so that recursion
    // takes less time
    if (n > m) {
        return multrecur(m, n);
    }
     
    // as long as m is not 0 recursively call multrecur for 
    // n and m-1 return sum of n and the product of n times m-1
    else if (m != 0) {
        return n + multrecur(n, m - 1);
    }
     
    // m=0 then return 0
    else {
        return 0;
    }
    }
 
    //Driver code
    public static void main(String[] args) {
         
        System.out.println("5 * 3 = " + multrecur(5, 3));
        System.out.println("5 * (-3) = " + multrecur(5, -3));
        System.out.println("(-5) * 3 = " + multrecur(-5, 3));
        System.out.println("(-5) * (-3) = " +multrecur(-5, -3));
    }
}

Python3




# Python 3 program to find product of two numbers
# using recursion
 
# Recursive function to calculate the product
# of 2 integers
def multrecur(n, m) :
 
    # case 1 : n<0 and m>0
    # swap the position of n and m to keep second
    # parameter positive
    if n > 0 and m < 0 :
        return multrecur(m,n)
 
    # case 2 : both n and m are less than 0
    # return the product of their absolute values
    elif n < 0 and m < 0 :
        return multrecur((-1 * n),(-1 * m))
 
    # if n>m , swap n and m so that recursion
    # takes less time
    if n > m :
        return multrecur(m, n)
 
    # as long as m is not 0 recursively call multrecur for 
    # n and m-1 return sum of n and the product of n times m-1
    elif m != 0 :
        return n + multrecur(n, m-1)
 
    # m=0 then return 0
    else :
        return 0
 
 
# Driver Code
if __name__ == "__main__" :
 
    print("5 * 3 =",multrecur(5, 3))
    print("5 * (-3) =",multrecur(5, -3))
    print("(-5) * 3 =",multrecur(-5, 3))
    print("(-5) * (-3) =",multrecur(-5, -3))
 
 
# This code is contributed by ANKITRAI1

C#




// C# program to find product of
// two numbers using recursion
using System;
class GFG
{
 
// Recursive function to calculate
// the product of 2 integers
static int multrecur(int n, int m)
{
// case 1 : n<0 and m>0
// swap the position of n and m
// to keep second parameter positive
if (n > 0 && m < 0)
{
    return multrecur(m, n);
}
 
// case 2 : both n and m are less than 0
// return the product of their absolute values
else if (n < 0 && m < 0)
{
    return multrecur((-1 * n), (-1 * m));
}
 
// if n>m , swap n and m so that
// recursion takes less time
if (n > m)
{
    return multrecur(m, n);
}
 
// as long as m is not 0 recursively
// call multrecur for n and m-1 return
// sum of n and the product of n times m-1
else if (m != 0)
{
    return n + multrecur(n, m - 1);
}
 
// m=0 then return 0
else
{
    return 0;
}
}
 
// Driver code
public static void Main()
{
    Console.WriteLine("5 * 3 = " +
                       multrecur(5, 3));
    Console.WriteLine("5 * (-3) = " +
                       multrecur(5, -3));
    Console.WriteLine("(-5) * 3 = " +
                      multrecur(-5, 3));
    Console.WriteLine("(-5) * (-3) = " +
                      multrecur(-5, -3));
}
}
 
// This code is contributed by anuj_67

PHP




<?php
// PHP program to find product of
// two numbers using recursion
 
// Recursive function to calculate
// the product of 2 integers
function multrecur($n, $m)
{
    // case 1 : n<0 and m>0
    // swap the position of n and m to keep second
    // parameter positive
    if ($n > 0 && $m < 0)
    {
        return multrecur($m, $n);
    }
     
    // case 2 : both n and m are less than 0
    // return the product of their absolute values
    else if ($n < 0 && $m < 0)
    {
        return multrecur((-1 * $n),
                        (-1 * $m));
    }
     
    // if n>m , swap n and m so that
    // recursion takes less time
    if ($n > $m)
    {
        return multrecur($m, $n);
    }
     
    // as long as m is not 0 recursively call multrecur for 
    // n and m-1 return sum of n and the product of n times m-1
    else if ($m != 0)
    {
        return $n + multrecur($n, $m - 1);
    }
     
    // m=0 then return 0
    else
    {
        return 0;
    }
}
 
// Driver code
echo "5 * 3 = " . multrecur(5, 3) . "\n";
echo "5 * (-3) = " . multrecur(5, -3) . "\n";
echo "(-5) * 3 = " . multrecur(-5, 3) . "\n";
echo "(-5) * (-3) = " . multrecur(-5, -3) . "\n";
     
// This code is contributed by mits
?>

Javascript




<script>
 
// Javascript program to find product
// of two numbers using recursion
 
// Recursive function to calculate the
// product of 2 integers
function multrecur(n, m)
{
     
    // case 1 : n<0 and m>0
    // Swap the position of n and m to
    // keep second parameter positive
    if (n > 0 && m < 0)
    {
        return multrecur(m, n);
    }
     
    // case 2 : both n and m are less than 0
    // return the product of their absolute values
    else if (n < 0 && m < 0)
    {
        return multrecur((-1 * n), (-1 * m));
    }
     
    // If n>m , swap n and m so that recursion
    // takes less time
    if (n > m)
    {
        return multrecur(m, n);
    }
     
    // As long as m is not 0 recursively
    // call multrecur for n and m-1
    // return sum of n and the product
    // of n times m-1
    else if (m != 0)
    {
        return n + multrecur(n, m - 1);
    }
     
    // m=0 then return 0
    else
    {
        return 0;
    }
}
 
// Driver code
document.write("5 * 3 = " + multrecur(5, 3) + "<br>");
document.write("5 * (-3) = " + multrecur(5, -3) + "<br>");
document.write("(-5) * 3 = " + multrecur(-5, 3) + "<br>");
document.write("(-5) * (-3) = " + multrecur(-5, -3) + "<br>");
 
// This code is contributed by rutvik_56
 
</script>
Output: 
5 * 3 = 15
5 * (-3) = -15
(-5) * 3 = -15
(-5) * (-3) = 15

 

Time Complexity: O(max(N, M)) 
Auxiliary Space: O(max(N, M)) 




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