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Product of 2 numbers using recursion | Set 2
• Difficulty Level : Easy
• Last Updated : 13 May, 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

```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 ``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

 `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

 ``
Output:
```5 * 3 = 15
5 * (-3) = -15
(-5) * 3 = -15
(-5) * (-3) = 15```

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