Day 14 - Recursion Series Part B
Today’s Problems - Sum of digits and product of numbers
Question 1
Given a number, write a program to find the sum of it’s digits using recursion
Example
input: 12345
output: 15
input: 91827
output: 27
Question 2
Given 2 numbers, write a program to find their product using recursion
Example
input: 10,5
output: 50
input: -8,4
output: -32
Hint: Multiplication is repeated addition!
Part A - Sum of Digits
JavaScript Implementation
Solution
/**
* @author MadhavBahl
* @date 08/01/2018
*/
function sumDigits (num) {
if (num/10 < 1)
return num;
else
return (num % 10) + sumDigits (parseInt(num/10));
}
let num1 = 12345, num2 = 91827;
console.log(`Sum of digits of ${num1} is ${sumDigits(num1)}`);
console.log(`Sum of digits of ${num2} is ${sumDigits(num2)}`);
Java Implementation
Solution
/**
* @author MadhavBahl
* @date 08/01//2018
*/
import java.util.Scanner;
public class SumDigits {
public static int sum (int num) {
if (num/10 < 1)
return num;
else
return (num%10) + sum(num/10);
}
public static void main(String[] args) {
Scanner input = new Scanner (System.in);
System.out.println("/* ===== Sum of digits using recursion ===== */");
System.out.print("\nEnter a number: ");
int num = input.nextInt();
System.out.println("Sum of digits of " + num + " is: " + sum(num));
}
}
Python Implementation
Solution by @vishalshirke7
"""
@author : vishalshirke7
@date : 08/01/2019
"""
def sum_of_digits(n):
if n <= 0:
return n
else:
return (n % 10) + sum_of_digits(n // 10)
print(sum_of_digits(int(input())))
#### [Solution](/day14/Java/Sumrec.java)
/**
- @date 08/01/19
- @author SPREEHA DUTTA
/
import java.util.;
public class Sumrec {
public static int sum(int n)
{
if(n==0)
return 0;
else
return n%10+sum(n/10);
}
public static void main(String []args)
{
int n,s;
Scanner sc=new Scanner(System.in);
System.out.println(“Enter a number”);
n=sc.nextInt();
s=sum(n);
System.out.println(s);
}
} ```
C++ Implementation
C++ Solution by @profgrammer
/*
*@author: profgrammer
*@date: 08-01-2019
*/
#include <bits/stdc++.h>
using namespace std;
int sum_digits(int n){
if(n < 10) return n;
return n%10 + sum_digits(n/10);
}
int main() {
int n;
cin>>n;
cout<<num_digits(n)<<endl;
}
C++ Solution by @divyakhetan
/**
* @author:divyakhetan
* @date: 10/1/2019
*/
#include<bits/stdc++.h>
using namespace std;
int sum(int n){
if(n < 10) return n;
else return n % 10 + sum(n /10);
}
int main(){
int n;
cin >> n;
cout << "The sum of digits is " << sum(n);
return 0;
}
Ruby Implementation
Solution
=begin
@author: aaditkamat
@date: 08/01/2019
=end
def sum_of_digits(num, sum)
if num < 0
return -1 * sum_of_digits(abs(num), sum)
end
if num === 0
return sum
end
sum_of_digits(num / 10, sum + num % 10)
end
def main
print "Enter a number: "
num = gets.chomp.to_i
puts "Sum of digits of #{num} is: #{sum_of_digits(num, 0)}"
end
main
C Implementation
Solution
/*
* @author: ashwek
* @date: 8/1/2019
*/
#include <stdio.h>
int sum(int num) {
if( num <= 0 ){
return 0;
}
return (num%10) + sum(num/10);
}
void main(){
int num;
printf("Enter a number = ");
scanf("%d", &num);
printf("Sum of digits = %d\n", sum(num));
}
Part B - Product of numbers
JavaScript Implementation
Solution
/**
* @author MadhavBahl
* @date 08/01/2018
* METHOD - We keep thte second argument (num2) positive and add the first arguement num2(second arg) times
*/
function recursiveProd (num1, num2) {
// If num 2 becomes 1, return num1
if (num2 === 1)
return num1;
// If any of the numbers is zero, return 0
if (num1 === 0 || num2 === 0)
return 0;
// If both numbers are less than zero negative signs can be removed
if (num1 < 0 && num2 < 0)
return recursiveProd (-1*num1, -1*num2);
else if (num2 < 0)
return recursiveProd (num2, num1);
else
return num1 + recursiveProd(num1, num2-1);
}
let n1 = 5, n2 = 10;
console.log (`${n1} x ${n2} = ${recursiveProd(n1, n2)}`);
let n3 = -8, n4 = 4;
console.log (`${n3} x ${n4} = ${recursiveProd(n3, n4)}`);
let n5 = 2, n6 = -7;
console.log (`${n5} x ${n6} = ${recursiveProd(n5, n6)}`);
let n7 = -4, n8 = -7;
console.log (`${n7} x ${n8} = ${recursiveProd(n7, n8)}`);
Java Implementation
Solution
/**
* @author MadhavBahl
* @date 08/01//2018
*/
import java.util.Scanner;
public class Product {
public static int recursiveProd (int num1, int num2) {
if (num2 == 1)
return num1;
// If any of the numbers is zero, return 0
if (num1 == 0 || num2 == 0)
return 0;
// If both numbers are less than zero negative signs can be removed
if (num1 < 0 && num2 < 0)
return recursiveProd (-1*num1, -1*num2);
else if (num2 < 0)
return recursiveProd (num2, num1);
else
return num1 + recursiveProd(num1, num2-1);
}
public static void main(String[] args) {
Scanner input = new Scanner (System.in);
System.out.println("/* ===== Product of numbers using recursion ===== */");
// Take input
System.out.print("\nEnter first number: ");
int num1 = input.nextInt();
System.out.print("Enter second number: ");
int num2 = input.nextInt();
// Print the result
System.out.println("Product of numbers " + num1 + " and " + num2 + " is: " + recursiveProd(num1, num2));
}
}
Solution
/**
* @date 08/01/19
* @author SPREEHA DUTTA
*/
import java.util.*;
public class Recmultiply {
public static int product(int m,int n)
{
if (n==0)
return 0;
else
return m+product(m,n-1);
}
public static void main(String []args)
{
int m,n,p;
Scanner sc=new Scanner(System.in);
System.out.println("Enter two numbers ");
m=sc.nextInt();
n=sc.nextInt();
p=product(Math.abs(m),Math.abs(n));
if((m<0&&n>0)||(m>0&&n<0))
p=p*-1;
System.out.println(p);
}
}
C++ Implementation
C++ Solution by @profgrammer
/*
*@author: profgrammer
*@date: 08-01-2019
*/
#include <bits/stdc++.h>
using namespace std;
int mul(int a, int b){
if(b == 0) return 0;
// a*b = (a*b/2) + (a*b/2) + (b%2 == 0)? 0:a;
int ans = mul(a, b/2);
ans = ans + ans;
if(b % 2 == 1) ans += a;
return ans;
}
int main() {
int a,b;
cin>>a>>b;
cout<<mul(a,b)<<endl;
}
C++ Solution by @divyakhetan
/**
* @author:divyakhetan
* @date: 10/1/2019
*/
#include<bits/stdc++.h>
using namespace std;
int sum(int n){
if(n < 10) return n;
else return n % 10 + sum(n /10);
}
int main(){
int n;
cin >> n;
cout << "The sum of digits is " << sum(n);
return 0;
}
Ruby Implementation
Solution
=begin
@author: aaditkamat
@date: 08/01/2019
=end
def product_of_two_numbers(first, second, product)
if first < 0 and second < 0
return product_of_two_numbers(-first, -second, product)
end
if first == 0 or second == 0
return product
end
if first < 0 or second < 0
new_first = [first, second].min
new_second = [first, second].max
return product_of_two_numbers(new_first, new_second - 1, product + new_first)
end
product_of_two_numbers(first, second - 1, product + first)
end
def main
puts "Enter two numbers: "
first = gets.chomp.to_i
second = gets.chomp.to_i
puts "#{first} * #{second} = #{product_of_two_numbers(first, second, 0)}"
end
main
C Implementation
Solution
/*
* @author: ashwek
* @date: 8/1/2019
*/
#include <stdio.h>
int product(int a, int b){
if( a == 0 || b == 0 )
return 0;
else if( a == 1 )
return b;
else if( b == 1 )
return a;
if( a < 0 && b < 0 )
return product(-a, -b);
else if( a < 0 )
return product(b, a);
else
return b + product(a-1, b);
}
void main(){
int a, b;
printf("Enter 1st number = ");
scanf("%d", &a);
printf("Enter 2nd number = ");
scanf("%d", &b);
printf("%d x %d = %d\n", a, b, product(a, b));
}