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)); }