Given a number 
The task is to print the multiples of the unit digit of N from the unit digit of N to N.
Note: If the unit digit is 0 then print the multiples of 10.
Examples:
Input : 39 Output : 9 18 27 36 Explanation : The unit digit of 39 is 9. So the multiples of 9 between 9 and 39 are: 9, 18, 27, 36 Input : 25 Output : 5 10 15 20 25
Recursive Approach:
We can use a recursive function that takes in the current number and the unit digit of N. The function first checks if the current number is greater than N, in which case it returns without doing anything. Otherwise, it checks if the current number is a multiple of the unit digit, and if it is, it prints the number and calls itself with the next number. If the current number is not a multiple of the unit digit, it simply calls itself with the next number.
Below is the implementation of the above approach:
C++
#include <iostream>using namespace std;// Recursive function to print the multiples of unit digitvoid printMultiples(int current, int n, int unitDigit){ if (current > n) { return; } if (current % unitDigit == 0) { cout << current << " "; printMultiples(current + unitDigit, n, unitDigit); } else { printMultiples(current + 1, n, unitDigit); }}// Driver Codeint main(){ int n = 39; int unitDigit = n % 10; if (unitDigit == 0) { unitDigit = 10; } printMultiples(unitDigit, n, unitDigit); return 0;} |
Java
import java.util.*;class GFG { // Recursive function to print the multiples of unit // digit static void printMultiples(int current, int n, int unitDigit) { if (current > n) { return; } if (current % unitDigit == 0) { System.out.print(current + " "); printMultiples(current + unitDigit, n, unitDigit); } else { printMultiples(current + 1, n, unitDigit); } } // Driver Code public static void main(String[] args) { int n = 39; int unitDigit = n % 10; if (unitDigit == 0) { unitDigit = 10; } printMultiples(unitDigit, n, unitDigit); }} |
Python3
# Recursive function to print the multiples of unit digitdef printMultiples(current, n, unitDigit): if current > n: return if current % unitDigit == 0: print(current, end=" ") printMultiples(current + unitDigit, n, unitDigit) else: printMultiples(current + 1, n, unitDigit)# Driver Coden = 39unitDigit = n % 10if unitDigit == 0: unitDigit = 10printMultiples(unitDigit, n, unitDigit) |
Output
9 18 27 36
Time Complexity: O(N)
Auxiliary Space: O(1)
Simple Approach:
- Find the unit digit of the input number. The unit digit of N will be (N%10), i.e., remainder upon dividing N by 10.
- Check if the unit digit is 0.
- If yes, then consider the multiple as 10.
- Print the multiples of unit digit until it is less than or equal to the input number.
Below is the implementation of the above approach:
C++
// C++ program to print multiples of // Unit Digit of Given Number#include <iostream>using namespace std;// Function to print the multiples// of unit digitvoid printMultiples(int n){ // Find the unit digit of // the given number int unit_digit = n % 10; // if the unit digit is 0 then // change it to 10 if (unit_digit == 0) unit_digit = 10; // print the multiples of unit digit // until the multiple is less than // or equal to n for (int i = unit_digit; i <= n; i += unit_digit) cout << i << " ";}// Driver Codeint main(){ int n = 39; printMultiples(n); return 0;} |
Java
// Java program to print multiples of // Unit Digit of Given Numberimport java.io.*;class GFG{// Function to print the multiples// of unit digitstatic void printMultiples(int n){ // Find the unit digit of // the given number int unit_digit = n % 10; // if the unit digit is 0 then // change it to 10 if (unit_digit == 0) unit_digit = 10; // print the multiples of unit digit // until the multiple is less than // or equal to n for (int i = unit_digit; i <= n; i += unit_digit) System.out.print( i + " ");} // Driver Code public static void main (String[] args) { int n = 39; printMultiples(n); }}// This code is contributed by inder_mca |
Python3
# Python3 program to print multiples # of Unit Digit of Given Number# Function to print the multiples# of unit digitdef printMultiples(n): # Find the unit digit of # the given number unit_digit = n % 10 # if the unit digit is 0 then # change it to 10 if (unit_digit == 0): unit_digit = 10 # print the multiples of unit digit # until the multiple is less than # or equal to n for i in range(unit_digit, n + 1, unit_digit): print(i, end = " ")# Driver Coden = 39printMultiples(n)# This code is contributed by Mohit Kumar |
C#
// C# program to print multiples of // Unit Digit of Given Number using System;class GFG { // Function to print the multiples // of unit digit static void printMultiples(int n) { // Find the unit digit of // the given number int unit_digit = n % 10; // if the unit digit is 0 then // change it to 10 if (unit_digit == 0) unit_digit = 10; // print the multiples of unit digit // until the multiple is less than // or equal to n for (int i = unit_digit; i <= n; i += unit_digit) Console.Write( i + " "); } // Driver Code public static void Main () { int n = 39; printMultiples(n); } } // This code is contributed by Ryuga |
Javascript
<script> // JavaScript program to print multiples of // Unit Digit of Given Number // Function to print the multiples // of unit digit function printMultiples(n) { // Find the unit digit of // the given number var unit_digit = parseInt(n % 10); // if the unit digit is 0 then // change it to 10 if (unit_digit == 0) unit_digit = 10; // print the multiples of unit digit // until the multiple is less than // or equal to n for (var i = unit_digit; i <= n; i += unit_digit) document.write(i + " "); } // Driver Code var n = 39; printMultiples(n); </script> |
Output
9 18 27 36
Time Complexity: O(N) //since one traversal from 1 to N is required to complete all operations hence the overall time needed for the algorithm is linear
Auxiliary Space: O(1) //since no extra array is used so the space taken by the algorithm is constant
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