Program for multiplication table

Try it on GfG Practice

Given a number n, we need to print its table. 

Examples : 

Input: 5
Output:
5 * 1 = 5
5 * 2 = 10
5 * 3 = 15
5 * 4 = 20
5 * 5 = 25
5 * 6 = 30
5 * 7 = 35
5 * 8 = 40
5 * 9 = 45
5 * 10 = 50

Input: 2
Output:
2 * 1 = 2
2 * 2 = 4
2 * 3 = 6
2 * 4 = 8
2 * 5 = 10
2 * 6 = 12
2 * 7 = 14
2 * 8 = 16
2 * 9 = 18
2 * 10 = 20

Iterative Approach

The iterative approach for printing a multiplication table involves using a loop to calculate and print the product of a given number and the numbers in range from 1 to 10. In this method, you begin with the number whose table you want to print and use a loop to multiply it with increasing values.

// CPP program to print table of a number
#include <iostream>
using namespace std;

void printTable(int n) { 
  for (int i = 1; i <= 10; ++i) 
        cout << n << " * " << i << " = "  
             << n * i << endl;
}

int main() {
    int n = 5;  
    printTable(n);
    return 0;
}

#include <stdio.h>

void printTable(int n) { 
    for (int i = 1; i <= 10; ++i) 
        printf("%d * %d = %d\n", n, i, n * i);
}

int main() {
    int n = 5;  
    printTable(n);
    return 0;
}

// Java program to print table of a number 
import java.io.*;

class GfG {

    public static void printTable(int n)  {
              
        for (int i = 1; i <= 10; ++i) 
            System.out.println(n + " * " + i +
                               " = " + n * i);
    }
  
    public static void main(String arg[]){   
        int n = 5; 
		printTable(n);
    }
}

# Python Program to print table of a number

def printTable(n):

    for i in range (1, 11): 
        
        # multiples from 1 to 10
        print ("%d * %d = %d" % (n, i, n * i))


if __name__ == "__main__":
  n = 5
  printTable(n)

// C# program to print table of a number 
using System;

class GfG {
    public static void printTable(int n) {
        
        for (int i = 1; i <= 10; ++i) 
            Console.Write(n + " * " + i +
                              " = " + n * 
                               i + "\n");
    }
  
    public static void Main()   { 
      int n = 5; 
      printTable(n);

    }
}

// Javascript program to print
// table of a number

function printTable(n) { 
for (let i = 1; i <= 10; ++i)
    console.log( n + " * " +i +
            " = " + n *
                i);
}

// Driver Code
let n = 5; 
printTable(n);

Output : 

5 * 1 = 5
5 * 2 = 10
5 * 3 = 15
5 * 4 = 20
5 * 5 = 25
5 * 6 = 30
5 * 7 = 35
5 * 8 = 40
5 * 9 = 45
5 * 10 = 50

Time Complexity - O(1)
Space Complexity - O(1)

Illustration

Step by step execution of loop for the multiplication table of n = 5.

We have n = 5, and the loop will iterate from i = 1 to i = 10.

First Iteration (i = 1):

• The loop multiplies n = 5 by i = 1.
• Result: 5 * 1 = 5.
• Output: 5 * 1 = 5.

Second Iteration (i = 2):

• The loop multiplies n = 5 by i = 2.
• Result: 5 * 2 = 10.
• Output: 5 * 2 = 10.

Third Iteration (i = 3):

• The loop multiplies n = 5 by i = 3.
• Result: 5 * 3 = 15.
• Output: 5 * 3 = 15.

....
....

Tenth Iteration (i = 10):

• The loop multiplies n = 5 by i = 10.
• Result: 5 * 10 = 50.
• Output: 5 * 10 = 50.

Recursive Approach

In this method, we pass i as an additional parameter with initial value as 1. We print n * i and then recursively call for i+1. We stop the recursion when i becomes 11 as we need to print only 10 multiples of given number and i.

#include <iostream>
using namespace std;

// printTable() prints table of number and takes
// 1 required value that is number of whose teble
// to be printed and an optional input i whose d
// efault value is 1
void printTable(int n, int i = 1)
{
    if (i == 11)
        return;
    cout << n << " * " << i << " = " << n * i << endl;
    i++;
    printTable(n, i);
}

int main()
{
    int n = 5;
    printTable(n);
}

#include <stdio.h>

// printTable() prints table of number and takes
// 1 required value that is number of whose table
// to be printed and an optional input i whose default value is 1
void printTable(int n, int i) {
    if (i == 11)
        return;
    printf("%d * %d = %d\n", n, i, n * i);
    i++;
    printTable(n, i);
}

int main() {
    int n = 5;
    printTable(n, 1);
    return 0;
}

import java.util.*;
 
class GfG {
 
    // printTable() prints table of number and takes
    // 1 required value that is number of whose teble to be
    // printed and an optional input i whose default value is 1
    static void printTable(int n, Integer... val)  {
          int i = 1;
        if (val.length != 0)
            i = val[0];
        if (i == 11) // base case
            return;
        System.out.println(n + " * " + i + " = " + n * i);
        i++; 
        printTable(n, i);
    }
 
    public static void main(String[] args) {
        int n = 5;
        printTable(n);
    }
}
 

# printTable() prints table of number and takes
# 1 required value that is number of whose teble to be printed
# and an optional input i whose default value is 1
def printTable(n, i=1):

    if (i == 11):  # base case
        return
    print(n, "*", i, "=", n * i)
    i += 1  
    printTable(n, i)

if __name__ == "__main__":
  n = 5
  printTable(n)

using System;
using System.Collections.Generic;

class GfG {

  // print_table() prints table of number and takes
  // 1 required value that is number of whose teble to be
  // printed and an optional input i whose default value is 1
  static void printTable(int n, int i = 1) {
    if (i == 11) // base case
      return;
    Console.WriteLine(n + " * " + i + " = " + n * i);
    i++;
    printTable(n, i);
  }
  
  public static void Main(string[] args) {
    int n = 5;
    printTable(n);
  }
}

// printTable() prints table of number and takes
//1 required value that is number of whose teble to be printed
//and an optional input i whose default value is 1

function printTable(n, i = 1) {
    if (i == 11)// base case
        return;
    console.log(n + " * " + i + " = " + n * i);
    i++;
    printTable(n,i);
}

// Driver Code
let n = 5;
printTable(n);

5 * 1 = 5
5 * 2 = 10
5 * 3 = 15
5 * 4 = 20
5 * 5 = 25
5 * 6 = 30
5 * 7 = 35
5 * 8 = 40
5 * 9 = 45
5 * 10 = 50

Time Complexity - O(1)
Space Complexity - O(1)

Table of A Number in Python

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