Given a number N, the task is to print N rows where every Kth row consists of the multiplication table of K up to Kth term.
Examples:
Input: N = 3
Output:
1
2 4
3 6 9
Explanation:
In the above series, in every Kth row, multiplication table of K upto K terms is printed.Input: N = 5
Output:
1
2 4
3 6 9
4 8 12 16
5 10 15 20 25
Approach: The idea is to use two for loops to print the multiplication table. The outer loop in ‘i’ serves as the value of ‘K’ and the inner loop in ‘j’ serves as the terms of the multiplication table of every ‘i’. Each term in the table of ‘i’ can be obtained with the formula ‘i * j’.
Below is the implementation of the above approach:
C++
// C++ program to print multiplication table// till N rows where every Kth row// is the table of K up to Kth term#include <iostream>using namespace std;// Function to print the multiplication table// upto K-th termvoid printMultiples(int N){ // For loop to iterate from 1 to N // where i serves as the value of K for (int i = 1; i <= N; i++) { // Inner loop which at every // iteration goes till i for (int j = 1; j <= i; j++) { // Printing the table value for i cout << (i * j) << " "; } // New line after every row cout << endl; }}// Driver codeint main(){ int N = 5; printMultiples(N); return 0;}// This code is contributed by Rajput-Ji |
Java
// Java program to print multiplication table// till N rows where every Kth row// is the table of K up to Kth termclass GFG { // Function to print the multiplication table // upto K-th term public static void printMultiples(int N) { // For loop to iterate from 1 to N // where i serves as the value of K for (int i = 1; i <= N; i++) { // Inner loop which at every // iteration goes till i for (int j = 1; j <= i; j++) { // Printing the table value for i System.out.print((i * j) + " "); } // New line after every row System.out.println(); } } // Driver code public static void main(String args[]) { int N = 5; printMultiples(N); }} |
Python3
# Python3 program to pr multiplication table# till N rows where every Kth row# is the table of K up to Kth term# Function to pr the multiplication table# upto K-th termdef prMultiples(N): # For loop to iterate from 1 to N # where i serves as the value of K for i in range(1, N + 1): # Inner loop which at every # iteration goes till i for j in range(1, i + 1): # Printing the table value for i print((i * j), end = " ") # New line after every row print()# Driver codeif __name__ == '__main__': N = 5 prMultiples(N)# This code is contributed by mohit kumar 29 |
C#
// C# program to print multiplication table// till N rows where every Kth row// is the table of K up to Kth termusing System;class GFG { // Function to print the multiplication table // upto K-th term public static void printMultiples(int N) { // For loop to iterate from 1 to N // where i serves as the value of K for (int i = 1; i <= N; i++) { // Inner loop which at every // iteration goes till i for (int j = 1; j <= i; j++) { // Printing the table value for i Console.Write((i * j) + " "); } // New line after every row Console.WriteLine(); } } // Driver code public static void Main(String []args) { int N = 5; printMultiples(N); }}// This code is contributed by Rajput-Ji |
Javascript
<script>// javascript program to print multiplication table// till N rows where every Kth row// is the table of K up to Kth term// Function to print the multiplication table// upto K-th termfunction printMultiples( N){ // For loop to iterate from 1 to N // where i serves as the value of K for (let i = 1; i <= N; i++) { // Inner loop which at every // iteration goes till i for (let j = 1; j <= i; j++) { // Printing the table value for i document.write((i * j) + " "); } // New line after every row document.write("<br/>"); }}// Driver code let N = 5; printMultiples(N); // This code is contributed by todaysgaurav </script> |
1 2 4 3 6 9 4 8 12 16 5 10 15 20 25
Time Complexity: O(N2), here there are two loops with one running from 1 to N and another from 1 to i which means it is also running N times so it is a N*N complexity
Auxiliary Space: O(1), no extra array is being used so the space required is constant
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