Given a number N denotes the total number of elements in a matrix, the task is to print all possible order of matrix. An order is a pair (m, n) of integers where m is number of rows and n is number of columns. For example, if the number of elements is 8 then all possible orders are:
(1, 8), (2, 4), (4, 2), (8, 1).
Examples:
Input: N = 8
Output: (1, 2) (2, 4) (4, 2) (8, 1)
Input: N = 100
Output:
(1, 100) (2, 50) (4, 25) (5, 20) (10, 10) (20, 5) (25, 4) (50, 2) (100, 1)
Approach:
A matrix is said to be of order m x n if it has m rows and n columns. The total number of elements in a matrix is equal to (m*n). So we start from 1 and check one by one if it divides N(the total number of elements). If it divides, it will be one possible order.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <iostream>using namespace std;// Function to print all possible ordervoid printAllOrder(int n){ // total number of elements in a matrix // of order m * n is equal (m*n) // where m is number of rows and n is // number of columns for (int i = 1; i <= n; i++) { // if n is divisible by i then i // and n/i will be the one // possible order of the matrix if (n % i == 0) { // print the given format cout << i << " " << n / i << endl; } }}// Driver codeint main(){ int n = 10; printAllOrder(n); return 0;} |
Java
// Java implementation of the above approachclass GFG { // Function to print all possible order static void printAllOrder(int n) { // total number of elements in a matrix // of order m * n is equal (m*n) // where m is number of rows and n is // number of columns for (int i = 1; i <= n; i++) { // if n is divisible by i then i // and n/i will be the one // possible order of the matrix if (n % i == 0) { // print the given format System.out.println( i + " " + n / i ); } } } // Driver code public static void main(String []args) { int n = 10; printAllOrder(n); }}// This code is contributed by ihritik |
Python
# Python implementation of the above approach# Function to print all possible orderdef printAllOrder(n): # total number of elements in a matrix # of order m * n is equal (m*n) # where m is number of rows and n is # number of columns for i in range(1,n+1): # if n is divisible by i then i # and n/i will be the one # possible order of the matrix if (n % i == 0) : # print the given format print( i ,n // i ) # Driver coden = 10printAllOrder(n)# This code is contributed by ihritik |
C#
// C# implementation of the above approachusing System;class GFG { // Function to print all possible order static void printAllOrder(int n) { // total number of elements in a matrix // of order m * n is equal (m*n) // where m is number of rows and n is // number of columns for (int i = 1; i <= n; i++) { // if n is divisible by i then i // and n/i will be the one // possible order of the matrix if (n % i == 0) { // print the given format Console.WriteLine( i + " " + n / i ); } } } // Driver code public static void Main() { int n = 10; printAllOrder(n); }}// This code is contributed by ihritik |
PHP
<?php// PHP implementation of the above approach // Function to print all possible order function printAllOrder($n) { // total number of elements in a matrix // of order m * n is equal (m*n) // where m is number of rows and n is // number of columns for ($i = 1; $i <= $n; $i++) { // if n is divisible by i then i // and n/i will be the one // possible order of the matrix if ($n % $i == 0) { // print the given format echo $i, " ", ($n / $i), "\n"; } } } // Driver code $n = 10; printAllOrder($n); // This code is contributed by Ryuga?> |
Javascript
<script>// Java Script implementation of the above approach // Function to print all possible order function printAllOrder( n) { // total number of elements in a matrix // of order m * n is equal (m*n) // where m is number of rows and n is // number of columns for (let i = 1; i <= n; i++) { // if n is divisible by i then i // and n/i will be the one // possible order of the matrix if (n % i == 0) { // print the given format document.write( i + " " + n / i+"<br>" ); } } } // Driver code let n = 10; printAllOrder(n); // This code is contributed by sravan</script> |
Output:
1 10 2 5 5 2 10 1
Time Complexity: O(n)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
