Print the pattern 2 2 1 1 $2 1

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Given the number N, the task is to print a pattern such that in each line all the digits from N to 1 are present in decreasing order and the frequency of the elements in i^th line is N-i (the lines are 0 based i.e., i varies in the range [0, N-1]).

Note: Instead of printing a new line print a “$” without quotes. After printing the total output, the end of the line is expected.

Examples:

Input: 2
Output: 2 2 1 1 $2 1 $

Input: 3
Output: 3 3 3 2 2 2 1 1 1 $3 3 2 2 1 1 $3 2 1 $

Approach: Follow the steps to solve this problem:

Run a loop from k = 0 to N-1:
- Run a nested loop from i = N to 1:
  - Run a loop from 0 to N-k and print i.
- After executing the i loop for every k, print ‘$‘.

Below is the implementation of the above approach.

C++

// C++ code to implement the approach 
  
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to print the pattern 
void printPat(int n) 
{ 
    int i, j, k; 
  
    for (k = 0; k < n; k++) { 
        for (i = n; i > 0; i--) { 
            for (j = 0; j < n - k; j++) { 
                cout << i << " "; 
            } 
        } 
        cout << '$'; 
    } 
} 
  
// Driver Code 
int main() 
{ 
    int N = 2; 
  
    // Function Call 
    printPat(2); 
  
    return 0; 
}

Java

// Java code to implement the approach 
  
import java.io.*; 
  
class GFG { 
  
    // Function to print the pattern 
    static void printPat(int n) 
    { 
        int i, j, k; 
  
        for (k = 0; k < n; k++) { 
            for (i = n; i > 0; i--) { 
                for (j = 0; j < n - k; j++) { 
                    System.out.print(i + " "); 
                } 
            } 
            System.out.print("$"); 
        } 
    } 
  
    public static void main(String[] args) 
    { 
        int N = 2; 
  
        // Function call 
        printPat(2); 
    } 
} 
  
// This code is contributed by lokeshmvs21.

Python3

# Python code to implement the approach 
  
# Function to print the pattern 
def printPat(n): 
    for k in range(n): 
        for i in range(n,0,-1): 
            for j in range(0,n-k): 
                print(i,end=" ") 
        print("$",end="") 
  
# Driver Code 
N=2
  
# Function Call 
printPat(N) 
  
# This code is contributed by Pushpesh Raj. 

C#

// C# code to implement the approach 
  
using System; 
  
public class GFG{ 
    
      // Function to print the pattern 
    static void printPat(int n) 
    { 
        int i, j, k; 
   
        for (k = 0; k < n; k++) { 
            for (i = n; i > 0; i--) { 
                for (j = 0; j < n - k; j++) { 
                    Console.Write(i + " "); 
                } 
            } 
            Console.Write("$"); 
        } 
    } 
  
    static public void Main (){ 
  
        // Code 
          int N = 2; 
   
        // Function call 
        printPat(2); 
    } 
} 
  
// This code is contributed by lokesh.

Javascript

// Javascript code to implement the approach 
  
// Function to print the pattern 
function printPat(n) 
{ 
    let i = 0, j = 0, k = 0; 
  
    for (k = 0; k < n; k++) { 
        for (i = n; i > 0; i--) { 
            for (j = 0; j < n - k; j++) { 
                console.log(i + " "); 
            } 
        } 
        console.log('$'); 
    } 
} 
  
// Driver Code 
let N = 2; 
  
// Function Call 
printPat(2); 
  
// This code is contributed by Samim Hossain Mondal. 

Output

2 2 1 1 $2 1 $

Time Complexity: O(N³)
Auxiliary Space: O(1)

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