Program to print number pattern

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We have to print a pattern where in middle column contains only 1, right side columns contain constant digit which is greater than 1 and left side columns contains constant digit which is greater than 1. Every row should look like a Palindrome.

Examples :

Input : 3
Output :
    1
 2  1  2
    1

Input : 5
Output :
     1
  2  1  2
3 2  1  2 3
  2  1  2
     1

Below is the implementation to print the following pattern :

C++

// CPP program to print pattern 
#include <bits/stdc++.h> 
using namespace std; 
  
void display() 
{ 
      
    int n = 5; 
    int space = n / 2, num = 1; 
       
    // Outer for loop for  
    // number of rows 
    for (int i = 1; i <= n; i++) 
    { 
        // Inner for loop for  
        // printing space 
        for (int j = 1; j <= space; j++)             
            cout<<" "; 
           
        // Logic for printing  
        // the pattern for everyline 
        int count = num / 2 + 1; 
          
        for (int k = 1; k <= num; k++) 
        { 
            cout<<count; 
              
            // Value of count decrements 
            // in every cycle 
            if (k <= num /2 ) 
                count--; 
  
            // Value of count will 
            // increment in every cycle 
            else
                count++; 
        } 
  
        cout<<"\n"; 
  
        // Before reaching half Space 
        // is decreased by 1 and num 
        // is increased by 2 
        if (i <= n / 2) 
        { 
            space = space - 1; 
            num = num + 2; 
        } 
  
        // After reaching to half 
        // space is increased by 1 
        // and num is decreased by 2 
        else
        { 
            space = space + 1; 
            num = num - 2; 
        } 
    } 
} 
  
// Driver Code 
int main() 
{ 
    display(); 
    return 0; 
} 
  
// This code is contributed by Nikita tiwari. 

Java

// Java program to print above pattern 
import java.util.Scanner; 
class Pattern 
{ 
    void display() 
    { 
        int n = 5; 
        int space = n / 2, num = 1; 
          
        // Outer for loop for  
        // number of rows 
        for (int i = 1; i <= n; i++) 
        { 
            // Inner for loop  
            // for printing space 
            for (int j = 1; j <= space; j++)             
                System.out.print(" "); 
              
            // Logic for printing  
            // the pattern for everyline 
            int count = num / 2 + 1; 
            for (int k = 1; k <= num; k++) 
            { 
                System.out.print(count); 
                // Value of count decrements  
                // in every cycle  
                if (k <= num /2 ) 
                    count--; 
  
                // Value of count will increment  
                // in every cycle 
                else
                    count++; 
            } 
  
            System.out.println(); 
  
            // Before reaching half Space is decreased 
            // by 1 and num is increased by 2 
            if (i <= n / 2) 
            { 
                space = space - 1; 
                num = num + 2; 
            } 
  
            // After reaching to half space is increased 
            // by 1 and num is decreased by 2 
            else
            { 
                space = space + 1; 
                num = num - 2; 
            } 
        } 
    } 
  
    // Driver Code 
    public static void main(String[] args) 
    { 
        Pattern p = new Pattern(); 
        p.display(); 
    } 
} 

Python3

# Python 3 program to  
# print above pattern 
  
def display() : 
    n = 5
    space = n // 2
    num = 1
      
    # Outer for loop for  
    # number of rows 
    for i in range(1, n+1) : 
          
        # Inner for loop for   
        # printing space 
        for j in range(1, space+1) : 
            print(" ", end = "")  
              
        # Logic for printing the 
        # pattern for everyline 
        count = num // 2 + 1
          
        for k in range(1, num+1) : 
            print(count, end = "") 
              
            # Value of count decrements 
            # in every cycle 
            if (k <= num // 2 ) : 
                count = count -1
  
            # Value of count will  
            # increment in every cycle 
            else : 
                count = count + 1
        print() 
  
        # Before reaching half Space 
        # is decreased by 1 and num 
        # is increased by 2 
        if (i <= n // 2) : 
            space = space - 1
            num = num + 2
          
        # After reaching to half  
        # space is increased by 1 
        # and num is decreased by 2 
        else : 
            space = space + 1
            num = num - 2
              
# Driver Code 
display() 
  
#This code is contributed by Nikita Tiwari. 

C#

// C# program to print above pattern 
using System; 
class Pattern 
{ 
    void display() 
    { 
        int n = 5; 
        int space = n / 2, num = 1; 
          
        // Outer for loop for 
        // number of rows 
        for (int i = 1; i <= n; i++) 
        { 
            // Inner for loop 
            // for printing space 
            for (int j = 1; j <= space; j++)         
                Console.Write(" "); 
              
            // Logic for printing 
            // the pattern for everyline 
            int count = num / 2 + 1; 
            for (int k = 1; k <= num; k++) 
            { 
                Console.Write(count); 
                // Value of count decrements 
                // in every cycle 
                if (k <= num /2 ) 
                    count--; 
  
                // Value of count will increment 
                // in every cycle 
                else
                    count++; 
            } 
  
            Console.WriteLine(); 
  
            // Before reaching half Space is decreased 
            // by 1 and num is increased by 2 
            if (i <= n / 2) 
            { 
                space = space - 1; 
                num = num + 2; 
            } 
  
            // After reaching to half space is increased 
            // by 1 and num is decreased by 2 
            else
            { 
                space = space + 1; 
                num = num - 2; 
            } 
        } 
    } 
  
    // Driver Code 
    public static void Main() 
    { 
        Pattern p = new Pattern(); 
        p.display(); 
    } 
} 
// This code is contributed by vt_m. 

PHP

<?php 
// php program to print  
// above pattern 
  
    function display($n) 
    { 
        $space = $n / 2; 
        $num = 1; 
          
        // Outer for loop for  
        // number of rows 
        for ($i = 1; $i <= $n; $i++) 
        { 
            // Inner for loop  
            // for printing space 
            for ($j = 1; $j <= $space; $j++)      
                echo " "; 
              
            // Logic for printing  
            // the pattern for everyline 
            $count = $num / 2 + 1; 
            for ($k = 1; $k <= $num; $k++) 
            { 
                echo floor($count); 
                // Value of count decrements  
                // in every cycle  
                if ($k <= $num /2 ) 
                    $count--; 
  
                // Value of count will increment  
                // in every cycle 
                else
                    $count++; 
            } 
  
            echo "\n"; 
  
            // Before reaching half Space 
            // is decreased by 1 and 
            // num is increased by 2 
            if ($i <= $n / 2) 
            { 
                $space = $space - 1; 
                $num = $num + 2; 
            } 
  
            // After reaching to half  
            // space is increased by 1 
            // and num is decreased by 2 
            else
            { 
                $space = $space + 1; 
                $num = $num - 2; 
            } 
        } 
    } 
  
// Driver Code 
$n = 5; 
display($n); 
  
// This code is contributed by mits  
?> 

Javascript

<script> 
  
 // JavaScript program to print pattern 
  
      function display() { 
        var n = 5; 
        var space = parseInt(n / 2), 
          num = 1; 
  
        // Outer for loop for 
        // number of rows 
          
        for (var i = 1; i <= n; i++) { 
          // Inner for loop for 
          // printing space 
          for (var j = 1; j <= space; j++)  
          document.write("  "); 
  
          // Logic for printing 
          // the pattern for everyline 
          var count = parseInt(num / 2 + 1); 
  
          for (var k = 1; k <= num; k++) { 
            document.write(count); 
  
            // Value of count decrements 
            // in every cycle 
            if (k <= num / 2) count--; 
              
            // Value of count will 
            // increment in every cycle 
            else count++; 
          } 
  
          document.write("<br>"); 
  
          // Before reaching half Space 
          // is decreased by 1 and num 
          // is increased by 2 
          if (i <= n / 2) { 
            space = space - 1; 
            num = num + 2; 
          } 
  
          // After reaching to half 
          // space is increased by 1 
          // and num is decreased by 2 
          else { 
            space = space + 1; 
            num = num - 2; 
          } 
        } 
      } 
  
      // Driver Code 
        
      display(); 
        
</script>

Output :

Time Complexity: O(n²), where n represents the given input.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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