Print concentric rectangular pattern in a 2d matrix (original) (raw)

Last Updated : 21 Dec, 2022

Given a positive integer n, print the matrix filled with rectangle pattern as shown below:
a a a a a
a b b b a
a b c b a
a b b b a
a a a a a
where a = n, b = n - 1,c = n - 2 and so on.

Examples:

Input : n = 4 Output : 4 4 4 4 4 4 4 4 3 3 3 3 3 4 4 3 2 2 2 3 4 4 3 2 1 2 3 4 4 3 2 2 2 3 4 4 3 3 3 3 3 4 4 4 4 4 4 4 4

Input : n = 3 Output : 3 3 3 3 3 3 2 2 2 3 3 2 1 2 3 3 2 2 2 3 3 3 3 3 3

For the given n, the number of rows or columns to be printed will be 2*n - 1. We will print the matrix in two parts. We will first print upper half from rows from 0 to floor((2*n - 1)/2) and then second half from floor((2*n - 1)/2) + 1 to 2*n - 2.

Now for each row, we will print it in three parts. First part is decreasing sequence which will start from n and decrease by 1 in each iteration. The number of iteration will be equal to row number, the second part is a constant sequence where constant is n - i and it will be print 2*n - 1 - 2 * row number, and the third part is increasing sequence which is nothing but opposite of the first sequence.

For lower half, observe, it is a mirror image of upper half (excluding middle row). So, simply run a loop of the upper half from (2*n - 1)/2 to 0.

Below is the basic implementation of this approach:

C++ `

// C++ program for printing // the rectangular pattern #include <bits/stdc++.h> using namespace std;

// Function to print the pattern void printPattern(int n) { // number of rows and columns to be printed int s = 2 * n - 1;

// Upper Half
for (int i = 0; i < (s / 2) + 1; i++) {
   
    int m = n;

    // Decreasing part
    for (int j = 0; j < i; j++) {
        cout << m << " ";
        m--;
    }

    // Constant Part
    for (int k = 0; k < s - 2 * i; k++) {
        cout << n - i << " ";
    }

    // Increasing part.
    m = n - i + 1;
    for (int l = 0; l < i; l++) {
        cout << m << " ";
        m++;
    }

    cout << endl;
}

// Lower Half
for (int i = s / 2 - 1; i >= 0; i--) {

    // Decreasing Part
    int m = n;
    for (int j = 0; j < i; j++) {
        cout << m << " ";
        m--;
    }

    // Constant Part.
    for (int k = 0; k < s - 2 * i; k++) {
        cout << n - i << " ";
    }

    // Decreasing Part
    m = n - i + 1;
    for (int l = 0; l < i; l++) {
        cout << m << " ";
        m++;
    }
    cout << endl;
}

} // Driven Program int main() { int n = 3;

printPattern(n);
return 0;

}

Java

// Java program for printing // the rectangular pattern import java.io.*;

class GFG {

// Function to // print the pattern static void printPattern(int n) { // number of rows and // columns to be printed int s = 2 * n - 1;

// Upper Half
for (int i = 0;
         i < (s / 2) + 1; i++) 
{
    int m = n;

    // Decreasing part
    for (int j = 0; j < i; j++) 
    {
        System.out.print(m + " ");
        m--;
    }

    // Constant Part
    for (int k = 0; 
             k < s - 2 * i; k++) 
    {
        System.out.print(n - i + " ");
    }

    // Increasing part.
    m = n - i + 1;
    for (int l = 0; l < i; l++) 
    {
        System.out.print(m + " ");
        m++;
    }

    System.out.println();
}

// Lower Half
for (int i = s / 2 - 1;
         i >= 0; i--) 
{

    // Decreasing Part
    int m = n;
    for (int j = 0; j < i; j++)
    {
        System.out.print(m + " ");
        m--;
    }

    // Constant Part.
    for (int k = 0; 
             k < s - 2 * i; k++)
    {
        System.out.print(n - i + " ");
    }

    // Decreasing Part
    m = n - i + 1;
    for (int l = 0; l < i; l++) 
    {
        System.out.print(m + " ");
        m++;
    }
    System.out.println();
}

} // Driver Code public static void main (String[] args) { int n = 3;

printPattern(n);

} }

// This code is contributed // by anuj_67.

Python3

Python3 program for printing

the rectangular pattern

Function to print the pattern

def printPattern(n) :

# number of rows and columns to be printed 
s = 2 * n - 1

# Upper Half 
for i in range(0, int(s / 2)): 
    
    m = n 

    # Decreasing part 
    for j in range(0, i): 
        print(m ,end= " " )
        m-=1
    
    # Constant Part 
    for k in range(0, s - 2 * i):
        print(n-i ,end= " " )
    
    # Increasing part. 
    m = n - i + 1
    for l in range(0, i): 
        print(m ,end= " " ) 
        m+=1
    
    print("") 

# Lower Half 
for i in range(int(s / 2),-1,-1): 

    # Decreasing Part 
    m = n 
    for j in range(0, i):
        print(m ,end= " " )
        m-=1
    

    # Constant Part. 
    for k in range(0, s - 2 * i):
        print(n-i ,end= " " )
    

    # Decreasing Part 
    m = n - i + 1
    for l in range(0, i): 
        print(m ,end= " " ) 
        m+=1
    
    print("")

Driven Program

if name=='main': n = 3 printPattern(n)

this code is contributed by Smitha Dinesh

Semwal

C#

// C# program for printing // the rectangular pattern using System;

class GFG {

// Function to // print the pattern static void printPattern(int n) { // number of rows and // columns to be printed int s = 2 * n - 1;

// Upper Half
for (int i = 0;
         i < (s / 2) + 1; i++) 
{
    int m = n;

    // Decreasing part
    for (int j = 0; j < i; j++) 
    {
        Console.Write(m + " ");
        m--;
    }

    // Constant Part
    for (int k = 0; 
             k < s - 2 * i; k++) 
    {
        Console.Write(n - i + " ");
    }

    // Increasing part.
    m = n - i + 1;
    for (int l = 0; l < i; l++) 
    {
        Console.Write(m + " ");
        m++;
    }

    Console.WriteLine();
}

// Lower Half
for (int i = s / 2 - 1;
         i >= 0; i--) 
{

    // Decreasing Part
    int m = n;
    for (int j = 0; j < i; j++)
    {
        Console.Write(m + " ");
        m--;
    }

    // Constant Part.
    for (int k = 0; 
             k < s - 2 * i; k++)
    {
        Console.Write(n - i + " ");
    }

    // Decreasing Part
    m = n - i + 1;
    for (int l = 0; l < i; l++) 
    {
        Console.Write(m + " ");
        m++;
    }
    Console.WriteLine();
}

} // Driver Code public static void Main () { int n = 3;

printPattern(n);

} }

// This code is contributed // by anuj_67.

PHP

s=2∗s = 2 * s=2∗n - 1; // Upper Half for ($i = 0; i<(int)(i < (int)(i<(int)(s / 2) + 1; $i++) { m=m = m=n; // Decreasing part for ($j = 0; j<j < j<i; $j++) { echo $m , " "; $m--; } // Constant Part for ($k = 0; k<k < k<s - 2 * i;i; i;k++) { echo ($n - $i) , " "; } // Increasing part. m=m = m=n - $i + 1; for ($l = 0; l<l < l<i; $l++) { echo $m , " "; $m++; } echo "\n"; } // Lower Half for ($i = (int)($s / 2 - 1); i>=0;i >= 0; i>=0;i--) { // Decreasing Part m=m = m=n; for ($j = 0; j<j < j<i; $j++) { echo $m, " "; $m--; } // Constant Part. for ($k = 0; k<k < k<s - 2 * i;i; i;k++) { echo n−n - n−i , " "; } // Decreasing Part m=m = m=n - $i + 1; for ($l = 0; l<l < l<i; $l++) { echo $m , " "; $m++; } echo "\n"; } } // Driver Code $n = 3; printPattern($n); // This code is contributed // by Sach_Code ?>

JavaScript

`

Output

3 3 3 3 3 3 2 2 2 3 3 2 1 2 3 3 2 2 2 3 3 3 3 3 3

Complexity Analysis:

Another Approach:

C++ `

// C++ program for printing // the rectangular pattern #include<bits/stdc++.h> using namespace std;

// Function to print the pattern void printPattern(int n) { int arraySize = n * 2 - 1; int result[arraySize][arraySize];

// Fill the values
for(int i = 0; i < arraySize; i++)
{
    for(int j = 0; j < arraySize; j++)
    {
        if(abs(i - arraySize / 2) > abs(j - arraySize / 2))
            result[i][j] = abs(i - arraySize / 2) + 1;
        else
            result[i][j] = (abs(j-arraySize / 2) + 1);
        
    }
}
    
// Print the array
for(int i = 0; i < arraySize; i++)
{
    for(int j = 0; j < arraySize; j++)
    {
        cout << result[i][j] << " ";
    }
    cout << endl;
}

}

// Driver Code int main() { int n = 3;

printPattern(n);
return 0;

}

// This code is contributed // by Rajput-Ji.

Java

// Java program for printing // the rectangular pattern import java.io.*;

class GFG {

// Function to print the pattern static void printPattern(int n) { int arraySize = n * 2 - 1; int[][] result = new int[arraySize][arraySize];

    //Fill the values
    for(int i = 0; i < arraySize; i++)
    {
        for(int j = 0; j < arraySize; j++)
        {
            result[i][j] = Math.max(Math.abs(i-arraySize/2),
                                Math.abs(j-arraySize/2))+1;
        }
    }
    
    //Print the array
    for(int i = 0; i < arraySize; i++)
    {
        for(int j = 0; j < arraySize; j++)
        {
        System.out.print(result[i][j]);
        }
        System.out.println();
    }

} // Driver Code public static void main (String[] args) { int n = 3;

printPattern(n);

} }

// This code is contributed // by MohitSharma23.

Python3

Python3 program for printing

the rectangular pattern

Function to print the pattern

def printPattern(n):

arraySize = n * 2 - 1;
result = [[0 for x in range(arraySize)] 
             for y in range(arraySize)];
    
# Fill the values
for i in range(arraySize):
    for j in range(arraySize):
        if(abs(i - (arraySize // 2)) > 
           abs(j - (arraySize // 2))):
            result[i][j] = abs(i - (arraySize // 2)) + 1;
        else:
            result[i][j] = abs(j - (arraySize // 2)) + 1;
        
# Print the array
for i in range(arraySize):
    for j in range(arraySize):
        print(result[i][j], end = " ");
    print("");

Driver Code

n = 3;

printPattern(n);

This code is contributed by mits

C#

// C# program for printing // the rectangular pattern using System;

class GFG {

// Function to print the pattern static void printPattern(int n) { int arraySize = n * 2 - 1; int[,] result = new int[arraySize,arraySize]; 

    // Fill the values 
    for(int i = 0; i < arraySize; i++) 
    { 
        for(int j = 0; j < arraySize; j++) 
        { 
            result[i,j] = Math.Max(Math.Abs(i-arraySize/2), 
                                Math.Abs(j-arraySize/2))+1; 
        } 
    } 
    
    // Print the array 
    for(int i = 0; i < arraySize; i++) 
    { 
        for(int j = 0; j < arraySize; j++) 
        { 
            Console.Write(result[i,j]+" "); 
        } 
        Console.WriteLine(); 
    }

}

// Driver Code public static void Main (String[] args) { int n = 3; 

printPattern(n);

} }

// This code has been contributed by 29AjayKumar

PHP

arraySize=arraySize = arraySize=n * 2 - 1; result=arrayfill(0,result=array_fill(0,result=arrayf​ill(0,arraySize,array_fill(0,$arraySize,0)); // Fill the values for($i = 0; i<i < i<arraySize; $i++) { for($j = 0; j<j < j<arraySize; $j++) { if(abs($i - (int)($arraySize / 2)) > abs($j - (int)($arraySize / 2))) result[result[result[i][$j] = abs($i - (int)($arraySize / 2)) + 1; else result[result[result[i][$j] = (abs($j-(int) ($arraySize / 2)) + 1); } } // Print the array for($i = 0; i<i < i<arraySize; $i++) { for($j = 0; j<j < j<arraySize; $j++) { echo result[result[result[i][$j]." "; } echo "\n"; } } // Driver Code $n = 3; printPattern($n); // This code is contributed by mits ?>

JavaScript

`

Output

3 3 3 3 3 3 2 2 2 3 3 2 1 2 3 3 2 2 2 3 3 3 3 3 3

Complexity Analysis:

Similar Reads