Print rectangular pattern with given center
Given 3 positive integer c1, c2 and n, where n is size of 2-D square matrix. The task is to print the matrix filled with rectangular pattern having center coordinates c1, c2 such that 0 <= c1, c2 < n.
Examples:
Input: c1 = 2, c2 = 2, n = 5
Output:
2 2 2 2 2
2 1 1 1 2
2 1 0 1 2
2 1 1 1 2
2 2 2 2 2Input: c1 = 3, c2 = 4, n = 7
Output:
4 3 3 3 3 3 3
4 3 2 2 2 2 2
4 3 2 1 1 1 2
4 3 2 1 0 1 2
4 3 2 1 1 1 2
4 3 2 2 2 2 2
4 3 3 3 3 3 3
Approach: This problem can be solved by using two nested loops. Follow the steps below to solve this problem:
- Iterate in the range[0, N-1], using a variable i and do the following steps:
- Iterate in the range[0, N-1], using a variable j and do the following steps:
- Print maximum of abs(c1 – i) and abs(c2 – j).
- Print new line.
- Iterate in the range[0, N-1], using a variable j and do the following steps:
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to print the matrix filled // with rectangle pattern having center // coordinates are c1, c2 void printRectPattern( int c1, int c2, int n) { // Iterate in the range[0, n-1] for ( int i = 0; i < n; i++) { // Iterate in the range[0, n-1] for ( int j = 0; j < n; j++) { cout << (max( abs (c1 - i), abs (c2 - j))) << " " ; } cout << endl; } } // Driver Code int main() { // Given Input int c1 = 2; int c2 = 2; int n = 5; // Function Call printRectPattern(c1, c2, n); // This code is contributed by Potta Lokesh return 0; } |
Java
// Java program for the above approach import java.io.*; class GFG{ // Function to print the matrix filled // with rectangle pattern having center // coordinates are c1, c2 static void printRectPattern( int c1, int c2, int n) { // Iterate in the range[0, n-1] for ( int i = 0 ; i < n; i++) { // Iterate in the range[0, n-1] for ( int j = 0 ; j < n; j++) { System.out.print((Math.max(Math.abs(c1 - i), Math.abs(c2 - j))) + " " ); } System.out.println(); } } // Driver code public static void main(String[] args) { // Given Input int c1 = 2 ; int c2 = 2 ; int n = 5 ; // Function Call printRectPattern(c1, c2, n); } } // This code is contributed by sanjoy_62 |
Python3
# Python3 program for the above approach # Function to print the matrix filled # with rectangle pattern having center # coordinates are c1, c2 def printRectPattern(c1, c2, n): # Iterate in the range[0, n-1] for i in range (n): # Iterate in the range[0, n-1] for j in range (n): print ( max ( abs (c1 - i), abs (c2 - j)), end = " " ) print ("") # Driver Code # Given Input c1 = 2 c2 = 2 n = 5 # Function Call printRectPattern(c1, c2, n) |
C#
// C# program for the above approach using System; class GFG{ // Function to print the matrix filled // with rectangle pattern having center // coordinates are c1, c2 static void printRectPattern( int c1, int c2, int n) { // Iterate in the range[0, n-1] for ( int i = 0; i < n; i++) { // Iterate in the range[0, n-1] for ( int j = 0; j < n; j++) { Console.Write((Math.Max(Math.Abs(c1 - i), Math.Abs(c2 - j))) + " " ); } Console.WriteLine(); } } // Driver Code public static void Main(String[] args) { // Given Input int c1 = 2; int c2 = 2; int n = 5; // Function Call printRectPattern(c1, c2, n); } } // This code is contributed by target_2 |
Javascript
<script> // Javascript program for the above approach // Function to print the matrix filled // with rectangle pattern having center // coordinates are c1, c2 function printRectPattern(c1, c2, n) { // Iterate in the range[0, n-1] for (let i = 0; i < n; i++) { // Iterate in the range[0, n-1] for (let j = 0; j < n; j++) { document.write(Math.max(Math.abs(c1 - i), Math.abs(c2 - j)) + " " ); } document.write( "<br>" ); } } // Driver Code // Given Input let c1 = 2; let c2 = 2; let n = 5; // Function Call printRectPattern(c1, c2, n); // This code is contributed by gfgking </script> |
Output:
2 2 2 2 2 2 1 1 1 2 2 1 0 1 2 2 1 1 1 2 2 2 2 2 2
Time Complexity: O(N ^2)
Auxiliary Space: O(1)