Print the following pattern for the given N number of rows Java 02 || Coding Ninjas Question

Question:

Pattern for N = 4

4444
333
22

Ans: package HelloWorld;

import java.util.Scanner;
public class LearningJava {
	
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		
		int n = scan.nextInt();
		for(int i = 1;i<=n;i++){
			for(int j=n; j>=i;j--) {
			System.out.print(n-i+1); 
		} 
			System.out.println();
	}
		
		
	
}
		
	
	
}

Yaha hamne loop me for (int i =1;i<=n;i++) { rkha hai , ye syntax outer loop ya row ke liye constant hota hai in maximum cases

}

second loop for(int j=n; j>=i;j--) { yaha j ko n ke barabr rkha hai mtlb agr n 4 hai to column me reverse print hoga 4 bar print hone se start hoga

j greater than equal to i ka mtlb column ko hamne i ke bdha aur i ke barabar rkha hai
j ko decrease kiya hai -- ke sath
print me hamne n-i+1 rkha hai jisme n hamara 4 hai to 4 se kam rkhega lekin 4 print nhi hoga kyon ki n-1 lga hua hai means if n = 4 then 4-1 then print hoga 3 se mgr hoga 4 bar hi so ham yaha +1 lga rhe

}

Comments

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