write a program in c/c++/java to print the pattern
1
2*2
3*3*3
4*4*4*4
4*4*4*4
3*3*3
2*2
1

Question

write a program in c/c++/java to print the pattern 
1
2*2
3*3*3
4*4*4*4
4*4*4*4
3*3*3
2*2
1

Atul Goel · Accepted Answer

#include

int main()
{
	int i,j,n;
	scanf("%d",&n);
	for(i=1;i=1;i--)
	{
		for(j=1;j<=i;j++)
		{
			printf("%d",i);
			if(j

sudhakar kumar · Answer

#include

 void printPattern(int n)
 {
 	int i,j;
 	for(i=1;i=1;i--)
 	{
 		j=1;
 		if(i==0)
 		printf("%d",i);
 		else 
 		{
 			printf("%d",i);
 		
 			while(j

Pavan Kumar · Answer

#include
int main()
{
	int n;
	int i, j;
	for(i = 1;i = 1;i--)
	{
		for(j = 1; j <= i; j++ )
		{
			printf("%d",i);
			if(j!=i)
			{
				printf("*");
			}
		}
		printf("
");
	}
}

Pavan Kumar · Answer

#include
int main()
{
	int n;
	int i, j;
	for(i = 1;i = 1;i--)
	{
		for(j = 1; j <= i; j++ )
		{
			printf("%d",i);
			if(j!=i)
			{
				printf("*");
			}
		}
		printf("
");
	}
}

Harsh bhardwaj · Answer

#include 

int main(void) {
	int i,j;
    	for(i=1;i=1;i--){
    		for(j=1;j<=(i*2)-1;j++){
    			
    			if(j%2==0){ printf("*");}
    			else{printf("%d",i);}
    			
    		}
    		printf("
");
    	}
	return 0;
}

Sarang Tayde · Answer

import java.io.BufferedReader;
import java.io.InputStreamReader;

public class Solution {
    public static void main(String [] args) throws Exception{
        int i , j  , k=1 , n;
        BufferedReader bufferedReader= new BufferedReader(new InputStreamReader(System.in));
        n=Integer.parseInt(bufferedReader.readLine());
        for (i=1;i<=n*2;i++){
            if (i<=n) {
                for (j = 1; j <= i * 2 - 1; j++) {
                    if (j % 2 == 0) {
                        System.out.print("*");
                    } else {
                        System.out.print("" + i);
                    }
                }
            }
            else {
                for (j = 1; j <= 2*(i-k)-1; j++) {
                    if (j % 2 == 0) {
                        System.out.print("*");
                    } else {
                        System.out.print("" + (i-k));
                    }
                }
                k+=2;
            }
            System.out.println("");
        }
    }
}

Mohit Sharma · Answer

public class JavaApplication4 {
    public static void main(String[] args) {
        // TODO code application logic here
        for(int i = 1; i  0) {
                    System.out.print("*");
                }
            }
            System.out.println();
        }
        for(int i = 4; i >= 1; i--){
            for(int j =1; j  0) {
                    System.out.print("*");
                }
            }
            System.out.println();
        }
    }
    }

sudhakar kumar · Answer

hi atul your answer is wrong it only printing 1 infinite time. i was able to solve that problem in exam and my program was working for all the test cases.

param deol · Answer

#include
void print_pattern(int);
int main()
{
	int n;
	scanf("%d",&n);
	print_pattern(n);
	return 0;
}
void print_pattern(int n)
{
	for(int i=1;i0;i--)
	{
		for(int j=1;j<=i;j++)
		{
			if(i==1)
			{
				printf("%d",i);
			}
			else
			{
			printf("%d",i);
			if(j!=i)
			{
				printf("*");
			}
		    }
		}
		printf("
");
	}
}

Nitesh Goyal · Answer

int main()
{
    int i, j, rows, k, x;

    printf("Enter number of rows: ");
    scanf("%d",&rows);

    for(i=1; i 0; i--)
	{
		k = i - 1;
		for(j = 1; j <= i; j++)
		{
			printf("%d", i);
			if(k != 0 && j != i)
			{
				printf("*");
			}
		}
		printf("
");
	}
	
	
    return 0;
}

Ankur Mishra · Answer

#include
int main()
{
    	int i,j,k;
	k=0;
	for(i=1;i=1;i--)
	{
		for(j=(i+k);j>=1;j--)
		{
			if(j%2==0)
				printf("*");
			else
				printf("%d",i);
		}
		k--;
		printf("
");
	}
    return 0;
}

Sigiloso · Answer

#include
void printPattren(int n);
int main(){
        int n;
        scanf("%d", &n);
        printPattren(n);
        return 0;
}
void printPattren(int n){
        int i, j;
        for(i = 1; i = 1; i--){
                j = 1;
                printf("%d", i);
                while(j < i){
                        printf("*");
                        printf("%d", i);
                        j++;
                }
                printf("
");
        }
}

Uday Pratap Maurya · Answer

import java.util.*;
public class qus1
{
    public static void main(String argc[])
    {
        int i,j;
        int n;
        Scanner sc=new Scanner(System.in);
        System.out.print("Enter value of n : ");
        n=sc.nextInt();
        for(i=1;i=1;i--)
        {
            for(j=i;j>=1;j--)
            {
                if(j==1)
                    System.out.print(i);
                else
                    System.out.print(i+"*");
            }
            System.out.println();
        }
    }
}

Proteep Banerjee · Answer

//THIS IS VERIFIED AND CORRECT CODE
import java.util.Scanner;

public class MyClass {

	public static void main(String[] args) {
		
		 int i,j;
	        int n;
	        Scanner sc=new Scanner(System.in);
	        System.out.print("Enter value of n : ");
	        n=sc.nextInt();
	        for(i=1;i=1;i--)
	        {
	            for(j=i;j>=1;j--)
	            {
	                while(j>1){
	                    System.out.print(i+"*");
	                    j--;
	                }
	                System.out.print(i);
	            }
	            System.out.println();
	        }

	    }

Sigiloso · Answer

#include
#include
void main()
{
    int i,j,k;

for(i=1;i=i;j--)
        {
            printf(" ");
        }
        for(k=1;k<=i;k++)
        {
            if(k==1)
            {
                // Notice the space before %d
                printf(" %d",i);
            }
              //Notice the* before %d
            else(printf("*%d",i));
        }
        printf("
");
    }
}

Devansh Singh · Answer

#n is the maximum number which would occur in the pattern.
n = int(input())

for i in range(1,n+1):
    s = []
    for j in range(i-1):
        s.append(str(i))
        s.append('*')
    s.append(str(i))
    print(''.join(s))

for i in range(n, 0, -1):
    s = []
    for j in range(i-1):
        s.append(str(i))
        s.append('*')
    s.append(str(i))
    print(''.join(s))

Proteep Banerjee · Answer

//THE SAME CODE IN C
#include 
void main(){
int n;
printf ("Enter the value of n");
scan ("%d",&n);
void printPattern(n);
}
void printPattern(int k){
int i,j;
for(i=1;i=1;i--)
{
for(j=i;j>=1;j--)
{
while(j>1)
{
printf ("%d",i);
printf ("*");
j--;
}
printf ("%d",i);
}
printf ("
");
}

Ankur Mishra · Answer

#include
int main()
{
    	int i,j,k;
	k=0;
	for(i=1;i=1;i--)
	{
		for(j=(i+k);j>=1;j--)
		{
			if(j%2==0)
				printf("*");
			else
				printf("%d",i);
		}
		k--;
		printf("
");
	}
    return 0;
}

Dev Spirit · Answer

#include
void printPattren(int n);
int main(){
        int n;
        scanf("%d", &n);
        printPattren(n);
        return 0;
}
void printPattren(int n){
        int i, j;
        for(i = 1; i = 1; i--){
                j = 1;
                printf("%d", i);
                while(j < i){
                        printf("*");
                        printf("%d", i);
                        j++;
                }
                printf("
");
        }
}

Bhavana Johri · Answer

there is a problem that only half of my code is getting displayed

Sigiloso · Answer

what is lt meaning

Nitesh Goyal · Answer

there is problem in the interface of this website, complete answer is not getting pasted

this will print full answer!!! · Answer

#include
void print_pattern(int);
int main()
{
	int n;
	scanf("%d",&n);
	print_pattern(n);
	return 0;
}
void print_pattern(int n)
{
	for(int i=1;i0;i--)
	{
		for(int j=1;j<=i;j++)
		{
			if(i==1)
			{
				printf("%d",i);
			}
			else
			{
			printf("%d",i);
			if(j!=i)
			{
				printf("*");
			}
		    }
		}
		printf("
");
	}
}

Sami · Answer

#include

int main() {
    // your code goes here

    int i,j,k=0,l;

	for(i=1;i1&&j=1;i--)
			{
				for(j=1;j1&&j

PRAYATNA JAIN · Answer

#include
//THIS CODE WORKS
void main()
{
    int i,n;
    for (n=1;n=1;n--)
    {
    for(i=1;i<=n;i++)
    {
        printf("%d",n);
        if(i!=n)
            printf("*");
        else
            printf("
");

    }}
}

Manu Gupta · Answer

for(int i = 1;i0;i--)
        {
            for(int j=0;j

Siddharth Verma · Answer

//Pattern Solution


import java.util.*;
import java.lang.*;
import java.io.*;


class Pattern
{
	public static void main (String[] args) throws java.lang.Exception
	{
		for(int j=1;j=1;k--)
	{
	    for(int l=1;l<=k;l++)
	    {
	        if(l==1)
		        System.out.print(k);
		        else
		        System.out.print("*"+k);
	    }
	    System.out.println();
	}
}}

Karan Sethi · Answer

C++ is way simpler than the rest of the code above

#include 
using namespace std;
int main() {
    int i,j,n;
    cout >n;
    for(i=1;i=1;n--&&i--)//starting from n till 1
    {
        for(j=i;j>=1;j--){//starting from i till 1
            cout<

Sigiloso · Answer

#include
#include
void main()
{
    int i,j,k;

for(i=1;i=i;j--)
        {
            printf(" ");
        }
        for(k=1;k<=i;k++)
        {
            if(k==1)
            {
                printf(" %d",i);
            }
            else(printf("*%d",i));
        }
        printf("
");
    }
}

Harsh bhardwaj · Answer

All codes are giving wrong output 
*** this is the correct one*****

#include 

int main(void) {
	int i,j;
    	for(i=1;i=1;i--){
    		for(j=1;j<=(i*2)-1;j++){
    			
    			if(j%2==0){ printf("*");}
    			else{printf("%d",i);}
    			
    		}
    		printf("
");
    	}
	return 0;
}

Nitesh Goyal · Answer

int main()
{
    int i, j, rows, k, x;

    printf("Enter number of rows: ");
    scanf("%d",&rows);

    for(i=1; i 0; i--)
	{
		k = i - 1;
		for(j = 1; j <= i; j++)
		{
			printf("%d", i);
			if(k != 0 && j != i)
			{
				printf("*");
			}
		}
		printf("
");
	}
	
	
    return 0;
}

Bhavana Johri · Answer

this code is running successfully and giving the output

#include 

int main(void) {
	// your code goes here
	
	int i,j,k=0,l;
	
	for(i=1;i=1;i--)
	{
	    l=i;
	    for(j=1;j<=i;j++)
	    {
	        printf("%d",i);
	        if(k

bhavana johri · Answer

#include 

int main(void) {
	// your code goes here
	
	int i,j,k=0,l;
	
	for(i=1;i=1;i--)
	{
	    l=i;
	    for(j=1;j<=i;j++)
	    {
	        printf("%d",i);
	        if(k

C++ Code with output · Answer

*This code could be improved on several levels but the immediate answer could be this.
Code :

#include 
using namespace std;
int main() {
    int i,j,n;
    cout >n;
    for(i=1;i=1;n--&&i--)//starting from n till 1
    {
        for(j=i;j>=1;j--){//starting from i till 1
            cout<

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