# Wipro NLTH Pattern Making Question: 2 (Trapezium)

## Trapezium Pattern in C, Java

Here you will learn how to print the Trapezium Pattern in C language and Trapezium Pattern program in Java, Trapezium Pattern print in C++ –

Ques. To print the trapezium pattern?

Please also post your code in the comments in different languages or same languages with short or better time complexity code.

`If N = 41*2*3*4*17*18*19*20  5*6*7*14*15*16    8*9*12*13      10*11If n = 51*2*3*4*5*26*27*28*29*30  6*7*8*9*22*23*24*25    10*11*12*19*20*21      13*14*17*18        15*16If N = 21*2*5*6  3*4`

`#include<stdio.h>int main (){    int n = 5, num = 1, i = 1, space = 0, k = 1, number = n;        for (i = 0; i < n; i++)    {        for (int j = 1; j <= space; j++)	    {    	    printf (" ");    	}        for (int m = 1; m < 2 * n - space; m++)	    {	        if (m % 2 == 0)	            printf ("%s", "*");	        else	            printf ("%d", num++);	   }              printf ("%s", "*");        for (int l = 1; l < 2 * n - space; l++)	    {	        if (l % 2 == 0)	            printf ("%s", "*");	        else	        {	            printf ("%d", k + number * number);	            k++;	        }	    }        number--;        space = space + 2;        printf ("\n");    }    return 0;}`
`#include<bits/stdc++.h>using namespace std;int main (){    int n = 5, num = 1, i = 1, space = 0, k = 1, number = n;    for (i = 0; i < n; i++)    {        for (int j = 1; j <= space; j++)	    {    	    cout << " ";    	}        for (int m = 1; m < 2 * n - space; m++)	    {	        if (m % 2 == 0)	            cout << "*";	        else	            cout << num++;	    }        cout << "*";              for (int l = 1; l < 2 * n - space; l++)	    {	        if (l % 2 == 0)	            cout << "*";	        else	        {	            cout << k + number * number;	            k++;	        }	    }        number--;        space = space + 2;        cout << endl;    }    return 0;}`
`class Main{    public static void main (String[]args)    {        int count1 = 0, count2 = 0;        int N = 2;        for (int i = N; i >= 1; i--)        {	        for (int j = N; j > i; j--)	            System.out.print (" ");            	for (int k = 1; k <= i; k++)	            System.out.print (++count1 + "*");            	for (int l = 1; l <= i; l++)	        {	            System.out.print (++count2 + i * i);	            if (l != i)	            System.out.print ("*");	        }	        System.out.println ();        }    }}`

NOTE:- Please do comment the code in other languages 🙂

### 9 comments on “Wipro NLTH Pattern Making Question: 2 (Trapezium)”

• satyam700770

n=int(input())
l=1
c=0
k=n**2+1
t=4*n-1
p=0
for i in range(n):
for m in range(p):
print(” “,end=”)
for j in range(t):
if j%2==0:
if j<=t//2:
print(l,end='')
l=l+1
else:
print(k,end='')
k=k+1
c=c+1
else:
print("*",end="")
k=k-2*c+1
c=0
p=p+2
t=t-4
print( )

• Nishant

n=int(input())
temp=n
count=1
for i in range(n):
for j in range(0,i):
print(” “,end=”)
for k in range(temp):
print(count,end=”*”)
count+=1
count=count-temp
for l in range(temp,2*temp):
if l==2*temp-1:
print(temp*temp+count,end=”)
else:
print(temp*temp+count,end=”*”)
count+=1
print()
temp-=1

• server

#include
using namespace std;

int main(int argc, char *argv[])
{
int n,m=1;
std::cin >> n;
for (int i = n; i > 0; –i) {
int k=(n*(n+1)/2)+(i*(i-1)/2)+1;
for (int y = 0; y < n-i; ++y) {
std::cout << "\t";
}
for (int j = 1; j <= i*2; ++j) {
if(j<=i)
{
std::cout << m << "\t";
++m;
}
else
{
std::cout << k++ << "\t";
}
}
std::cout << std::endl;
}
return 0;
}

• Shahla

#include
#include
using namespace std;

int main() {
int n;
cin>>n;

int cnt=n,l=1,r=n*n+1;

while(cnt)
{
for(int i=0;i<n-cnt;i++)cout<<" "; //space
for(int i=l; i<l+cnt ;i++) // first n numbers
cout<<i<<"*";

for(int j=r ; j<r+cnt-1 ;j++) //last n numbers
cout<<j<<"*";
cout<<r+cnt-1<<endl;

l+=cnt;
r-=cnt-1;
cnt–;

}

return 0;
}

• Gourav

In Python:-

input1 = int(input(“Enter the range:”))
sum = 0
total = (input1*(input1+1))
count = total
for i in range(input1):
print(” “*(i),end=””)
for k in range(input1,i,-1):
sum += 1
print(sum,end=”*”)
count = total – sum
for j in range(input1,i,-1):
count += 1
if(j==i+1):
print(count,end=””)
else:
print(count,end=”*”)
print()

• venu

venuabi python code
n=int(input())
t=(n*(n+1))+1
z=1
space=0

for i in range(0,n):
y=(t-(n-i))
for j in range(space):
print(‘ ‘,end=”)

for k in range(1,(n+1)-i):
print(z,end=’*’)
z=z+1
for l in range(y,t):
if l==(t-1):
print(l)
break
print(l,end=’*’)
space+=2
t=y

• Gaurav

# include
# include

void main()
{
int n,i,j,k,temp,count1,count2,count3=-1,temp2;
printf(“Enter number n “);
scanf (“%d”,&n);
count1=1;
count2=n*(n+1);
temp=2*n;
temp2=count2;
for(i=1;i<=n;i++)
{
for(j=2;j<=i;j++)
{
printf(" ");
}
for(k=1;k<=temp;k++)
{
if(k<=temp/2)
{
printf("%d*",count1);
count1++;
count3++;
if (k== temp/2) count2=count2-count3;
}
else
{
if(k==temp)
{
printf("%d",count2);

}
else
{
printf("%d*",count2);

}
temp2–;
count2++;
}
}
temp=temp-2;
count3=-1;
count2=temp2;
printf("\n");
}
getch();
}

• akshay achz

n=int(input())
rval=1
m=n
k=1
for i in range(n):
lval=m*m+k
for j in range(i*2):
print(“-“,end=””)
for j in range((n-i)):
print(str(rval)+”*”,end=””)
rval=rval+1
for j in range(n-i):
if(j==(n-i-1)):
print(str(lval),end=””)
else:
print(str(lval)+”*”,end=””)
lval=lval+1
m=m-1
k=k+(n-i)
print()