Pattern Question (Trapezium)

October 11, 2019

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++ – 

Table of Contents

  1. Trapezium Pattern in C
  2. Trapezium Pattern in C++
  3. Trapezium Pattern in Java

 

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 = 4
1*2*3*4*17*18*19*20
  5*6*7*14*15*16
    8*9*12*13
      10*11
If n = 5
1*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*16
If N = 2
1*2*5*6
  3*4

 

In C

#include
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;
}

In C++

#include
using namespace std;
int main(){
int n=4,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;
}

Trapezium Pattern program in Java

public class Pattern {

public static void main(String[] args) {

int count1=0,count2=0;
int N=4;
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();
}
}

}

3 comments on “Pattern Question (Trapezium)”


  • 2BA18CS009_Anand_kumbar

    n=int(input())
    m=n*(n+1)
    num=1
    l=[]
    for i in range(n):
    k=[]
    for j in range(n-i):
    k.append(num)
    num+=1
    l.append(k)
    for i in range(n,-1,-1):
    for j in range(n-i):
    l[i].append(num)
    num+=1
    for i in range(len(l)):
    for j in range(i*2):
    print(” “,end=””)
    for j in range(len(l[i])):
    if(j==len(l[i])-1):
    print(l[i][j],end=””)
    else:
    print(l[i][j],”*”,end=””,sep=””)
    print()


  • Alok

    #include
    using namespace std;

    int main() {
    int n=5;

    int start=1, end=(n*(n+1)), space;

    for(int i=0; i<n; i++)
    {
    space=i*2;

    for(int j=0; j<space; j++)
    cout<<" ";
    for(int j=0; j<n-i; j++)
    {
    if(j==0)
    cout<<start;
    else
    cout<<"*"<<start;
    start++;
    }

    end=end-(n-i)+1;
    for(int j=0; j<n-i; j++)
    {
    cout<<"*"<<end++;
    }
    end=end-(n-i+1);
    cout<<endl;
    }
    return 0;
    }


  • Homemade

    public static void main(String[] args)
    {
    int n = scan.nextInt();
    int s1 = 1;
    int s2 = 2*((n*(n+1))/2);
    int s3 = n;
    int spa = 0;
    for(int i = 0;i<n;i++) {
    for(int j = 0;j<spa;j++)
    System.out.print(" ");
    spa+=2;
    for(int j = 0;j<s3;j++) {
    System.out.print(s1+"*");
    s1++;
    }
    int s4 = s2 – s3 +1;
    for(int j = 0;j<s3;j++) {
    System.out.print(s4+"");
    if(j!=s3-1)
    System.out.print("*");
    s4++;
    s2–;
    }
    s3–;
    System.out.println();
    }