Miscellaneous Star Patterns
If you have gone through the previous articles, I can confidently say that now we are ready to solve most of the problems of star patterns. We have covered various approaches and techniques required to solve these problems. In this article, we will do mixed problems.
"Developing the attitude and basic logical aptitude for problem solving really helps in the long run."
Now, Let's start with the questions with our weapon. (A pen and paper)
Problem 1: Diamond Star Pattern
Write a C++ program to print diamond star pattern for N rows.
Input
Input number of rows: 5Output
*
***
*****
*******
*********
*******
*****
***
*Approach
Before you read further have a close look at the above pattern. And note down your observations. Try to solve it on your own as we covered the concept in previous articles.🙂
Intuition
The given pattern is a combination of pyramid star pattern and inverted pyramid star pattern. It consists of 2*N-1 rows (for this case N=5). Each row contains spaces and stars printed in increasing and decreasing order.
Stars are printed in increasing order till the Nth row. After Nth row stars are printed in decreasing order.
Spaces are printed in decreasing order till the Nth row. After the Nth row spaces are printed in increasing order. Point your mouse cursor over the pattern to count total spaces.
Now give it a try guys we have already solved various problems on similar lines in previous articles. I am very confident you will be able to solve it.😀
Explanation/Steps
- Input number of rows to print from user (in real number of rows/2). Store it in a variable say N.
- Declare two variables to keep track of total columns to print each row, say stars=1 and spaces=N-1.
- To iterate through rows, run an outer loop from 1 to 2*N-1. The loop structure should look like for(i=1; i<2*N; i++).
- To print spaces, run an inner loop from 1 to spaces. The loop structure should look like for(j=1; j<=spaces; j++). Inside this loop print single space.
- To print stars, run another inner loop from 1 to 2*stars-1. The loop structure should look like for(j=1; j<=stars; j++). Inside this loop print star.
- After printing all columns of a row, move to next line i.e. print new line.
- Check if(i < rows) then increment stars and decrement spaces. Otherwise increment spaces and decrement stars.
Solution
#include <iostream>
using namespace std;
int main()
{
int i, j, N;
int stars, spaces;
// Input number of rows from user
cout<<"Enter number of rows to print: "<<endl;
cin>>N;
stars = 1;
spaces = N - 1;
/* Iterate through rows */
for(i=1; i<2*N; i++)
{
/* Print spaces */
for(j=1; j<=spaces; j++)
cout<<" ";
/* Print stars */
for(j=1; j<stars*2; j++)
cout<<"*";
/* Move to next line */
cout<<endl;
if(i<N)
{
spaces--;
stars++;
}
else
{
spaces++;
stars--;
}
}
return 0;
}Problem 2: Right Arrow Star Pattern
Write a C++ program to print right arrow star pattern.
Input
Input number of rows: 5Output
*****
****
***
**
*
**
***
****
*****
Approach
Notice the pattern. What do you think?? Try to break down the pattern into further patterns and then try to approach the question. This pattern is a combination of two patterns hence, let’s first divide it into two parts.
*****
****
***
**
Part 1
*
**
***
****
*****Part 2
Intuition
This pattern is a combination of two patterns. We will be solving both the parts separately.
If we remove leading spaces, then the upper part star pattern is similar to inverted right triangle star pattern and bottom part to simple right triangle star pattern which we have already covered in previous articles.
For upper pattern,
For N=5
Row 1: 0 spaces // 2*current row number-2=?(2*1-2=0)
Row 2: 2 spaces // 2*current row number-2=?(2*2-2=2)
Row 3: 4 spaces // 2*current row number-2=?(2*3-2=4)
Row 4: 6 spaces // 2*current row number-2=?(2*4-2=6)
Total number of leading spaces in upper part is 2*i – 2. ( where i is current row number)
For lower pattern,
For N=5
Row 1: 8 spaces // 2*N-2*current row number=?(2*5-2*1=8)
Row 2: 6 spaces // 2*N-2*current row number=?(2*5-2*2=6)
Row 3: 4 spaces // 2*N-2*current row number=?(2*5-2*3=4)
Row 4: 2 spaces // 2*N-2*current row number=?(2*5-2*4=2)
Row 5: 0 spaces // 2*N-2*current row number=?(2*5-2*5=0)
Total number of leading spaces in lower part is 2*N – 2*i. ( where i is current row number)
Explanation/Steps
- Input number of rows to print from user. Store it in a variable say N.
- To iterate through rows for upper part, run an outer loop from 1 to N. The loop structure should look like for(i=1; i<N; i++).
- To print spaces, run an inner loop from 1 to 2*i-2. The loop structure should look like for(j=1; j<=(2*i-2); j++). Inside this loop print single space.
- To print star, run another inner loop from 1 to N-i+1. The loop structure should look like for(j=1; j<=N-i+1; j++). Inside this loop print star.
- After printing stars for current row, move to next line i.e. print new line.
- Now to iterate through rows for lower part, run an outer loop from 1 to N. The loop structure should look like for(i=1; i<=N; i++).
- To print spaces, run an inner loop from 1 to (2*N - 2*i). The loop structure should look like for(j=1; j<=(2*N - 2*i); j++). Inside this loop print single space.
- To print star, run another inner loop from 1 to i. The loop structure should look like for(j=1; j<=i; j++). Inside this loop print star.
- After printing stars for current row, move to next line i.e. print new line.
Solution
#include <iostream>
using namespace std;
int main()
{
int i, j, N; // i--> rows , j-->columns , N-->user input
// Input total number of rows from user
cout<<"Enter number of rows: "<<endl;
cin>>N;
// Print the upper part of the arrow
for(i=1; i<N; i++)
{
// Print leading (2*rownumber-2) spaces
for(j=1; j<=(2*i-2); j++)
{
cout<<" ";
}
// Print inverted right triangle star pattern
for(j=1; j<=N-i+1; j++)
{
cout<<"*";
}
cout<<endl;
}
// Print lower part of the arrow
for(i=1; i<=N; i++)
{
// Print leading (2*N - 2*rownumber) spaces
for(j=1; j<=(2*N - 2*i); j++)
{
cout<<" ";
}
// Print simple right triangle star pattern
for(j=1; j<=i; j++)
{
cout<<"*";
}
cout<<endl;
}
return 0;
}Problem 3: Plus Sign Pattern
Write a C++ program to print plus sign pattern.
Input
Input number of rows: 5Output
+
+
+
+
+++++++++
+
+
+
+Approach
Notice the pattern. What do you think?? Note down your observations and then try to solve the problem.
Intuition
- The pattern consists of 2 * N - 1 rows.
- When you look to the centre horizontal plus line i.e. +++++++++ this line. It also consists of 2 * N - 1 columns.
- For every other row, single plus symbol is printed after N - 1 blank spaces.
Now give it a shot !! It's not a difficult one.✌️
Explanation/Steps
- Input number of rows to print from user. Store it in a variable say N.
- To iterate through rows, run an outer loop from 1 to 2 * N - 1. The loop structure should look like for(i=1; i<=(2 * N - 1); i++).
- To print centre horizontal plus if(i==N), run a inner for loop inside if condition from 1 to 2 * N - 1. The loop structure should look like for(j=1; j<=(2 * N - 1); j++). Inside this loop print plus sign.
- To print spaces before single plus sign, run another loop inside the else condition from 1 to N - 1. The loop structure should look like for(j=1; j<=N-1; j++). Inside this loop print single space. After the loop ends, print a single plus sign.
- After printing all plus sign for each row, move to next line i.e. print new line.
Solution
#include <iostream>
using namespace std;
int main()
{
int i, j, N; // i--> rows , j-->columns , N-->total rows
// Input total number of rows from user
cout<<"Enter number of rows: "<<endl;
cin>>N;
// Run an outer loop from 1 to 2 * N - 1
for(i=1; i<=(2 * N - 1); i++)
{
// For the center horizontal plus
if(i == N)
{
for(j=1; j<=(2 * N - 1); j++)
{
cout<<"+";
}
}
else
{
// For spaces before single plus sign
for(j=1; j<=N-1; j++)
{
cout<<" ";
}
cout<<"+";
}
cout<<endl;
}
return 0;
}Problem 4: Cross Star Pattern
Write a C++ program to print cross star pattern.
Input
Input number of rows: 5Output
* *
* *
* *
* *
*
* *
* *
* *
* *Approach
Notice the pattern carefully. We have done enough number of questions. Can we give it a honest try without looking at the solution??
Intuition
Essentially in the given pattern, there are two diagonals crossing each other.
The pattern consists of exactly 2 * N - 1 rows and 2 * N - 1 columns. Each row contains exactly 2 * N - 1 columns.
Can you notice that stars are printed for the below two cases, otherwise print space?
- For the first diagonal, when row and column numbers both are equal.
- For the second diagonal, stars are printed if column number == 2 * N- row number.
Explanation/Steps
- The pattern consists of exactly 2 * N - 1 rows and 2 * N - 1 columns. Hence run an outer loop to iterate through rows with structure for(i=1; i<= 2 * N - 1; i++).
- Since each row contains exactly 2 * N - 1 columns. Therefore, run inner loop as for(j=1; j<=2 * N - 1; j++).
- Inside this loop, you can notice that stars are printed for the below two cases, otherwise print space.
- For the first diagonal i.e. when row and column numbers both are equal. Print star if(i == j).
- For the second diagonal, stars are printed if(j == 2 * N - i).
Solution
#include <iostream>
using namespace std;
int main()
{
int i, j, N; // i--> rows , j-->columns , N-->total rows
// Input total number of rows from user
cout<<"Enter number of rows: "<<endl;
cin>>N;
for(i=1; i<=2*N-1; i++)
{
for(j=1; j<=2*N-1; j++)
{
if(j==i || (j==2*N - i ))
{
cout<<"*";
}
else
{
cout<<" ";
}
}
cout<<endl;
}
return 0;
}Problem 5: 8 Star Pattern
Write a C++ program to print 8 star pattern of N rows.
Input
Input number of columns: 5Output
***
* *
* *
* *
***
* *
* *
* *
***Approach
Tired?? Take a break and then come back with a fresh mind.
Are we ready now? Notice the pattern carefully, this question is more about observation. Try to do it on your own at least give it a fair try.
Intuition
Take into account the below three things and we are good to go.
- Print a hollow square star pattern with N columns and 2*N- 1 rows.
- Modify hollow square pattern logic(We have already done this problem guys). Instead of star, print space at top and bottom corners and in the center intersection edge.
- Finally, modify hollow square logic to print star instead of space for the Nth row.
Explanation/Steps
- Input number of columns to print from user. Store it in a variable say N.
- Iterate through (N*2)-1 rows. Run an outer loop with structure for(i=1; i<N*2; i++).
- Iterate through N columns. Run an inner loop with structure for(j=1; j<=N; j++).
- Inside the inner loop first check conditions for corner and center intersection spaces. Print space for:
1st row and 1st or Nth column i.e. if(i==1 && (j==1 || j==N))
Nth row and 1st or Nth column i.e. if(i==N && (j==1 || j==N))
N*2-1 row and 1st or Nth column i.e. if(i==N*2-1 && (j==1 || j==N)) - Print stars for 1st row or 1st column i.e. if(i==1 || j==1), Nth row or Nth column i.e. if(i==N || j==N), N*2-1th row i.e. if(i==N*2-1).
- Otherwise, print spaces if the above conditions do not match.
Solution
#include <iostream>
using namespace std;
int main()
{
int i, j, N; // i--> rows , j-->columns , N-->size
// Input total number of rows from user
cout<<"Enter size: "<<endl;
cin>>N;
for(i=1; i<N*2; i++)
{
for(j=1; j<=N; j++)
{
// Condition for corner and center intersection space
if((i==1 && (j==1 || j==N)) ||
(i==N && (j==1 || j==N)) ||
(i==N*2-1 && (j==1 || j==N)))
{
cout<<" ";
}
else if(i==1 || i==N || i==(N*2)-1 || j==1 || j==N)
{
cout<<"*";
}
else
{
cout<<" ";
}
}
cout<<endl;
}
return 0;
}How was this exercise? Did you enjoy doing it?? I am assuming a positive response. (Am I very optimistic XD 😅)
I think now you must be comfortable with the star pattern problems as we have covered maximum number of variations. I guess now we are also able to find the pattern and similarities between different problems which is a very good thing.
"Remember Practice leads to Perfection".
Happy Learning !! 😀See you in the next article with more interesting concepts.