C Program To Print Continuous Character Pyramid Pattern
There are 2 ways to print continuous character patterns in C i.e:
- Using for loop
- Using while loop
Input:
rows = 5
Output:
A B C D E F G H I J K L M N O
1. Using for loop
Approach 1: Using Character
- Assign any character to one variable for the printing pattern.
- The first for loop is used to iterate the number of rows.
- The second for loop is used to repeat the number of columns.
- Then print the character based on the number of columns and increment the character value at each column to print a continuous character pattern.
Below is the C program to print continuous character patterns using character using for loop:
C
// C program to print continuous // character pattern using character #include <stdio.h> int main() { int i, j; // Number of rows int rows = 3; // Taking first character of alphabet // which is useful to print pattern char character = 'A' ; // This loop is used to identify // number rows for (i = 0; i < rows; i++) { // This for loop is used to // identify number of columns // based on the rows for (j = 0; j <= i; j++) { // Printing character to get // the required pattern printf ( "%c " , character); // Incrementing character value so // that it will print the next character character++; } printf ( "\n" ); } return 0; } |
A B C D E F
Time complexity: O(R*R) where R is the given number of rows.
Auxiliary space: O(1), as constant space is being used.
Approach 2: Converting a given number into a character
- Assign any number to one variable for the printing pattern.
- The first for loop is used to iterate the number of rows.
- The second for loop is used to repeat the number of columns.
- After entering into the loop convert the given number into character to print the required pattern based on the number of columns and increment the character value at each column to print a continuous character pattern.
Below is the C program to print continuous character patterns by converting numbers into a character:
C
// C program to print continuous // character pattern by converting // number in to character #include <stdio.h> // Driver code int main() { int i, j; // Number of rows int rows = 5; // Given a number int number = 65; // This loop is used to identify // number of rows for (i = 0; i < rows; i++) { // This loop is used to identify number // of columns based on the rows for (j = 0; j <= i; j++) { // Converting number in to character char character = ( char )(number); // Printing character to get the // required pattern printf ( "%c " , character); // Incrementing number value so // that it will print the next // character number++; } printf ( "\n" ); } return 0; } |
A B C D E F G H I J K L M N O
Time complexity: O(R*R) where R is the given number of rows.
Auxiliary space: O(1), as constant space is being used.
2. Using while loop:
Approach 1: Using character
The while loops check the condition until the condition is false. If the condition is true then enter into a loop and execute the statements. Below is the C program to print continuous character patterns using character:
C
// C program to print the continuous // character pattern using while loop #include <stdio.h> int main() { int i = 1, j = 0; // Number of rows int rows = 5; // Given a character char character = 'A' ; while (i <= rows) { while (j <= i - 1) { // Printing character to get // the required pattern printf ( "%c " , character); j++; // Incrementing character value // so that it will print the next // character character++; } printf ( "\n" ); j = 0; i++; } return 0; } |
A B C D E F G H I J K L M N O
Time complexity: O(R*R) where R is the given number of rows.
Auxiliary space: O(1), as constant space is being used.
Approach 2: Converting a given number into a character
Below is the C program to print a continuous character pattern by converting a given number into a character using a while loop:
C
// C program to print continuous // character pattern by converting // number in to character #include <stdio.h> int main() { int i = 1, j = 0; // Number of rows int rows = 5; // Given a number int number = 65; while (i <= rows) { while (j <= i - 1) { // Converting number in to character char character = ( char )(number); // Printing character to get the // required pattern printf ( "%c " , character); j++; // Incrementing number value so // that it will print the next // character number++; } printf ( "\n" ); j = 0; i++; } return 0; } |
A B C D E F G H I J K L M N O
Time complexity: O(R*R) where R is the given number of rows.
Auxiliary space: O(1), as constant space is being used.
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