C Program To Print Character Pyramid Pattern
Here, we will build a C Program To Print Character Pyramid Pattern using 2 approaches i.e.
- Using for loop
- Using while loop
Input:
rows = 5
Output:
A B B C C C D D D D E E E E E
1. Using for loop
Approach 1: Assign any character to one variable for the printing pattern
The first for loop is used to iterate the number of rows and the second for loop is used to repeat the number of columns. Then print the character based on the number of columns.
Example:
C
// C program to print character pattern using character #include <stdio.h> int main() { int i, j; // input entering number of rows int rows = 5; // taking first character of alphabet // which is useful to print pattern char character = 'A' ; // first for loop is used to identify number rows for (i = 0; i < rows; i++) { // second 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); } printf ( "\n" ); // incrementing character value so that it // will print the next character character++; } return 0; } |
A B B C C C D D D D E E E E E
Time complexity: O(R*R)
Here R is given no of rows.
Auxiliary space: O(1)
As constant extra space is used.
Approach 2: Printing pattern by 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 and 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.
Example:
C
// C program to print character pattern by // converting number in to character #include <stdio.h> int main() { int i, j; // input entering number of rows int rows = 5; // given a number int number = 65; // first for loop is used to identify number rows for (i = 0; i < rows; i++) { // second for 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); } printf ( "\n" ); // incrementing number value so that it // will print the next character number++; } return 0; } |
A B B C C C D D D D E E E E E
Time complexity: O(R*R)
Here R is given no of rows.
Auxiliary space : O(1)
As constant extra space is used.
2. Using while loops:
Approach 1: The while loops check the condition until the condition is false. If the condition is true then enter into the loop and execute the statements.
Example:
C
// C program to print character pattern by // converting number into character #include <stdio.h> int main() { int i = 1, j = 0; // input entering number of rows int rows = 5; // given a character char character = 'A' ; // while loops checks the conditions until the // condition is false if condition is true then enters // in to the loop and executes the statements while (i <= rows) { while (j <= i - 1) { // printing character to get the required // pattern printf ( "%c " , character); j++; } printf ( "\n" ); // incrementing character value so that it // will print the next character character++; j = 0; i++; } return 0; } |
A B B C C C D D D D E E E E E
Time complexity: O(R*R)
Here R is given no of rows.
Auxiliary space : O(1)
As constant extra space is used.
Approach 2: Printing pattern by converting given number into character using while loop
Example:
C
// C program to print character pattern by // converting number in to character #include <stdio.h> int main() { int i = 1, j = 0; // input entering number of rows int rows = 5; // given a number int number = 65; // while loops checks the conditions until the // condition is false if condition is true then enters // in to the loop and executes the statements 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++; } printf ( "\n" ); // incrementing number value so that it // will print the next character number++; j = 0; i++; } return 0; } |
A B B C C C D D D D E E E E E
Time complexity: O(R*R) where R is given no of rows
Auxiliary space : O(1)
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