Last Updated : 10 Aug, 2022
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Here we will build a C++ Program To Print Character patterns using 2 Approaches i.e.
- Using for loop
- Using while loop
Printing 1 character pattern in C++ using different approaches.
1. Using for loop
Input:
rows = 5
Output:
AB BC C CD D D DE E E E E
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.
C++
// C++ program to print character
// pattern using character
#include <iostream>
using namespace std;
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
cout << character << " ";
}
cout << "\n";
// incrementing character value so that it
// will print the next character
character++;
}
return 0;
}
Output
A B B C C C D D D D E E E E E
Approach 2:
Printing pattern by converting given number into 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.
C++
// C++ program to print character pattern by
// converting number in to character
#include <iostream>
using namespace std;
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
cout << character << " ";
}
cout << "\n";
// incrementing number value so that it
// will print the next character
number++;
}
return 0;
}
Output
A B B C C C D D D D E E E E E
2. Using while loops
Input:
rows = 5
Output:
AB BC C CD D D DE E E E E
Approach 1:
The while loops check the condition until the condition is false. If the condition is true then it enters into the loop and executes the statements.
C++
// C++ program to print character pattern by
// converting number in to character
#include <iostream>
using namespace std;
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
cout << character << " ";
j++;
}
cout << "\n";
// incrementing character value so that it
// will print the next character
character++;
j = 0;
i++;
}
return 0;
}
Output
A B B C C C D D D D E E E E E
Approach 2:
Printing pattern by converting given number into character using while loop.
C++
// C++ program to print character pattern by
// converting number in to character
#include <iostream>
using namespace std;
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
cout << character << " ";
j++;
}
cout << "\n";
// incrementing number value so that it
// will print the next character
number++;
j = 0;
i++;
}
return 0;
}
Output
A B B C C C D D D D E E E E E
Time complexity: O(n2) where n is given no of rows
Space complexity: O(1)
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