Odd-Length Alphabet Triangle in C

What You’ll Learn

Each row is a block of letters from A through the next “odd step” in the alphabet: A, ABC, ABCDE, ABCDEFG, ABCDEFGHI. The outer loop uses i += 2 (A, C, E, G, I). Compare Program 1 (step 1) and Program 13 (running counter). Includes a live preview, worked C examples, edge cases, and complexity.

📝 Problem & Approach

Given a row count r (or a fixed odd-step ending letter like 'I'), print a left-aligned triangle of alphabet prefixes with odd lengths.

// First 5 rows (conceptual shape)
// A
// ABC
// ABCDE
// ABCDEFG
// ABCDEFGHI

Inputs & Outputs

Minimal workflow

for end in A, C, E, ... up to last:
    for ch from A to end:
        print ch (no newline)
    print newline

Approach comparison

🔮 Live Preview

Choose a row count between 1 and 13 and draw the odd-length alphabet triangle in the browser.

Press "Draw pattern".

Examples Gallery

Three complete C programs — fixed through I, ending-letter input, and a row-count formula. Click View Output to reveal sample console results.

#include <stdio.h>

int main() {
    int i, j;

    for (i = 'A'; i <= 'I'; i += 2) {
        for (j = 'A'; j <= i; ++j) {
            printf("%c", j);
        }
        printf("\n");
    }

    return 0;
}

A
ABC
ABCDE
ABCDEFG
ABCDEFGHI

#include <stdio.h>

int main() {
    int i, j;
    char endChar;

    printf("Enter the ending letter (e.g. E for A,C,E rows): ");
    scanf(" %c", &endChar);

    for (i = 'A'; i <= endChar; i += 2) {
        for (j = 'A'; j <= i; ++j) {
            printf("%c", j);
        }
        printf("\n");
    }

    return 0;
}

Enter the ending letter (e.g. E for A,C,E rows): E
A
ABC
ABCDE

#include <stdio.h>

int main() {
    int rows = 5;
    int row, j;
    char end;

    for (row = 1; row <= rows; ++row) {
        end = 'A' + 2 * (row - 1);
        for (j = 'A'; j <= end; ++j) {
            printf("%c", j);
        }
        printf("\n");
    }

    return 0;
}

A
ABC
ABCDE
ABCDEFG
ABCDEFGHI

🔎 Worked Walkthrough — through 'I'

Trace each outer-loop value of i and count how many letters the inner loop prints.

Total letter prints: 1 + 3 + 5 + 7 + 9 = 25 = 5².

🎯 Practice Problems

Try these variations to lock in the pattern.

Notes

⏱️ Time and Space Complexity

Because 1+3+…+(2r-1)=r², the letter count is exactly a perfect square.

💡 Best Practices