Increasing Start Letter Alphabet Triangle in C

What You’ll Learn

Print a triangle where each row starts one letter later but always ends at the same letter (E in the 5-row example): ABCDE, BCDE, CDE, DE, E. Widths go 5, 4, 3, 2, 1. Contrast Program 5 (rows restart at A) and Program 3 (reverse starting letter but still ends at E). Includes a live preview, worked C examples, edge cases, and complexity.

📝 Problem & Approach

Given a row count (or fixed end E), print a shrinking triangle where each row is a forward suffix ending at the same letter.

// Five rows (end = E)
// ABCDE
// BCDE
// CDE
// DE
// E

Inputs & Outputs

Minimal workflow

end = 'A' + rows - 1
for i from 'A' to end:          // start letter advances
    for j from i to end:        // forward suffix each row
        print j
    print newline

Approach comparison

🔮 Live Preview

Choose 1–10 rows and draw the increasing-start alphabet triangle in the browser.

Press "Draw pattern".

Examples Gallery

Three complete C programs — fixed A–E, scanf row count, and a spaced-letter variant. Click View Output to reveal sample console results.

#include <stdio.h>

int main() {
    char i, j;

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

    return 0;
}

ABCDE
BCDE
CDE
DE
E

#include <stdio.h>

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

    printf("Enter the number of rows: ");
    scanf("%d", &rows);

    endChar = (char)('A' + rows - 1);
    for (i = 'A'; i <= endChar; i++) {
        for (j = i; j <= endChar; j++) {
            printf("%c", j);
        }
        printf("\n");
    }

    return 0;
}

Enter the number of rows: 4
ABCD
BCD
CD
D

#include <stdio.h>

int main() {
    char endChar = 'E';
    char i, j;

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

    return 0;
}

A B C D E 
B C D E 
C D E 
D E 
E 

🔎 Worked Walkthrough — End = E (5 rows)

Trace each start letter and the resulting forward suffix up to E.

Row lengths are 5, 4, 3, 2, 1. The right edge is always E.

🎯 Practice Problems

Try these variations to lock in the pattern.

Notes

⏱️ Time and Space Complexity

For n rows you print n+(n-1)+…+1 = n(n+1)/2 characters, so total work is O(n²).

💡 Best Practices