Decreasing & Increasing Alphabet Rows in C

What You’ll Learn

Build fixed-width alphabet rows from two parts: a short descending prefix (row letter down to B) plus an ascending suffix (A up to a computed cap) — ABCDE, BABCD, CBABC, DCBAB, EDCBA. Compare Program 24 (palindrome split) and Program 26 (rotations). Includes a live preview, worked C examples, edge cases, and complexity.

📝 Problem & Approach

Given an end letter (or fixed E), print n+1 fixed-width rows where a descending prefix grows and an ascending suffix shrinks.

// Five rows (end = E, width 5)
// ABCDE
// BABCD
// CBABC
// DCBAB
// EDCBA

Inputs & Outputs

Minimal workflow

n = end - 'A'
for i from 0 to n:
    for j from i down to 1:      // descending prefix (skip A)
        print letter[j]
    for k from 0 to n - i:       // ascending suffix
        print letter[k]
    print newline

Approach comparison

🔮 Live Preview

Choose an end letter from A to F and draw the decreasing/increasing alphabet rows in the browser.

Press "Draw pattern".

Examples Gallery

Three complete C programs — fixed A–E, scanf end letter, and a print_row helper. Click View Output to reveal sample console results.

#include <stdio.h>

int main() {
    char alpha[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    int i, j, k;

    for (i = 0; i <= 4; i++) {
        for (j = i; j > 0; j--)
            printf("%c", alpha[j]);

        for (k = 0; k <= 4 - i; k++)
            printf("%c", alpha[k]);

        printf("\n");
    }

    return 0;
}

ABCDE
BABCD
CBABC
DCBAB
EDCBA

#include <stdio.h>

int main() {
    char alpha[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    char end;
    int n, i, j, k;

    printf("Enter end letter (like E): ");
    scanf(" %c", &end);

    n = end - 'A';

    for (i = 0; i <= n; i++) {
        for (j = i; j > 0; j--)
            printf("%c", alpha[j]);

        for (k = 0; k <= n - i; k++)
            printf("%c", alpha[k]);

        printf("\n");
    }

    return 0;
}

Enter end letter (like E): C
ABC
BAB
CBA

#include <stdio.h>

void print_row(char alpha[], int n, int i) {
    int j, k;

    for (j = i; j > 0; j--)
        printf("%c", alpha[j]);

    for (k = 0; k <= n - i; k++)
        printf("%c", alpha[k]);

    printf("\n");
}

int main() {
    char alpha[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    int n = 4;
    int i;

    for (i = 0; i <= n; i++)
        print_row(alpha, n, i);

    return 0;
}

ABCDE
BABCD
CBABC
DCBAB
EDCBA

🔎 Worked Walkthrough — End = E (n = 4)

Trace each row index, both parts, and the joined 5-letter line.

Width is always 4 + 1 = 5. The last row is a full reverse run ending at a single A.

🎯 Practice Problems

Try these variations to lock in the pattern.

Notes

⏱️ Time and Space Complexity

For n+1 letters (indexes 0..n) there are n+1 rows and each row prints n+1 characters, so total work is O(n²).

💡 Best Practices