Right-Aligned Reverse Pyramid in C

What You’ll Learn

Each row is a reverse suffix (A, BA, CBA, …) padded on the left so the letters line up on the right in a fixed-width column. This is the same j > i idea as the left half of Program 19, but without the second mirror loop. Compare Program 2 (reverse, left-aligned). Includes a live preview, worked C examples, edge cases, and complexity.

📝 Problem & Approach

Given a top letter (like E), print a right-aligned pyramid of reverse alphabet suffixes.

// Classic sample (top = E; leading spaces matter)
//     A
//    BA
//   CBA
//  DCBA
// EDCBA

Inputs & Outputs

Minimal workflow

for i from 'A' to top:
    for j from top down to 'A':
        if j > i: print space
        else: print j
    print newline

Approach comparison

🔮 Live Preview

Enter a top letter from A to J and draw the right-aligned reverse pyramid in the browser.

Press "Draw pattern".

Examples Gallery

Three complete C programs — fixed top E, scanf top letter, and explicit pad + suffix. . Click View Output to reveal sample console results.

#include <stdio.h>

int main() {
    int i, j;

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

    return 0;
}

    A
   BA
  CBA
 DCBA
EDCBA

#include <stdio.h>

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

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

    for (i = 'A'; i <= top; ++i) {
        for (j = top; j >= 'A'; --j) {
            if (j > i) {
                printf(" ");
            } else {
                printf("%c", j);
            }
        }
        printf("\n");
    }

    return 0;
}

Enter the top letter (like E): D
   A
  BA
 CBA
DCBA

#include <stdio.h>

int main() {
    char top = 'E';
    int width = top - 'A' + 1;
    char i, ch;
    int s, letters, pad;

    for (i = 'A'; i <= top; ++i) {
        letters = i - 'A' + 1;
        pad = width - letters;

        for (s = 0; s < pad; ++s) {
            printf(" ");
        }
        for (ch = i; ch >= 'A'; --ch) {
            printf("%c", ch);
        }
        printf("\n");
    }

    return 0;
}

    A
   BA
  CBA
 DCBA
EDCBA

🔎 Worked Walkthrough — top E

Trace each peak i and how many pads vs letters print.

Pad count is top - i (as char distance). Each row still scans 5 columns.

🎯 Practice Problems

Try these variations to lock in the pattern.

Notes

⏱️ Time and Space Complexity

With n = top − ‘A’ + 1, each of n rows scans n columns (or pads + letters totaling n), so work is O(n²).

💡 Best Practices