Right-Aligned Reverse Alphabet Pyramid in JavaScript

What You’ll Learn

Companion to Program 22: same fixed-width grid (two spaces for padding + String.fromCharCode(j).padStart(2, " ") for letters), but each row prints a reverse alphabet slice from the top letter down to the current row letter. Use a monospace terminal so columns stay aligned. Compare Program 20 (right-aligned reverse without fixed-width cells). Includes a live preview, worked JavaScript examples, edge cases, and complexity.

📝 Problem & Approach

Given a top letter (or fixed E), print a right-aligned pyramid of reverse alphabet slices ending at A.

// Five rows (monospace; each cell is width 2)
//         E
//       E D
//     E D C
//   E D C B
// E D C B A

Inputs & Outputs

Minimal workflow

for i from top down to base:
    line = ""
    for j from base to (i - 1):
        line += "  "
    for j from top down to i:
        line += String.fromCharCode(j).padStart(2, " ")
    console.log(line)

Approach comparison

🔮 Live Preview

Choose a top letter from A to F and draw the right-aligned reverse pyramid in the browser (monospace cells).

Press "Draw pattern".

Examples Gallery

Three complete JavaScript programs - fixed A–E, user-chosen top letter, and an int-index rewrite. Click View Output to reveal sample console results, or Try it Yourself to run in the browser editor.

const base = "A".charCodeAt(0);
const top = "E".charCodeAt(0);

for (let i = top; i >= base; i--) {
  let line = "";
  for (let j = base; j < i; j++) {
    line += "  ";
  }
  for (let j = top; j >= i; j--) {
    line += String.fromCharCode(j).padStart(2, " ");
  }
  console.log(line);
}

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

const raw = (prompt("Enter the top letter (like E):") || "").trim().toUpperCase();
if (raw.length !== 1 || raw < "A" || raw > "Z") {
  console.log("Please enter a single letter.");
} else {
  const top = raw.charCodeAt(0);
  const base = "A".charCodeAt(0);

  for (let i = top; i >= base; i--) {
    let line = "";
    for (let j = base; j < i; j++) {
      line += "  ";
    }
    for (let j = top; j >= i; j--) {
      line += String.fromCharCode(j).padStart(2, " ");
    }
    console.log(line);
  }
}

Enter the top letter (like E): C
    C
  C B
C B A

const top = "E".charCodeAt(0);
const base = "A".charCodeAt(0);
const n = top - base + 1;

for (let row = 1; row <= n; row++) {
  let line = "";
  for (let s = 0; s < n - row; s++) {
    line += "  ";
  }
  for (let k = 0; k < row; k++) {
    line += String.fromCharCode(top - k).padStart(2, " ");
  }
  console.log(line);
}

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

🔎 Worked Walkthrough — Top = E

Trace each row’s pads, reverse slice, and printed line.

Row count = top - 'A' + 1 (5 for E). Each cell is 2 columns wide.

🎯 Practice Problems

Try these variations to lock in the pattern.

Notes

⏱️ Time and Space Complexity

Each of n rows does O(n) pad + letter work, so total work is O(n²).

💡 Best Practices