JavaScript Reverse Alphabet Pyramid Pattern (Slice from E)

What Is This Pattern?

A right-aligned reverse alphabet pyramid prints a reverse slice each row (E, then E D, then E D C…) with empty width-2 cells on the left so columns line up.

Rule: pad left; each row prints top down to current letter

         E
       E D
     E D C
   E D C B
 E D C B A     ← top = E

Companion to Program 22: same fixed-width grid (" " pads + padStart(2, " ") letters), but each row is a reverse slice from the top letter — not a running counter across rows.

How to Solve It

Walk the row letter from top down to A. Pad first, then append letters from top down to that row letter.

Pseudocode

base = code of 'A'
top  = code of 'E'   // or chosen top
for i from top down to base:     // row letter
    line = ""
    for j from base to i - 1:    // pad cells
        append "  "
    for j from top down to i:    // reverse slice
        append fromCharCode(j).padStart(2, " ")
    print line

Cheat sheet

Printing Letters vs Starting a New Line

Append without a newline, then end the row once.

Live Preview

Change the size (top letter) and the reverse pyramid updates instantly — including pad count on the first row.

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

Worked Walkthrough — top = D

Width is 4 cells. Trace pads and the reverse slice for each row letter.

Each of n rows does O(n) pad + letter work → O(n²) total.

JavaScript Programs

Three complete programs: fixed A–E, top-letter prompt, and an int row-index style. Use View Output for sample results, or Try It Yourself to edit and run in the playground.

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 (!/^[A-Z]$/.test(raw)) {
  console.log("Please enter a single letter A-Z.");
} 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);
  }
}

     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

Time and Space Complexity

Each of n rows does O(n) pad + letter work, so total time is quadratic in the number of letters.

Key Takeaways