Reverse Alphabet Pattern in JavaScript

What You’ll Learn

The reverse alphabet pattern prints descending letters on each row, from a fixed top letter down to a row-specific stop. This tutorial covers the shape rule, fixed top formula, reverse range step, a live preview, algorithm steps, worked JavaScript examples, edge cases, and complexity.

📝 Problem & Approach

Given a positive integer rows, print a left-aligned reverse alphabet pattern: each row prints descending letters from a fixed top letter down to a row-specific stop (EDCBA when rows = 5).

# First 5 rows (conceptual shape)
# EDCBA
# EDCB
# EDC
# ED
# E

Inputs & Outputs

Minimal workflow

top = "A".charCodeAt(0) + rows - 1
for i from 0 to rows - 1:
    stop = base + i
    for code from top down to stop (code--):
        append letter to line
    log line

Approach comparison

🔮 Live Preview

Choose a row count between 1 and 26 and draw the fixed-top reverse alphabet pattern in the browser.

Press "Draw pattern".

Examples Gallery

Three complete JavaScript programs — fixed row count, prompt input, and a letters.slice(i, rows) reverse shortcut. Click View Output to reveal sample console results, or Try it Yourself to run in the browser editor.

const rows = 5;

const base = "A".charCodeAt(0);
const top = base + rows - 1; // 'E' when rows = 5

for (let i = 0; i < rows; i++) {
  const stop = base + i;
  let line = "";
  for (let code = top; code >= stop; code--) {
    line += String.fromCharCode(code);
  }
  console.log(line);
}

EDCBA
EDCB
EDC
ED
E

const rowsInput = prompt("Enter the number of rows (max 26):");
let rows = parseInt(rowsInput, 10);
rows = Math.max(1, Math.min(rows, 26));

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

for (let i = 0; i < rows; i++) {
  const stop = base + i;
  let line = "";
  for (let code = top; code >= stop; code--) {
    line += String.fromCharCode(code);
  }
  console.log(line);
}

Enter the number of rows (max 26): 4
DCBA
DCB
DC
D

const rows = 5;
const clampedRows = Math.max(1, Math.min(rows, 26));
const letters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

for (let i = 0; i < clampedRows; i++) {
  console.log(letters.slice(i, clampedRows).split("").reverse().join(""));
}

EDCBA
EDCB
EDC
ED
E

🔎 Worked Walkthrough — rows = 5

Trace each outer-loop value i (0-based) and see what the inner loop prints from fixed top down to row stop.

Total letter prints: 5 + 4 + 3 + 2 + 1 = 15 = 5×6/2. Same triangular total as Programs 1, 4, 5, and 7 — only the letter bounds per row differ.

🎯 Practice Problems

Try these variations to lock in the fixed-top reverse pattern.

Notes

⏱️ Time and Space Complexity

💡 Best Practices