Shifting-Start Alphabet Pattern in JavaScript

What You’ll Learn

The shifting-start alphabet pattern keeps a fixed end letter on every row while the start letter moves right each line. This tutorial covers the shape rule, fixed end formula, a live preview, algorithm steps, worked JavaScript examples, edge cases, and complexity.

📝 Problem & Approach

Given a positive integer rows, print a left-aligned shifting-start alphabet pattern: each row starts one letter later, but every row ends at the same fixed letter (E when rows = 5).

# First 5 rows (conceptual shape)
# ABCDE
# BCDE
# CDE
# DE
# E

Inputs & Outputs

Minimal workflow

end = "A".charCodeAt(0) + rows - 1
for i from 0 to rows - 1:
    start = "A".charCodeAt(0) + i
    for code from start to end:
        append letter to line
    log line

Approach comparison

🔮 Live Preview

Choose a row count between 1 and 26 and draw the shifting-start 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) 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 end = base + rows - 1; // 'E' when rows = 5

for (let i = 0; i < rows; i++) {
  const start = base + i;
  let line = "";
  for (let code = start; code <= end; code++) {
    line += String.fromCharCode(code);
  }
  console.log(line);
}

ABCDE
BCDE
CDE
DE
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 end = base + rows - 1;

for (let i = 0; i < rows; i++) {
  const start = base + i;
  let line = "";
  for (let code = start; code <= end; code++) {
    line += String.fromCharCode(code);
  }
  console.log(line);
}

Enter the number of rows (max 26): 4
ABCD
BCD
CD
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));
}

ABCDE
BCDE
CDE
DE
E

🔎 Worked Walkthrough — rows = 5

Trace each outer-loop value i (0-based) and see what the inner loop prints from start through fixed end = E.

Total letter prints: 5 + 4 + 3 + 2 + 1 = 15 = 5×6/2. Same triangular total as Programs 1 and 5 — only which letter is fixed differs.

🎯 Practice Problems

Try these variations to lock in the shifting-start pattern.

Notes

⏱️ Time and Space Complexity

💡 Best Practices