Odd-Length Alphabet Triangle in JavaScript

What You’ll Learn

Each row is a block of letters from A through the next “odd step” in the alphabet: A, ABC, ABCDE, ABCDEFG, ABCDEFGHI. The outer loop steps the end letter by 2 (A, C, E, G, I) with i += 2. Compare Program 1 (step 1) and Program 13 (running counter). Includes a live preview, worked JavaScript examples, edge cases, and complexity.

📝 Problem & Approach

Given a row count r (or a fixed odd-step ending letter like 'I'), print a left-aligned triangle of alphabet prefixes with odd lengths.

// First 5 rows (conceptual shape)
// A
// ABC
// ABCDE
// ABCDEFG
// ABCDEFGHI

Inputs & Outputs

Minimal workflow

for i from "A".charCodeAt(0) to end step 2:
    for j from "A".charCodeAt(0) to i:
        print String.fromCharCode(j) (no newline)
    console.log the row

Approach comparison

🔮 Live Preview

Choose a row count between 1 and 13 and draw the odd-length alphabet triangle in the browser.

Press "Draw pattern".

Examples Gallery

Three complete JavaScript programs - fixed ending letter, prompt for ending letter, and a row-count driven variant. Click View Output to reveal sample console results, or Try it Yourself to run in the browser editor.

const base = "A".charCodeAt(0);
const last = "I".charCodeAt(0);

for (let i = base; i <= last; i += 2) {
  let line = "";
  for (let j = base; j <= i; j++) {
    line += String.fromCharCode(j);
  }
  console.log(line);
}

A
ABC
ABCDE
ABCDEFG
ABCDEFGHI

const raw = (prompt("Enter the ending letter (odd step like I):") || "").trim().toUpperCase();
const end = raw ? raw.charCodeAt(0) : "I".charCodeAt(0);
const base = "A".charCodeAt(0);

for (let i = base; i <= end; i += 2) {
  let line = "";
  for (let j = base; j <= i; j++) {
    line += String.fromCharCode(j);
  }
  console.log(line);
}

Enter the ending letter (odd step like I): E
A
ABC
ABCDE

const rows = 5;
const base = "A".charCodeAt(0);

for (let row = 1; row <= rows; row++) {
  const end = base + 2 * (row - 1);
  let line = "";
  for (let j = base; j <= end; j++) {
    line += String.fromCharCode(j);
  }
  console.log(line);
}

A
ABC
ABCDE
ABCDEFG
ABCDEFGHI

🔎 Worked Walkthrough — through 'I'

Trace each outer-loop value of i and count how many letters the inner loop prints.

Total letter prints: 1 + 3 + 5 + 7 + 9 = 25 = 5².

🎯 Practice Problems

Try these variations to lock in the pattern.

Notes

⏱️ Time and Space Complexity

Because 1+3+…+(2r-1)=r², the letter count is exactly a perfect square.

💡 Best Practices