Palindromic Alphabet Pyramid (BAB) in JavaScript

What You’ll Learn

Build each row as a palindrome by stitching a descending run (i down to B) with an ascending run (A up to i), so a single A sits at the center. This version prints letters with no extra spacing (unlike right-aligned patterns that use 2-column cells). Compare Program 18 (another palindrome style) and Program 16 (centering with pads). Includes a live preview, worked JavaScript examples, edge cases, and complexity.

📝 Problem & Approach

Given a row count n (or fixed A–E), print a left-aligned palindromic alphabet pyramid with peak letters growing each row.

// Five rows (left-aligned; no cell padding)
// A
// BAB
// CBABC
// DCBABCD
// EDCBABCDE

Inputs & Outputs

Minimal workflow

for i from base to peak:
    line = ""
    for j from i down while j > base:
        line += String.fromCharCode(j)
    for j from base to i:
        line += String.fromCharCode(j)
    console.log(line)

Approach comparison

🔮 Live Preview

Choose between 1 and 8 rows and draw the palindromic alphabet pyramid in the browser.

Press "Draw pattern".

Examples Gallery

Three complete JavaScript programs - fixed A–E, user-chosen row count, and a spaced 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 top = "E".charCodeAt(0);

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

A
BAB
CBABC
DCBABCD
EDCBABCDE

let n = parseInt(prompt("Enter number of rows (1..26):"), 10);
if (!Number.isFinite(n)) {
  console.log("Please enter a whole number.");
} else {
  n = Math.max(1, Math.min(n, 26));
  const base = "A".charCodeAt(0);

  for (let r = 0; r < n; r++) {
    const peak = base + r;
    let line = "";

    for (let ch = peak; ch > base; ch--) {
      line += String.fromCharCode(ch);
    }
    for (let ch = base; ch <= peak; ch++) {
      line += String.fromCharCode(ch);
    }

    console.log(line);
  }
}

Enter number of rows (1..26): 4
A
BAB
CBABC
DCBABCD

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

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

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

🔎 Worked Walkthrough — A–E

Trace each row’s left wing, center+right wing, and full line.

Lengths: 1+3+5+7+9 = 25 = 5².

🎯 Practice Problems

Try these variations to lock in the pattern.

Notes

⏱️ Time and Space Complexity

Total printed characters are 1+3+…+(2n-1) = n², so time is O(n²).

💡 Best Practices