Right-Aligned Sequential Pyramid in JavaScript

What You’ll Learn

Print letters in a running sequence (A, then B C, then D E F…) while keeping the triangle right-aligned by printing empty 2-column cells first. View output in a monospace terminal because alignment relies on fixed-width cells. Compare Program 13 (sequential, left-aligned) and Program 20 (right-aligned reverse). Includes a live preview, worked JavaScript examples, edge cases, and complexity.

📝 Problem & Approach

Given a row count n (or fixed 5), print a right-aligned pyramid of continuous alphabet letters in 2-column cells.

// Five rows (monospace; each cell is width 2)
//         A
//       B C
//     D E F
//   G H I J
// K L M N O

Inputs & Outputs

Minimal workflow

k = "A".charCodeAt(0)
for i in 1..n:
    line = ""
    for j from n down to 1:
        if j > i: line += "  "
        else: line += String.fromCharCode(k++).padStart(2, " ")
    console.log(line)

Approach comparison

🔮 Live Preview

Choose between 1 and 6 rows and draw the right-aligned sequential pyramid in the browser (monospace cells).

Press "Draw pattern".

Examples Gallery

Three complete JavaScript programs - fixed 5 rows, user-chosen row count, and an explicit pad-then-letters rewrite. Click View Output to reveal sample console results, or Try it Yourself to run in the browser editor.

let k = "A".charCodeAt(0);

for (let i = 1; i <= 5; i++) {
  let line = "";
  for (let j = 5; j >= 1; j--) {
    if (j > i) {
      line += "  ";
    } else {
      line += String.fromCharCode(k++).padStart(2, " ");
    }
  }
  console.log(line);
}

        A
      B C
    D E F
  G H I J
K L M N O

let n = parseInt(prompt("Enter number of rows (like 5):"), 10);
if (!Number.isFinite(n)) {
  console.log("Please enter a whole number.");
} else {
  n = Math.max(1, Math.min(n, 6));
  let k = "A".charCodeAt(0);

  for (let i = 1; i <= n; i++) {
    let line = "";
    for (let j = n; j >= 1; j--) {
      if (j > i) {
        line += "  ";
      } else {
        line += String.fromCharCode(k++).padStart(2, " ");
      }
    }
    console.log(line);
  }
}

Enter number of rows (like 5): 3
    A
  B C
D E F

const n = 5;
let k = "A".charCodeAt(0);

for (let i = 1; i <= n; i++) {
  let line = "";
  for (let s = 0; s < n - i; s++) {
    line += "  ";
  }
  for (let L = 0; L < i; L++) {
    line += String.fromCharCode(k++).padStart(2, " ");
  }
  console.log(line);
}

        A
      B C
    D E F
  G H I J
K L M N O

🔎 Worked Walkthrough — 5 rows

Trace each row’s pads, letters, and the running counter range.

Total letters: 1+2+3+4+5 = 15 (A through O). Each cell is 2 columns wide.

🎯 Practice Problems

Try these variations to lock in the pattern.

Notes

⏱️ Time and Space Complexity

Each of n rows scans n cells (or pads + letters totaling n), so total work is O(n²).

💡 Best Practices