Triangle with Reverse Starting Letter in JavaScript

What You’ll Learn

Print an alphabet triangle where each row starts one letter earlier, but letters still run forward to a fixed top — E, DE, CDE, BCDE, ABCDE. Mixes ideas from Program 1 (forward run) and Program 2 (moving start). 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 triangle of forward alphabet suffixes ending at a fixed top.

// Five rows (top = E)
# E
# DE
# CDE
# BCDE
# ABCDE

Inputs & Outputs

Minimal workflow

top = "A".charCodeAt(0) + rows - 1
for i from top down to base:      // start letter
    line = ""
    for j from i up to top:      // forward run
        line += String.fromCharCode(j)
    console.log(line)

Approach comparison

🔮 Live Preview

Choose 1–10 rows and draw the reverse-starting-letter alphabet triangle in the browser.

Press "Draw pattern".

Examples Gallery

Three complete JavaScript programs - fixed A–E, user-chosen row count, and a spaced-letter variant. Click View Output to reveal sample console results, or Try it Yourself to run in the browser editor.

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

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

E
DE
CDE
BCDE
ABCDE

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

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

Enter the number of rows: 4
D
CD
BCD
ABCD

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

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

E 
DE 
CDE 
BCDE 
ABCDE 

🔎 Worked Walkthrough — Top = E (5 rows)

Trace each start letter and the resulting forward suffix.

Row lengths are 1, 2, 3, 4, 5. The right edge is always E.

🎯 Practice Problems

Try these variations to lock in the pattern.

Notes

⏱️ Time and Space Complexity

For n rows you print 1+2+…+n = n(n+1)/2 letters, so total work is O(n²).

💡 Best Practices