Reverse Alphabet Right-Angled Triangle Pattern in JavaScript

What You’ll Learn

Print a reverse alphabet right-angled triangle: each row has one more character than the previous, and letters go from a top letter down toward A — E, ED, EDC, EDCB, EDCBA. Same geometry as Program 1, but descending along the alphabet. Includes a live preview, worked JavaScript examples, edge cases, and complexity.

📝 Problem & Approach

Given a positive integer rows, print a left-aligned reverse alphabet right-angled triangle of letters with rows lines.

# First 5 rows (conceptual shape)
# E
# ED
# EDC
# EDCB
# EDCBA

Inputs & Outputs

Minimal workflow

top = "A".charCodeAt(0) + rows - 1
for i from 1 to rows:
    line = ""
    for code from top down to top - i + 1:
        line += letter
    console.log(line)

Approach comparison

🔮 Live Preview

Choose between 1 and 10 rows and draw the reverse alphabet triangle in the browser.

Press "Draw pattern".

Examples Gallery

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

const rows = 5;
const top = "A".charCodeAt(0) + rows - 1;

for (let i = 1; i <= rows; i++) {
  let line = "";
  for (let code = top; code > top - i; code--) {
    line += String.fromCharCode(code);
  }
  console.log(line);
}

E
ED
EDC
EDCB
EDCBA

let rows = parseInt(prompt("Enter the number of rows:"), 10);
if (!Number.isFinite(rows)) rows = 5;
rows = Math.max(1, Math.min(rows, 26));
const top = "A".charCodeAt(0) + rows - 1;

for (let i = 1; i <= rows; i++) {
  let line = "";
  for (let code = top; code > top - i; code--) {
    line += String.fromCharCode(code);
  }
  console.log(line);
}

Enter the number of rows: 4
D
DC
DCB
DCBA

const rows = 5;
const top = "A".charCodeAt(0) + rows - 1;

for (let i = 1; i <= rows; i++) {
  let line = "";
  for (let code = top; code > top - i; code--) {
    line += String.fromCharCode(code) + " ";
  }
  console.log(line);
}

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

🔎 Worked Walkthrough — rows = 5

Trace each outer value of i and the descending codes the inner loop prints from top = E.

*range stops before the end value, so top - 5 is below A and all five letters print. Total letter prints: 1 + 2 + 3 + 4 + 5 = 15 = 5×6/2.

🎯 Practice Problems

Try these variations to lock in the pattern.

Notes

⏱️ Time and Space Complexity

Row k prints k letters; summing 1..n gives n(n+1)/2 character writes.

💡 Best Practices