A reverse-starting-letter alphabet triangle grows leftward: each row begins one letter earlier, but letters along the row still run forward to a fixed top.
Remember
Rule: start letter moves top→A; print forward to top
E
DE
CDE
BCDE
ABCDE ← top = 'E' (5 rows)
Mixes Program 1 (forward run) with Program 2 (moving start). The right edge stays fixed at top on every row.
Approach
How to Solve It
Two ways to emit the same shape — nested charCodeAt loops, or a growing suffix of a fixed alphabet string.
Method
Idea
Best for
Nested code loops
Outer = start code; inner = start up to top
Learning, interviews, exams
Alphabet suffix
Slice the last r letters of "ABC…top"
Shorter demos once loops click
Pseudocode
Pseudocode
top = code of 'A' + rows - 1
for start from top down to 'A':
line = ""
for code from start up to top:
append fromCharCode(code) to line
print line
Cheat sheet
Goal
Pattern
Top from rows
const top = "A".charCodeAt(0) + rows - 1;
Move the start
for (let i = top; i >= base; i--)
Print forward
for (let j = i; j <= top; j++) line += String.fromCharCode(j);
Append letters without a newline, then end the row once.
Try it
Live Preview
Change the row count and the reverse-start forward triangle updates instantly — every row ends at the same top letter.
Whole number from 1 to 10. Tap a chip or type a count — the preview redraws as you go.
Live result5 rows · top E · 15 letters
E
DE
CDE
BCDE
ABCDE
Trace
Worked Walkthrough — Top = E (5 rows)
Trace each start letter and the resulting forward suffix.
i (start)
Inner range
Printed row
E
E..E
E
D
D..E
DE
C
C..E
CDE
B
B..E
BCDE
A
A..E
ABCDE
Row lengths are 1, 2, 3, 4, 5. The right edge is always E.
Code
JavaScript Programs
Three complete programs: fixed top E, row-count prompt, and spaced letters. Use View Output for sample results, or Try It Yourself to edit and run in the playground.
Example 1 — Fixed Top E
Outer loop chooses the first letter on the row; inner loop appends forward up to E.
JavaScript
const top = "E".charCodeAt(0);
const base = "A".charCodeAt(0);
for (let i = top; i >= base; i--) {
let line = "";
for (let j = i; j <= top; j++) {
line += String.fromCharCode(j);
}
console.log(line);
}
1. Outer loop moves the start.i runs from the code of E down to A — one start letter per row.
2. Inner loop appends forward. For each row, j runs from i up to top. When i is C, that builds CDE.
3. Break the line.console.log(line) after the inner loop ends the row.
4. Full last row. When i is A, the inner loop builds the full forward run ABCDE.
Example 2 — Row Count Input
Read the number of rows and compute top = "A".charCodeAt(0) + rows - 1. Validate with parseInt and clamp to 26.
JavaScript
let rows = parseInt(prompt("Enter the number of rows:"), 10);
const base = "A".charCodeAt(0);
if (!Number.isFinite(rows) || rows < 1) {
console.log("Please enter a whole number of rows >= 1.");
} else {
if (rows > 26) rows = 26;
const top = base + rows - 1;
for (let i = top; i >= base; i--) {
let line = "";
for (let j = i; j <= top; j++) {
line += String.fromCharCode(j);
}
console.log(line);
}
}
1. Scale the top. For 4 rows, top becomes the code of D. Cap rows at 26 so top stays within A–Z.
2. Same core loops. Only the stop letter changes; the start-moves-backward rule is unchanged.
3. Safer input tip. Always check Number.isFinite(rows) before the loops — bare parseInt(prompt()) yields NaN on letters.
Example 3 — Spaced Letters
Append a trailing space after each letter so columns are easier to scan.
JavaScript
const top = "E".charCodeAt(0);
const base = "A".charCodeAt(0);
for (let i = top; i >= base; i--) {
let line = "";
for (let j = i; j <= top; j++) {
line += String.fromCharCode(j) + " ";
}
console.log(line);
}
1. Same bounds. Loop limits are unchanged — only the appended unit becomes fromCharCode(j) + " ".
2. Optional trim. Use line.trimEnd() later if you need a compact line for grading scripts.
Edge Cases & Pitfalls
Check these before calling the solution done.
Inner j--
Wrong direction on the row
If the inner loop counts down, you get Program 2 (E, ED, EDC). Keep j++ from i to top.
Outer ++
Growing from A instead
If the outer loop runs A..top with the same inner rule, you print Program 1 upside-down. Start at top and go down to A.
log inside
Column of letters
If console.log is inside the inner loop, each letter lands on its own line. Append with +=; call console.log only after the row.
rows = 1
Single A
When top is A, you print one row: A. A good sanity check.
rows > 26
Beyond Z
Cap at 26 (or reject) so top stays within A–Z.
Bad prompt
Use Number.isFinite
Bare parseInt(prompt()) yields NaN on empty or non-numeric text. Validate before the outer loop.
Analysis
Time and Space Complexity
Program
Time
Extra space
Nested loops (Examples 1–2)
O(n²)
O(n) for the current line string
Spaced letters (Example 3)
O(n²)
O(n) for the current line string
For n rows, total letters are 1 + 2 + … + n = n(n + 1)/2 — still quadratic in n.
Remember
Key Takeaways
Rule: start letter moves top → A; print forward to top.
Fixed right edge: every row ends at the same top letter.
Break the row: call console.log(line) only after the inner loop.
Complexity:O(n²) time from the triangular letter count.
One line: for each start from top down to A, append that letter forward up to top, then console.log.
Frequently Asked Questions
The outer loop moves the starting letter from E down to A. The inner loop always appends forward from that start letter up to E (or top), so the last character remains E on every row.
Program 2 prints letters descending along the row (E, ED, EDC). This program prints forward along the row (E, DE, CDE) while only the row’s first letter moves backward.
Program 1 always starts at A and grows the end letter (A, AB, ABC). This pattern grows the start letter backward while keeping the right edge fixed at top.
Yes. Read rows with prompt(), set top = "A".charCodeAt(0) + rows - 1, then loop code from top down to A and append j from code up to top.
O(n²) for n rows, because the total printed letters are 1+2+...+n = n(n+1)/2.
Use parseInt(prompt(...), 10), check Number.isFinite(rows), require n ≥ 1, and cap at 26 so the top letter stays within A–Z.
Because the inner loop always stops at top (E in the fixed example), so the last printed character is always that same letter.
Yes. Use "a".charCodeAt(0) as the base: top = "a".charCodeAt(0) + rows - 1, then loop the same way with j++ up to top.
🤔
Did you know?
This triangle changes only the starting letter of each row (E, D, C, …), while letters along the row still increase forward. In the 5-row example, every row ends at E, producing E, DE, CDE, BCDE, ABCDE.
