JavaScript Descending Number Pattern (Diagonal Asterisk)
Beginner
5 min read
Updated: Sep 2026
3 programs
Live preview
Definition
What Is This Pattern?
Each row prints digits n..1, but when i === j the digit is replaced with * — so the star slides left as the row index grows.
Remember
Rule: for i = 1..n
for j = n..1
print * if i === j, else j
5432*
543*1
54*21
5*321
*4321 ← n = 5
A classic i === j diagonal trick — a natural step after the bidirectional triangle in Program 25.
Approach
How to Solve It
Outer i = 1..n; reverse j = n..1; append * when indices match, otherwise append j.
Method
Idea
Best for
i === j swap
Star on the descending diagonal
Learning, interviews
Custom symbol
Same loops; replace * with #
Variant practice
prompt input
Same loops; read size at runtime
Interactive practice
Pseudocode
Pseudocode
for i from 1 to n:
line = ""
for j from n down to 1:
if i equals j:
line += "*"
else:
line += j
print line (with newline)
Cheat sheet
Goal
Pattern
Set size
const n = 5;
Outer loop
for (let i = 1; i <= n; i++)
Reverse digits
for (let j = n; j >= 1; j--)
Diagonal star
if (i === j) line += "*"; else line += j;
End row
console.log(line);
Custom symbol
line += "#"; (or any char) in the if branch
Printing Numbers vs Starting a New Line
API
Effect
Use for
line += j / line += "*"
Stays on the same row
Each digit or star
console.log(line)
Ends the line
After the inner loop
Build the row with +=, then break once with console.log. Putting console.log inside the inner loop prints one character per line.
Try it
Live Preview
Change size n and the diagonal asterisk pattern updates instantly.
Whole numbers from 1 to 9. Tap a chip or type a value — the preview redraws as you go.
Live resultn = 5 · chars = 25
5432*
543*1
54*21
5*321
*4321
Trace
Worked Walkthrough — n = 3
Trace each row i, where i === j fires, and the printed characters.
i
Star at j
Sequence
Printed row
1
1 (right)
3 2 *
32*
2
2 (middle)
3 * 1
3*1
3
3 (left)
* 2 1
*21
Because j counts down, a larger i matches earlier in the scan — the star drifts left.
Code
JavaScript Programs
Three complete programs: fixed n = 5, custom diagonal symbol, and prompt input. Use View Output for sample results, or Try It Yourself to edit and run in the playground.
Example 1 — Fixed n = 5
Hard-coded size — reverse inner loop prints n..1 with one * where i === j.
JavaScript
const n = 5;
for (let i = 1; i <= n; i++) {
let line = "";
for (let j = n; j >= 1; j--) {
if (i === j) {
line += "*";
} else {
line += j;
}
}
console.log(line);
}
1. Outer rows.i runs from 1 to n — also the value that becomes a star.
2. Reverse scan.j counts down from n to 1 so digits appear in descending order.
3. Swap on match. When i === j, append *; otherwise append j. Then console.log(line).
Example 2 — Custom Symbol #
Keep n = 5 but use hash instead of asterisk on the diagonal.
JavaScript
const n = 5;
const mark = "#";
for (let i = 1; i <= n; i++) {
let line = "";
for (let j = n; j >= 1; j--) {
if (i === j) {
line += mark;
} else {
line += j;
}
}
console.log(line);
}
1. Same structure. Outer and reverse inner loops match Example 1.
2. Swap the mark. Only the if-branch string changes — mark instead of "*".
3. Same diagonal. The star still lands where i === j; only the glyph differs.
Example 3 — prompt Input
Read size at runtime; both loops use the same n.
JavaScript
const n = parseInt(prompt("Enter size:"), 10);
if (!Number.isFinite(n) || n < 1) {
console.log("Please enter a positive integer.");
} else {
for (let i = 1; i <= n; i++) {
let line = "";
for (let j = n; j >= 1; j--) {
if (i === j) {
line += "*";
} else {
line += j;
}
}
console.log(line);
}
}
1. Prompt and parse. Convert the answer with parseInt(..., 10).
2. Validate first. Reject NaN or non-positive values before looping.
3. Same diagonal. The i === j swap scales automatically with the chosen n.
Edge Cases & Pitfalls
Check these before calling the solution done.
j++
Ascending digits
If j runs 1 to n, you get ascending digits and a different diagonal. Keep j counting down.
i == j
Wrong comparison
Use strict i === j. Comparing to a fixed column (e.g. j === 1) pins the star in one place.
log inside
Vertical output
If console.log is inside the inner loop, each character lands on its own line. Append with +=; log only after the loop.
n = 1
Single star
Output is just * — i and j are both 1.
n ≤ 0
Empty output
The outer loop never runs. Guard prompt input with n >= 1.
NaN input
Validate parseInt
Letters or empty prompt yield NaN — check Number.isFinite(n) && n >= 1.
Analysis
Time and Space Complexity
Program
Time
Extra space
Examples 1–3
O(n²)
O(n) for the current line string
Each of n rows appends exactly n characters → n² total → O(n²).
Remember
Key Takeaways
Rule: print n..1; swap to * when i === j.
Reverse j: counting down makes digits descend and the star drift left.
Write vs log:line += … builds; console.log(line) breaks.
Complexity:O(n²) for size n.
One line: scan digits downward, and replace the cell where row equals column with a star.
Frequently Asked Questions
Because i increases from 1 to n while j decreases from n to 1. The condition i === j becomes true at a different position each row.
j holds the descending column digit (n … 1). When i !== j, append that digit to fill the row.
Yes. Replace line += "*" with any character or string — see Example 2 with #.
line += j appends each digit or star on the same line. console.log(line) ends the row after the inner loop finishes.
j runs from n down to 1 so each row prints digits in descending order with the star at position i.
Replace 5 with n in both loops — see Example 3 and the live preview.
O(n²) for n rows because each row prints n characters using a nested loop.
Use parseInt(prompt(...), 10) and check Number.isFinite(n) && n >= 1 so bad input does not produce NaN.
Only one row prints — a single *.
🤔
Did you know?
This pattern prints descending numbers from n to 1 on each row. When the row index equals the current column value (i === j), it appends * instead of the number, creating a diagonal asterisk that moves left each row.
