An increasing triangle from 11 grows one value per row — each cell is base + i + j (with base 9), so the first row is 11.
Rule: for i = 1..rows
for j = 1..i
print (9 + i + j) + " "
11
12 13
13 14 15
14 15 16 17
15 16 17 18 19 ← rows = 5A natural step after Program 31’s number-star diamond — here you practice a formula-driven triangle (compare Program 33’s i + j - 1).
Walk i from 1 to rows; on each row append i values using 9 + i + j.
| Method | Idea | Best for |
|---|---|---|
| Nested loops | Outer i; inner j = 1..i; value 9 + i + j | Learning, interviews |
| prompt base + rows | Same loops; replace 9 with baseVal | Interactive practice |
| Compact rows = 3 | Same formula; easier to trace by hand | Paper walkthrough |
for i from 1 to rows:
line = ""
for j from 1 to i:
line += (9 + i + j) + " "
print line (with newline)| Goal | Pattern |
|---|---|
| Set rows | const rows = 5; |
| Outer loop | for (let i = 1; i <= rows; i++) |
| Inner loop | for (let j = 1; j <= i; j++) |
| Value | line += (9 + i + j) + " "; |
| End row | console.log(line); |
| Custom base | line += (baseVal + i + j) + " "; |
| First cell | 9 + 1 + 1 = 11 |
| API | Effect | Use for |
|---|---|---|
line += value + " " | Stays on the same row | Each number |
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 number per line.
Change the row count and the triangle from 11 updates instantly.
rows = 3Trace each outer i, the formula 9 + i + j, and the printed row.
i | Values 9+i+j | Printed row |
|---|---|---|
1 | 9+1+1 = 11 | 11 |
2 | 12, 13 | 12 13 |
3 | 13, 14, 15 | 13 14 15 |
Each next row starts one higher and adds one more value.
Three complete programs: fixed rows = 5, custom base + rows via prompt, and compact rows = 3. Use View Output for sample results, or Try It Yourself to edit and run in the playground.
rows = 5Hard-coded height — nested loops with formula 9 + i + j.
const rows = 5;
for (let i = 1; i <= rows; i++) {
let line = "";
for (let j = 1; j <= i; j++) {
line += (9 + i + j) + " ";
}
console.log(line);
}1. Outer loop. i runs from 1 to 5 — one growing row per iteration.
2. Inner loop. j runs from 1 to i; append (9 + i + j) + " ".
3. Newline. console.log(line) after the inner loop prints the row.
Read rows and baseVal at runtime; use baseVal + i + j instead of hard-coded 9.
const rows = parseInt(prompt("Enter rows:"), 10);
const baseVal = parseInt(prompt("Enter base:"), 10);
if (!Number.isFinite(rows) || rows < 1 || !Number.isFinite(baseVal)) {
console.log("Please enter valid integers.");
} else {
for (let i = 1; i <= rows; i++) {
let line = "";
for (let j = 1; j <= i; j++) {
line += (baseVal + i + j) + " ";
}
console.log(line);
}
}1. Prompt twice. Parse rows and base; reject invalid values before looping.
2. Same shape. Inner loop still runs j = 1..i — only the formula offset changes.
3. Shift freely. Base 0 starts at 2; base 9 starts at 11.
rows = 3Same nested-loop formula with a smaller row count — easy to trace on paper.
const rows = 3;
for (let i = 1; i <= rows; i++) {
let line = "";
for (let j = 1; j <= i; j++) {
line += (9 + i + j) + " ";
}
console.log(line);
}1. Same structure. Outer and inner loops match Example 1.
2. Only rows changes. Total values = 1+2+3 = 6.
3. Trace first. Walk i = 1, 2, 3 on paper before scaling to 5 or more.
Check these before calling the solution done.
If console.log is inside the inner loop, each value lands on its own line. Log only after the row is built.
Omitting + " " prints 1213 instead of 12 13.
Using j <= rows makes every row the same width. Keep j <= i.
Output is just 11.
Using 10 + i + j starts at 12; i + j - 1 is Program 33’s formula.
Letters or empty prompt yield NaN — check Number.isFinite for both rows and base.
| Program | Time | Extra space |
|---|---|---|
| Examples 1–3 | O(n²) | O(n) for the current line string |
Total values = 1 + 2 + … + n = n(n+1)/2 → O(n²).
9 + i + j for j = 1..i on each row.11; change the base to shift the triangle.line += … builds; console.log(line) breaks.O(n²) for n rows.One line: grow the row with j = 1..i and compute each cell as base + i + j.
Each printed value is computed as 9 + i + j. Row i = 1 appends 11; row i = 2 appends 12 and 13 — a left-shifted increasing triangle.
Continue with the next pattern in the JavaScript number-pattern series.
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