An increasing triangle from 1 grows one value per row — each cell is i + j - 1, so row i starts at i and counts upward.
Rule: for i = 1..rows
for j = 1..i
print (i + j - 1) + " "
1
2 3
3 4 5
4 5 6 7
5 6 7 8 9 ← rows = 5Twin of Program 32 (starts at 11 with 9 + i + j): same nested shape, different formula — next is Program 34’s i + j from 0.
Walk i from 1 to rows; on each row append i values using i + j - 1.
| Method | Idea | Best for |
|---|---|---|
| Nested loops | Outer i; inner j = 1..i; value i + j - 1 | Learning, interviews |
| prompt rows | Same loops; read rows at runtime | 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 += (i + j - 1) + " "
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 += (i + j - 1) + " "; |
| End row | console.log(line); |
| First cell of row | i + 1 - 1 = i |
| vs Program 32 | 9 + i + j (starts at 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 i + j - 1 triangle updates instantly.
rows = 3Trace each outer i, the formula i + j - 1, and the printed row.
i | Values i+j-1 | Printed row |
|---|---|---|
1 | 1+1-1 = 1 | 1 |
2 | 2, 3 | 2 3 |
3 | 3, 4, 5 | 3 4 5 |
Each next row starts at i and adds one more consecutive value.
Three complete programs: fixed rows = 5, prompt rows, 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 i + j - 1.
const rows = 5;
for (let i = 1; i <= rows; i++) {
let line = "";
for (let j = 1; j <= i; j++) {
line += (i + j - 1) + " ";
}
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 (i + j - 1) + " ".
3. Newline. console.log(line) after the inner loop prints the row.
Read rows at runtime with prompt and parseInt.
const rows = parseInt(prompt("Enter rows:"), 10);
if (!Number.isFinite(rows) || rows < 1) {
console.log("Please enter a positive integer.");
} else {
for (let i = 1; i <= rows; i++) {
let line = "";
for (let j = 1; j <= i; j++) {
line += (i + j - 1) + " ";
}
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 loops. Only the source of rows changes — still print i + j - 1 on each row.
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 += (i + j - 1) + " ";
}
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 23 instead of 2 3.
Using j <= rows makes every row the same width. Keep j <= i.
Output is just 1.
Using i + j starts at 2; 9 + i + j is Program 32. Keep i + j - 1.
Letters or empty prompt yield NaN — check Number.isFinite(rows) && rows >= 1.
| 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).
i + j - 1 for j = 1..i on each row.j = 1, the value equals i.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 i + j - 1.
Each printed value is computed as i + j - 1. Row i = 1 prints 1; row i = 4 prints 4, 5, 6, 7 — a left-shifted increasing triangle starting at 1.
Continue with the next pattern in the JavaScript number-pattern series.
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