A right-angled triangle star pattern prints a left-aligned staircase of * characters: row i has exactly i stars.
Remember
Rule: on row i, print i stars
*
**
***
****
***** ← 5 rows
In C you solve it with two nested for loops: the outer loop picks the row, the inner loop prints stars on that row, then printf("\n") moves to the next line. Once this clicks, inverted triangles, pyramids, and hollow shapes become much easier.
Approach
How to Solve It
Two ways to emit the same shape — start with nested loops, then optionally shorten with putchar.
Method
Idea
Best for
Nested loops + printf
Outer = rows, inner = stars via printf("*")
Learning, interviews, exams
Nested loops + putchar
Same loops; putchar('*') per star
Slightly leaner demos once loops click
Pseudocode
Pseudocode
for i from 1 to rows:
for j from 1 to i:
print "*" (no newline)
print newline
Change the row count and the triangle updates instantly — including the triangular star total.
Whole numbers from 1 to 20. Tap a chip or type a value — the preview redraws as you go.
Live result5 rows · 15 stars
*
**
***
****
*****
Trace
Worked Walkthrough — rows = 4
Trace each outer-loop value of i and count how many times the inner loop runs.
i
Inner j
Printed row
Stars
1
1..1
*
1
2
1..2
**
2
3
1..3
***
3
4
1..4
****
4
Total star prints: 1 + 2 + 3 + 4 = 10 = 4×5/2. That triangular sum is why time is O(n²).
Code
C Programs
Three complete programs: fixed rows, scanf input, and a putchar shortcut. Use View Output to reveal sample results.
Example 1 — Fixed rows = 5
Hard-coded height — ideal for first demos and screenshots.
C
#include <stdio.h>
int main(void)
{
int rows = 5;
int i, j;
for (i = 1; i <= rows; i++)
{
for (j = 1; j <= i; j++)
{
printf("*");
}
printf("\n");
}
return 0;
}
Output
*
**
***
****
*****
How It Works
1. Set height.rows = 5 means the triangle has five lines.
2. Outer loop picks the row.i runs from 1 to rows.
3. Inner loop prints stars. For each i, j runs from 1 to i, so row i gets exactly i stars via printf("*").
4. Break the line.printf("\n") after the inner loop starts the next row.
When i = 1 you get *; when i = 2 you get **; and so on up to five stars.
Example 2 — User Input Version
Read the row count at runtime with scanf. Always check its return value in real apps (shown in the tip below).
C
#include <stdio.h>
int main(void)
{
int rows;
int i, j;
printf("Enter the number of rows: ");
scanf("%d", &rows);
for (i = 1; i <= rows; i++)
{
for (j = 1; j <= i; j++)
{
printf("*");
}
printf("\n");
}
return 0;
}
Output (when user enters 7)
Enter the number of rows: 7
*
**
***
****
*****
******
*******
How It Works
1. Prompt and read. Ask for a row count, then store it with scanf("%d", &rows).
2. Same nested-loop core. Only the source of rows changes — the print logic matches Example 1.
3. Safer input tip. Unchecked scanf leaves rows uninitialized on bad input. Prefer:
Safer input
if (scanf("%d", &rows) != 1 || rows < 1)
{
printf("Enter a positive whole number.\n");
return 1;
}
Example 3 — putchar('*')
Same nested loops — write each star with putchar instead of printf.
C
#include <stdio.h>
int main(void)
{
int rows = 5;
int i, j;
for (i = 1; i <= rows; i++)
{
for (j = 1; j <= i; j++)
{
putchar('*');
}
putchar('\n');
}
return 0;
}
Output
*
**
***
****
*****
How It Works
1. Same outer and inner bounds. Still walk i from 1 to rows and j from 1 to i.
2. One character at a time.putchar('*') writes a star without a format string.
3. End the row.putchar('\n') advances to the next line.
Learn the printf version first (Examples 1–2) so you can explain both bounds in an interview; treat putchar as a polish shortcut afterward.
Edge Cases & Pitfalls
Check these before calling the solution done.
\n inside
Column of stars
If printf("\n") is inside the inner loop, each star lands on its own line. Print stars without a newline; call printf("\n") only after the inner loop.
j <= rows
Rectangle, not triangle
Inner bound must be j <= i. j <= rows prints a filled rectangle.
No newline
One endless line
Omitting the row break glues every star onto a single line.
rows = 1
Single star
Output is just * on one line — a good sanity check.
rows ≤ 0
Empty output
Outer loop never runs. Validate and re-prompt for interactive programs.
scanf
Check the return value
If scanf fails, rows may be uninitialized — always test scanf(...) == 1.
Analysis
Time and Space Complexity
Program
Time
Extra space
Nested loops + printf (Examples 1–2)
O(rows²)
O(1)
putchar (Example 3)
O(rows²)
O(1)
Total stars printed = 1 + 2 + … + n = n(n+1)/2, which is still quadratic in n.
Remember
Key Takeaways
Rule: row i prints exactly i stars.
Two loops: outer = rows, inner = stars with printf("*").
Break the row: call printf("\n") only after the inner loop.
Complexity:O(n²) time from the triangular star count; O(1) extra space.
One line: for each row i, print i stars with printf("*"), then printf("\n").
Frequently Asked Questions
The outer loop runs i from 1 to rows. For each row i, the inner loop runs j from 1 to i and prints a star with printf. Row 1 prints 1 star, row 2 prints 2 stars, and so on.
You need one loop for which row you are on and another for how many characters belong on that row. Nested for loops express that directly.
printf("*") stays on the same line. printf("\n") ends the current line. Stars use printf("*"); the row break uses printf("\n") after the inner loop.
Reverse the outer loop so i runs from rows down to 1, for example for (i = rows; i >= 1; i--). The first line then has rows stars. See Program 2.
O(n²) where n is the number of rows. Total printf star calls equal 1+2+…+n = n(n+1)/2.
Yes. putchar('*') writes one character without a format string. Nested loops with putchar are a common, slightly leaner style; keep printf if your course examples use it.
Check scanf's return value: if (scanf("%d", &rows) != 1) handle bad input. Unchecked scanf leaves rows uninitialized on failure.
The outer loop never runs, so nothing is printed. Validate and prompt again if you want a clear user message.
🤔
Did you know?
Row i prints exactly i stars. Total stars for n rows is the triangular number n(n+1)/2 — the same count that makes this pattern O(n²).
