Start with fixed 5 levels (i <= 9), then generalize with scanf — optionally drop the indent loop for a left-aligned trace.
Method
Idea
Best for
Centered pyramid
Odd i, indent spaces, print %4d of m*m
Learning, interviews, exams
Left-aligned
Same squares; skip the indent loop
Paper tracing of m
Pseudocode
Pseudocode
m = 1
maxWidth = 2 * levels - 1
for i from 1 to maxWidth step 2:
for j from i to maxWidth - 1:
print two spaces
for k from 1 to i:
print m*m in width 4; m++
print newline
Cheat sheet
Goal
Pattern
Odd row widths
for (i = 1; i <= maxWidth; i += 2)
Center the row
for (j = i; j < maxWidth; j++) printf(" ");
Print next square
printf("%4d", m * m); m++;
End the row
printf("\n");
Levels → width
maxWidth = 2 * levels - 1
Printing Numbers vs Starting a New Line
API
Effect
Use for
printf(" ") / printf("%4d", m * m)
Stays on the same line
Indent groups and each square
printf("\n")
Ends the current line
After indent + print loops
Print spaces and squares without a newline, then end the row once.
Try it
Live Preview
Change the level count and the centered square pyramid updates instantly — each square uses a width-4 field like %4d.
Whole numbers from 1 to 5 so every square still fits in %4d. Tap a chip or type a value — the preview redraws as you go.
With maxWidth = 5, trace each odd i, indent groups, and the m values used for m*m.
i
Indent groups
m used
Squares
Printed row
1
4
1
1
1
3
2
2, 3, 4
4, 9, 16
4 9 16
5
0
5..9
25..81
25 36 49 64 81
Total squares for n levels = 1 + 3 + … + (2n - 1) = n² (here 9). That square sum is why time is O(n²).
Code
C Programs
Three complete programs: fixed 5 levels, scanf levels, and a left-aligned variant. Use View Output to reveal sample results.
Example 1 — Fixed 5 Levels
Hard-coded pyramid — i <= 9, indent with " ", print %4d of m*m.
C
#include <stdio.h>
int main(void)
{
int i, j, k;
int m = 1;
for (i = 1; i <= 9; i += 2)
{
for (j = i; j < 9; j++)
printf(" ");
for (k = 1; k <= i; k++)
{
printf("%4d", m * m);
m++;
}
printf("\n");
}
return 0;
}
1. Odd widths.i steps 1, 3, 5, 7, 9 — that is how many squares print on each row.
2. Indent then print. The j loop prints two spaces per remaining gap; the k loop prints m*m with %4d and bumps m.
3. End the line.printf("\n") runs only after both inner loops finish.
When i = 1 you get 1; when i = 3 you get 4, 9, 16.
Example 2 — User Input Levels
Read levels with scanf, set maxWidth = 2 * levels - 1, and reject invalid input.
C
#include <stdio.h>
int main(void)
{
int levels;
int i, j, k;
int m = 1;
int maxWidth;
printf("Enter number of levels: ");
if (scanf("%d", &levels) != 1 || levels < 1)
return 0;
maxWidth = 2 * levels - 1;
for (i = 1; i <= maxWidth; i += 2)
{
for (j = i; j < maxWidth; j++)
printf(" ");
for (k = 1; k <= i; k++)
{
printf("%4d", m * m);
m++;
}
printf("\n");
}
return 0;
}
Output (when user enters 3)
Enter number of levels: 3
1
4 9 16
25 36 49 64 81
How It Works
1. Prompt and validate. Exit early if scanf fails or levels < 1.
2. Derive width.maxWidth = 2 * levels - 1 replaces the hard-coded 9 in both loops.
3. Safer input tip. Prefer an explicit message instead of a silent exit:
Safer input
if (scanf("%d", &levels) != 1 || levels < 1 || levels > 5)
{
printf("Enter a whole number from 1 to 5.\n");
return 1;
}
Example 3 — Left-Aligned Pyramid
Same squares and counter — no leading spaces, easier to trace m on paper.
C
#include <stdio.h>
int main(void)
{
int i, k;
int m = 1;
for (i = 1; i <= 9; i += 2)
{
for (k = 1; k <= i; k++)
{
printf("%4d", m * m);
m++;
}
printf("\n");
}
return 0;
}
Output (first 3 rows)
1
4 9 16
25 36 49 64 81
How It Works
1. Same squares. Odd i, m*m, and %4d stay identical to Example 1.
2. Drop the indent. Removing the j loop leaves rows flush left — useful while learning the counter.
3. Add centering next. Once the sequence is clear, restore the indent loop for the full pyramid.
Edge Cases & Pitfalls
Check these before calling the solution done.
no m++
Repeated squares
Forgetting m++ prints the same square every column. Increment after each printed value.
i++
Even row widths
If the outer loop steps by 1, you get even counts and break the classic pyramid. Keep i += 2.
indent mismatch
Off-center pyramid
The indent bound must match the outer maximum (9 or maxWidth). A smaller bound leaves rows skewed.
\n inside
Broken rows
If printf("\n") sits inside the print loop, each square lands on its own line. Call the newline only after the row finishes.
%4d overflow
Wide squares
Squares at 1000+ need a wider field. Cap demos at 5 levels, or switch to %5d / long for larger pyramids.
scanf
Check the return value
If scanf fails, levels may be uninitialized — always test scanf(...) == 1.
Analysis
Time and Space Complexity
Program
Time
Extra space
Fixed / input (Examples 1–2)
O(levels²)
O(1)
Left-aligned (Example 3)
O(levels²)
O(1)
Total squares = n² because odd numbers sum to a square. For 5 levels that is 25 printed values (plus indent work also proportional to n²).
Remember
Key Takeaways
Rule: odd i, indent spaces, print %4d of m*m, then m++.
Continuous squares: never reset m between rows — the sequence keeps climbing.
Match bounds: indent loop and outer loop share the same maxWidth.
Complexity:O(n²) time from n² squares; O(1) extra space.
One line: for odd i, indent, print i squares of m*m with %4d and m++, then printf("\n").
Frequently Asked Questions
A centered pyramid of perfect squares: row 1 prints 1 (1²), row 2 prints 4 9 16 (2², 3², 4²), and so on.
The outer loop increases i by 2 each time (i += 2), so i takes odd values — each becomes the count of squares printed on that row.
An indentation loop prints two spaces before each row while j runs from i to maxWidth-1. As i grows, fewer spaces print, so wider rows stay centered.
m starts at 1 and increments after every printed square. Each value printed is m*m — the next perfect square in sequence.
Fixed-width columns keep the pyramid aligned as squares grow from 1 to 625. Without it, columns drift apart.
Increase the outer-loop maximum (e.g. i <= 11) and match the indent bound — or use levels input as in Example 2.
Program 40 alternates 1 and 0 with shrinking rows. Program 41 prints perfect squares in a centered pyramid with growing odd-width rows.
O(n²) for n levels — total prints are 1+3+5+...+(2n-1) = n².
🤔
Did you know?
Each printed value is m² from a running counter m. Row widths are odd (1, 3, 5, 7, 9) — total prints for n levels = n².
