An increasing odd-length number rows pattern prints digits 1..i on each row, where i steps by 2 — so row lengths are 1, 3, 5, 7, 9 with no spaces between digits.
Remember
Rule: for odd i = 1, 3, 5, …, max
print digits 1 through i (no spaces)
1
123
12345
1234567
123456789 ← max = 9
The key idea is the outer step i += 2. Same nested-loop idea as a growing triangle, but only odd widths. Follows the jump triangle in Program 21; next up is Program 23.
Approach
How to Solve It
Start with consecutive digits 1..i, then optionally print only odd digits with an inner j += 2.
Method
Idea
Best for
Outer i += 2, inner 1..i
Odd row lengths; print every digit up to i
Learning, interviews, exams
Inner j += 2
Same outer step; print only odd digits
Variant once the core loops click
Pseudocode
Pseudocode
for i from 1 to max step 2:
for j from 1 to i:
print j (no newline, no space)
print newline
Cheat sheet
Goal
Pattern
Odd row lengths
for (i = 1; i <= max; i += 2)
Print 1..i
for (j = 1; j <= i; j++) printf("%d", j);
End the row
printf("\n");
Keep max odd
if (max % 2 == 0) max -= 1;
Odd digits only
for (j = 1; j <= i; j += 2)
Printing Numbers vs Starting a New Line
API
Effect
Use for
printf("%d", j)
Stays on the same line
Each digit
printf("\n")
Ends the current line
After the inner loop
Print digits without a newline, then end the row once.
Try it
Live Preview
Change the maximum odd length and the pattern updates instantly — even values snap down to the previous odd.
Whole numbers from 1 to 19. Even inputs become the previous odd (e.g. 8 → 7).
Live resultmax 9 · 5 rows · 25 digits
1
123
12345
1234567
123456789
Trace
Worked Walkthrough — max = 7
Trace each outer-loop value of i (1, 3, 5, 7) and list the digits the inner loop prints.
i
Inner j
Printed row
Digits
1
1..1
1
1
3
1..3
123
3
5
1..5
12345
5
7
1..7
1234567
7
Total digit prints: 1 + 3 + 5 + 7 = 16 = ((7 + 1) / 2)². That odd-sum square is why time is O(n²).
Code
C Programs
Three complete programs: fixed max, scanf input, and an odd-digits-only variant. Use View Output to reveal sample results.
Example 1 — Fixed max = 9
Hard-coded maximum — outer step 2; inner loop concatenates digits with printf("%d", j).
C
#include <stdio.h>
int main(void)
{
int i, j;
for (i = 1; i <= 9; i += 2)
{
for (j = 1; j <= i; j++)
{
printf("%d", j);
}
printf("\n");
}
return 0;
}
Output
1
123
12345
1234567
123456789
How It Works
1. Outer loop picks odd lengths.i runs 1, 3, 5, 7, 9 because of i += 2.
2. Inner loop prints 1..i. For each i, j runs from 1 to i and printf("%d", j) concatenates digits.
3. Break the line.printf("\n") after the inner loop starts the next (longer) row.
When i = 1 you get 1; when i = 5 you get 12345; when i = 9 you get 123456789.
Example 2 — User Input Version
Read max with scanf. If the user enters an even value, drop to the previous odd so the last row stays odd-length.
C
#include <stdio.h>
int main(void)
{
int max;
int i, j;
printf("Enter the maximum value: ");
scanf("%d", &max);
if (max % 2 == 0)
max -= 1;
for (i = 1; i <= max; i += 2)
{
for (j = 1; j <= i; j++)
{
printf("%d", j);
}
printf("\n");
}
return 0;
}
Output (when user enters 7)
Enter the maximum value: 7
1
123
12345
1234567
How It Works
1. Prompt and read. Ask for a maximum, then store it with scanf("%d", &max).
2. Force an odd max.if (max % 2 == 0) max -= 1; — entering 8 becomes 7.
3. Same nested-loop core. Only the source of max changes — the print logic matches Example 1.
4. Safer input tip. Unchecked scanf leaves max uninitialized on bad input. Prefer:
Safer input
if (scanf("%d", &max) != 1 || max < 1)
{
printf("Enter a positive whole number.\n");
return 1;
}
Example 3 — Odd digits only (j += 2)
Keep the outer odd lengths, but print only odd digits — 1, 13, 135, …
C
#include <stdio.h>
int main(void)
{
int i, j;
for (i = 1; i <= 9; i += 2)
{
for (j = 1; j <= i; j += 2)
{
printf("%d", j);
}
printf("\n");
}
return 0;
}
Output
1
13
135
1357
13579
How It Works
1. Same outer step.i still walks 1, 3, 5, 7, 9.
2. Inner step skips evens.j += 2 prints 1, 3, 5, … up to i instead of every digit.
3. Learn consecutive digits first. Use Examples 1–2 for the standard pattern; treat this as a small variation afterward.
Edge Cases & Pitfalls
Check these before calling the solution done.
i++
Every length, not odd
If the outer loop uses i++ instead of i += 2, you get all lengths 1..max. Keep the step at 2 for this pattern.
printf space
Spaced digits
printf("%d ", j) inserts spaces. Use printf("%d", j) with no trailing space for 12345.
\n inside
Column of digits
If printf("\n") is inside the inner loop, each digit lands on its own line. Call the newline only after the inner loop.
even max
Last row even
Without max -= 1 on even input, the last printed length is still odd (loop stops at the largest odd ≤ max), but clamping makes the intent explicit for students.
max = 1
Single digit
Output is just 1 — a good sanity check.
scanf
Check the return value
If scanf fails, max may be uninitialized — always test scanf(...) == 1.
Analysis
Time and Space Complexity
Program
Time
Extra space
Consecutive digits (Examples 1–2)
O(max²)
O(1)
Odd digits only (Example 3)
O(max²)
O(1)
Total digits for odd max n = 1 + 3 + … + n = ((n + 1) / 2)² — still quadratic in n.
Remember
Key Takeaways
Rule: for each odd i, print digits 1 through i with no spaces.
Step size:i += 2 is what makes lengths 1, 3, 5, …
Break the row: call printf("\n") only after the inner loop.
Complexity:O(n²) time from the odd-sum digit count; O(1) extra space.
One line: for odd i up to max, print 1..i with printf("%d", j), then printf("\n").
Frequently Asked Questions
Because the outer loop uses i += 2, so i becomes 1, 3, 5, 7, 9 — each row prints one more odd count of digits than the previous.
printf("%d", j) prints each digit immediately after the previous one on the same line — no space character is added.
Yes. Change the inner loop to j += 2 and print j to output 1, 13, 135, 1357, 13579 (see Example 3).
printf("%d", j) stays on the same line. printf("\n") ends the current line. Digits use printf("%d", j); the row break uses printf("\n") after the inner loop.
Because the outer loop condition is i <= 9. Change 9 to any odd maximum to extend the pattern.
Subtract 1 to make it odd (see Example 2) so the last row still has an odd length.
O(n²) for maximum row length n because total prints are 1+3+5+...+n = ((n+1)/2)².
Check scanf's return value: if (scanf("%d", &max) != 1) handle bad input. Unchecked scanf leaves max uninitialized on failure.
🤔
Did you know?
Each row length increases by 2 because the outer loop uses i += 2 (1, 3, 5, 7, 9). The inner loop prints 1..i with no spaces — total prints grow as O(n²) for maximum row length n.
