C++ Descending Number Triangle Pattern (Odd Length)
Beginner
5 min read
Updated: Sep 2026
3 programs
Live preview
Definition
What Is This Pattern?
An odd-length descending number triangle prints 1..i only for odd widths, stepping the outer loop by 2 — 1234567, 12345, 123, 1.
Remember
Rule: for i from max down to 1 step 2
for j from 1 to i: print j
1234567
12345
123
1 ← max = 7 (odd widths only)
Follows the alternating zigzag triangle in Program 13; next is the alternating binary triangle in Program 15.
Approach
How to Solve It
Outer loop visits odd lengths with i -= 2. Inner loop always prints ascending digits 1..i.
Method
Idea
Best for
Step-by-2 outer loop
Outer i = max..1 with i -= 2; inner 1..i
Learning, interviews, exams
Full descending triangle
Change step to i-- to include even widths
Comparing loop steps side by side
Pseudocode
Pseudocode
for i from max down to 1 step 2:
for j from 1 to i:
print j
print newline
Cheat sheet
Goal
Pattern
Odd lengths only
for (i = max; i >= 1; i -= 2)
Print 1..i
for (j = 1; j <= i; j++) cout << j;
End of row
cout << "\n";
Force odd max
if (max % 2 == 0) max -= 1;
Include even widths
for (i = max; i >= 1; i--)
Printing Numbers vs Starting a New Line
API
Effect
Use for
cout << j
Stays on the same line
Each digit on the row
cout << "\n"
Ends the current line
After the inner loop finishes
Print digits without a newline, then end the row once.
Try it
Live Preview
Change the odd maximum width and the pattern updates instantly — even values are adjusted down to the nearest odd width.
Whole numbers from 1 to 9. Prefer odd widths; even input is reduced by 1 for this demo.
Live resultmax = 7 · 16 digits
1234567
12345
123
1
Trace
Worked Walkthrough — max = 7
Trace how i -= 2 visits only odd lengths while the inner loop always prints 1..i.
i
Inner j
Prints
7
1..7
1234567
5
1..5
12345
3
1..3
123
1
1..1
1
Even widths 6, 4, 2 never appear. Digit total for odd n is ((n+1)/2)² → still O(n²).
Code
C++ Programs
Three complete programs: fixed max = 7, cin input (forced odd), and a compact max = 5 demo. Use View Output to reveal sample results.
Example 1 — Fixed max = 7
Hard-coded odd maximum — outer loop steps by 2; inner loop prints 1..i.
C++
#include <iostream>
using namespace std;
int main()
{
int max = 7;
int i, j;
for (i = max; i >= 1; i -= 2)
{
for (j = 1; j <= i; j++)
cout << j;
cout << "\n";
}
return 0;
}
Output
1234567
12345
123
1
How It Works
1. Outer skips even widths.i visits 7, 5, 3, 1 because of i -= 2.
2. Inner prints ascending digits.j always runs from 1 to i on that row.
3. Newline. Call cout << "\n" only after the inner loop finishes the row.
Example 2 — User Input Max
Read max with cin, validate, force odd if needed, then run the same nested loops.
C++
#include <iostream>
using namespace std;
int main()
{
int max;
int i, j;
cout << "Enter an odd maximum (e.g. 7): ";
if (!(cin >> max) || max <= 0)
{
cout << "Please enter a positive integer.\n";
return 1;
}
if (max % 2 == 0)
max -= 1;
for (i = max; i >= 1; i -= 2)
{
for (j = 1; j <= i; j++)
cout << j;
cout << "\n";
}
return 0;
}
Output (when user enters 5)
Enter an odd maximum (e.g. 7): 5
12345
123
1
How It Works
1. Prompt and validate. Reject failed reads and non-positive values before printing.
2. Force odd. Even input loses one so the first row stays an odd width.
3. Same core. Outer step-by-2 / inner 1..i matches Example 1 — only max comes from the user.
Example 3 — Compact max = 5
Same structure with a smaller odd maximum — easy to confirm the skipped even widths.
C++
#include <iostream>
using namespace std;
int main()
{
int max = 5;
int i, j;
for (i = max; i >= 1; i -= 2)
{
for (j = 1; j <= i; j++)
cout << j;
cout << "\n";
}
return 0;
}
Output
12345
123
1
How It Works
1. Three odd rows.i = 5 → 12345; i = 3 → 123; i = 1 → 1.
2. Trace on paper. If you use i-- instead, you also get 1234 and 12 — the even widths return.
3. Scale up next. Once the small demo is clear, use Examples 1–2 for max = 7 or user input.
Edge Cases & Pitfalls
Check these before calling the solution done.
i--
Even widths return
Using i-- instead of i -= 2 prints every length. Keep the step of 2 for odd-only rows.
even max
First row is even
Starting at 8 with i -= 2 yields even widths. Force odd first: if (max % 2 == 0) max -= 1;.
\n inside
Broken rows
If cout << "\n" sits inside the inner loop, each digit lands on its own line. Call it only after the row finishes.
max = 1
Single digit
Output is just 1 — a good sanity check for input validation.
max > 9
Multi-digit width
Digits 10+ make tight rows hard to read. Cap demos at 9 or switch to spaced cout << j << " ".
cin
Check the stream
Validate cin >> max before looping — a failed read leaves max unset or stale.
Analysis
Time and Space Complexity
Program
Time
Extra space
Fixed / input (Examples 1–2)
O(n²)
O(1)
Compact max = 5 (Example 3)
O(n²)
O(1)
For odd n, digits printed = 1 + 3 + … + n = ((n+1)/2)² → still O(n²) time. Only a few loop variables are needed.
Remember
Key Takeaways
Step by 2: outer i -= 2 visits only odd row lengths.
Always 1..i: inner loop prints ascending digits on every odd-width row.
Break the row:cout << j in the inner loop; cout << "\n" after it.
Complexity:O(n²) time; O(1) extra space. Prefer an odd max.
One line: for each odd i from max down to 1, print 1..i, then newline.
Frequently Asked Questions
An odd-length descending triangle: for max=7 you get 1234567, 12345, 123, 1 — even widths like 6, 4, 2 are skipped.
The outer loop uses i -= 2, so it visits only odd widths: 7, 5, 3, 1.
Because i -= 2 skips even row lengths. After 1234567 (7 digits), the next row is 5 digits (12345), not 6.
Program 13 alternates ascending/descending with i % 2. Program 14 always prints 1..i but only for odd row lengths using i -= 2.
Change the outer loop step to i--. Then you print every width: 7, 6, 5, 4, 3, 2, 1.
cout << j stays on the same line. cout << "\n" ends the current line. Digits use cout << j; the row break uses cout << "\n" after the inner loop.
Force it odd first: if (max % 2 == 0) max -= 1; so the first row stays odd-length.
O(n²) for maximum width n. You print about 1+3+5+…+n digits, which is still O(n²).
🤔
Did you know?
Only odd-length rows print. The outer loop uses i −= 2 (7, 5, 3, 1) and the inner loop prints 1..i — still O(n²) total digit prints for maximum width n.
