A palindrome number triangle prints ascending digits 1..i, then mirrors with i-1..1 — each row reads the same forwards and backwards.
Remember
Rule: for i from 1 to rows
print j from 1 to i
print k from i-1 down to 1
1
121
12321
1234321
123454321 ← rows = 5
Follows the diagonal asterisk pattern in Program 26; next is the 0-centered descending mirror in Program 28.
Approach
How to Solve It
Outer loop grows i. First inner loop prints 1..i; second prints i-1..1 so the peak digit appears once.
Method
Idea
Best for
Two inner loops
Ascend 1..i, then descend i-1..1
Learning, interviews, exams
Spaced digits
Same loops with cout << j << " "
Easier reading for larger peaks
Pseudocode
Pseudocode
for i from 1 to rows:
for j from 1 to i:
print j
for k from i - 1 down to 1:
print k
print newline
Cheat sheet
Goal
Pattern
Grow each row
for (i = 1; i <= rows; i++)
Left ascending
for (j = 1; j <= i; j++) cout << j;
Right mirror
for (k = i - 1; k >= 1; k--) cout << k;
Avoid double peak
Start the mirror at i - 1, not i
End of row
cout << "\n";
Printing Numbers vs Starting a New Line
API
Effect
Use for
cout << j / k
Stays on the same line
Each digit on the row
cout << "\n"
Ends the current line
After both inner loops finish
Print digits without a newline, then end the row once.
Try it
Live Preview
Change the row count and the palindrome triangle updates instantly — capped at 9 for readable demos.
Whole numbers from 1 to 9. Tap a chip or type a value — the preview redraws as you go.
Live resultrows = 5 · 25 digits
1
121
12321
1234321
123454321
Trace
Worked Walkthrough — rows = 5
Trace how ascending then descending halves build each palindrome row.
i
Ascending
Mirror
Prints
1
1
(none)
1
2
12
1
121
3
123
21
12321
4
1234
321
1234321
5
12345
4321
123454321
Row i prints 2i - 1 digits. Total = 1 + 3 + … + (2n-1) = n² → O(n²).
Code
C++ Programs
Three complete programs: fixed rows = 5, cin input, and a compact rows = 3 demo. Use View Output to reveal sample results.
Example 1 — Fixed rows = 5
Hard-coded height — ascend 1..i, then mirror i-1..1.
C++
#include <iostream>
using namespace std;
int main()
{
int rows = 5;
int i, j, k;
for (i = 1; i <= rows; i++)
{
for (j = 1; j <= i; j++)
cout << j;
for (k = i - 1; k >= 1; k--)
cout << k;
cout << "\n";
}
return 0;
}
Output
1
121
12321
1234321
123454321
How It Works
1. Outer grows the peak.i is both the row index and the middle digit.
2. Ascend then mirror. Print 1..i, then i-1..1 so the peak is not repeated.
3. Newline once. Call cout << "\n" only after both inner loops finish.
Example 2 — User Input Rows
Read rows with cin, validate, then use the same two-loop core.
C++
#include <iostream>
using namespace std;
int main()
{
int rows;
int i, j, k;
cout << "Enter rows: ";
if (!(cin >> rows) || rows <= 0)
{
cout << "Please enter a positive integer.\n";
return 1;
}
for (i = 1; i <= rows; i++)
{
for (j = 1; j <= i; j++)
cout << j;
for (k = i - 1; k >= 1; k--)
cout << k;
cout << "\n";
}
return 0;
}
Output (when user enters 4)
Enter rows: 4
1
121
12321
1234321
How It Works
1. Prompt and validate. Reject failed reads and non-positive values before printing.
2. Same core. Ascend + mirror matches Example 1 — only rows comes from the user.
3. Safer input tip. Cap demos for readable output:
Safer input
if (!(cin >> rows) || rows < 1 || rows > 9)
{
cout << "Enter a whole number from 1 to 9.\n";
return 1;
}
Example 3 — Compact rows = 3
Same structure with only three rows — easy to confirm the mirror start at i - 1.
C++
#include <iostream>
using namespace std;
int main()
{
int rows = 3;
int i, j, k;
for (i = 1; i <= rows; i++)
{
for (j = 1; j <= i; j++)
cout << j;
for (k = i - 1; k >= 1; k--)
cout << k;
cout << "\n";
}
return 0;
}
Output
1
121
12321
How It Works
1. Three rows.i = 1 → 1; i = 2 → 121; i = 3 → 12321.
2. Trace on paper. If the mirror starts at i instead of i - 1, you get a doubled peak (123321).
3. Scale up next. Once the small demo is clear, use Examples 1–2 for five rows or user input.
Edge Cases & Pitfalls
Check these before calling the solution done.
k = i
Doubled peak
Starting the mirror at i prints the middle digit twice (123321). Use k = i - 1.
no mirror
Missing descending loop
Without the second loop you get 1, 12, 123 — not palindromes. Keep both inners.
\n inside
Broken rows
If cout << "\n" sits inside either inner loop, the palindrome splits across lines. Call it only after both finish.
rows = 1
Single digit
Output is just 1 — the mirror loop never runs. A good sanity check for input validation.
rows > 9
Multi-digit glue
Digits run together (…91011…). Cap demos or add spaces with cout << j << " ".
cin
Check the stream
Validate cin >> rows before looping — a failed read leaves rows unset or stale.
Analysis
Time and Space Complexity
Program
Time
Extra space
Fixed / input (Examples 1–2)
O(n²)
O(1)
Compact rows = 3 (Example 3)
O(n²)
O(1)
Row i prints 2i - 1 digits. Total = 1 + 3 + … + (2n-1) = n² → O(n²) time. Only a few loop variables are needed.
Remember
Key Takeaways
Two halves: print 1..i, then mirror with i-1..1.
Start at i-1: the descending loop must not repeat the peak digit.
Break the row: digits in both inners; cout << "\n" after both finish.
Complexity:O(n²) time from n² digits; O(1) extra space.
One line: for each row i, print 1..i then i-1..1, then newline.
Frequently Asked Questions
A palindrome number triangle: for rows=5 you get 1 / 121 / 12321 / 1234321 / 123454321 — each row reads the same forwards and backwards.
Starting from i-1 avoids repeating the middle number. For i=3, printing 123 then 21 makes 12321.
The ascending loop prints 1..i and the descending loop prints i-1..1 — together they read the same forwards and backwards.
cout << j prints digits on the same line. cout << "\n" ends the row after both inner loops finish.
One loop handles the ascending half and the other handles the mirror half — clearer than a single complex loop.
Use cout << j << " " and cout << k << " " in the loops instead of cout << j.
After prompting, use if (!(cin >> rows) || rows <= 0) to reject bad input before printing.
O(n²) for n rows because row i prints 2i-1 digits and the total is 1+3+5+…+(2n-1) = n².
🤔
Did you know?
This palindrome triangle prints 1..i and then i-1..1 on each row. The second loop mirrors the first, producing outputs like 12321 and 123454321.
