A bidirectional number triangle repeats one digit per row while the row shortens. Digits rise (1, 2, 3) then mirror down (2, 1).
Remember
Rule: for i from 1 to rows
val = i if i < rows - 1
else rows + 1 - i
print val, (rows - i + 1) times
11111
2222
333
22
1 ← rows = 5
Follows the centered continuous pyramid in Program 24; next is the diagonal asterisk pattern in Program 26.
Approach
How to Solve It
Outer loop grows i. Inner loop starts at i (shrinking). Map the digit with if/else or a ternary.
Method
Idea
Best for
If/else mapping
Print i while i < rows - 1; else print rows + 1 - i
Learning, interviews, exams
Ternary val
val = (i < rows - 1) ? i : (rows + 1 - i);
Same logic in one line
Pseudocode
Pseudocode
for i from 1 to rows:
if i < rows - 1:
val = i
else:
val = rows + 1 - i
for j from i to rows:
print val
print newline
Cheat sheet
Goal
Pattern
Walk rows
for (i = 1; i <= rows; i++)
Shrink length
for (j = i; j <= rows; j++)
Pick digit
val = (i < rows - 1) ? i : (rows + 1 - i);
Print digit
cout << val;
End of row
cout << "\n";
Printing Numbers vs Starting a New Line
API
Effect
Use for
cout << val
Stays on the same line
Each repeated 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 row count and the bidirectional 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 · 15 digits
11111
2222
333
22
1
Trace
Worked Walkthrough — rows = 5
Trace how shrinking length and the digit map build each row.
i
val
Count
Prints
1
1
5
11111
2
2
4
2222
3
3
3
333
4
2
2
22
5
1
1
1
Total digits = n + (n-1) + … + 1 = n(n+1)/2 → O(n²). Mirror starts when i >= rows - 1.
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 — shrink with j = i..5; map the digit with if/else.
C++
#include <iostream>
using namespace std;
int main()
{
int rows = 5;
int i, j;
for (i = 1; i <= rows; i++)
{
for (j = i; j <= rows; j++)
{
if (i < rows - 1)
cout << i;
else
cout << rows + 1 - i;
}
cout << "\n";
}
return 0;
}
Output
11111
2222
333
22
1
How It Works
1. Outer walks rows.i is both the row index and the early-row digit.
2. Inner shrinks.for (j = i; j <= rows; j++) prints fewer digits each line.
3. Mirror late rows. When i >= rows - 1, print rows + 1 - i instead of i.
Example 2 — User Input Rows
Read rows with cin, validate, then use a ternary for the digit map.
C++
#include <iostream>
using namespace std;
int main()
{
int rows;
int i, j, val;
cout << "Enter rows: ";
if (!(cin >> rows) || rows <= 0)
{
cout << "Please enter a positive integer.\n";
return 1;
}
for (i = 1; i <= rows; i++)
{
val = (i < rows - 1) ? i : (rows + 1 - i);
for (j = i; j <= rows; j++)
cout << val;
cout << "\n";
}
return 0;
}
Output (when user enters 5)
Enter rows: 5
11111
2222
333
22
1
How It Works
1. Prompt and validate. Reject failed reads and non-positive values before printing.
2. Same core. Ternary digit map + shrinking loop 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 rise then mirror.
C++
#include <iostream>
using namespace std;
int main()
{
int rows = 3;
int i, j, val;
for (i = 1; i <= rows; i++)
{
val = (i < rows - 1) ? i : (rows + 1 - i);
for (j = i; j <= rows; j++)
cout << val;
cout << "\n";
}
return 0;
}
Output
111
22
1
How It Works
1. Three rows.i = 1 → 111; i = 2 → 22; i = 3 → 1.
2. Trace on paper. If you always print i, the last rows become 22/3 instead of mirroring down.
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.
no mirror
Always printing i
Skipping the else branch makes late rows print 444/5 instead of 22/1. Keep the rows + 1 - i map.
j = 1
Wrong inner start
Starting at j = 1 keeps every row full width. For shrinking rows use j = i.
hard-code
Fixed 4 and 6
Hard-coding i < 4 and 6 - i only works for rows = 5. Prefer rows - 1 and rows + 1 - i.
\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.
rows = 1
Single digit
Output is just 1 — a good sanity check for input validation.
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)
Total digits printed = n + (n-1) + … + 1 = n(n+1)/2 → O(n²) time. Only a few loop variables are needed.
Remember
Key Takeaways
Shrink with j = i: each row prints rows - i + 1 digits.
Rise then mirror: use i early, then rows + 1 - i for the last rows.
Break the row:cout << val in the inner loop; cout << "\n" after it.
Complexity:O(n²) time from n(n+1)/2 digits; O(1) extra space.
One line: for each row i, pick the mapped digit, print it rows - i + 1 times, then newline.
Frequently Asked Questions
A bidirectional shrinking triangle: for rows=5 you get 11111 / 2222 / 333 / 22 / 1 — repeated digits that rise then mirror down while each row shortens.
For i = 5, the condition i < rows - 1 is false, so the program prints rows + 1 - i which becomes 1.
Because the inner loop runs from j = i to rows. As i increases, the inner loop executes fewer times.
Digits rise (1, 2, 3) on early rows then mirror down (2, 1) on the last rows via the rows + 1 - i mapping.
cout << val repeats the digit on the same line. cout << "\n" ends the row after the inner loop finishes.
Use val = (i < rows - 1) ? i : (rows + 1 - i) instead of hard-coding 4 and 6.
After prompting, use if (!(cin >> rows) || rows <= 0) to reject bad input before printing.
O(n²) for n rows because total prints are n+(n-1)+…+1 = n(n+1)/2.
🤔
Did you know?
This pattern prints repeated digits per row. The inner loop runs from j = i to rows, shrinking each row. The row digit is i for the first half, then switches to rows + 1 - i to produce 22 and 1.
