A fill-with-5 number triangle prints an ascending sequence on each row, then pads the rest with the maximum value so every row has the same width n.
Remember
Rule: for i from n down to 1
print j from i to n
then print n, (i - 1) times
5 5 5 5 5
4 5 5 5 5
3 4 5 5 5
2 3 4 5 5
1 2 3 4 5 ← n = 5
Follows the alternating odd/even triangle in Program 18; next is the continuous number triangle in Program 20.
Approach
How to Solve It
Outer loop walks i from n down to 1. First inner loop prints the sequence; second pads with n.
Method
Idea
Best for
Sequence + fill
Print i..n, then pad i - 1 times with n
Learning, interviews, exams
Custom fill
Same sequence loop; second loop prints a separate fill value
Variants where padding ≠ n
Pseudocode
Pseudocode
for i from n down to 1:
for j from i to n:
print j and a space
for j from 1 to i - 1:
print n and a space
print newline
Cheat sheet
Goal
Pattern
Walk rows top → bottom
for (i = n; i >= 1; i--)
Print sequence
for (j = i; j <= n; j++) cout << j << " ";
Pad with n
for (j = 1; j < i; j++) cout << n << " ";
End of row
cout << "\n";
Custom fill
cout << fill << " "; in the second loop
Printing Numbers vs Starting a New Line
API
Effect
Use for
cout << j << " "
Stays on the same line
Each number on the row
cout << "\n"
Ends the current line
After both inner loops finish
Print numbers without a newline, then end the row once.
Try it
Live Preview
Change the width n and the fill-with-n 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 resultn = 5 · 25 numbers
5 5 5 5 5
4 5 5 5 5
3 4 5 5 5
2 3 4 5 5
1 2 3 4 5
Trace
Worked Walkthrough — n = 5
Trace how each outer i builds a sequence, then pads to width 5.
i
Sequence
Fill count
Prints
5
5
4
5 5 5 5 5
4
4 5
3
4 5 5 5 5
3
3 4 5
2
3 4 5 5 5
2
2 3 4 5
1
2 3 4 5 5
1
1 2 3 4 5
0
1 2 3 4 5
Every row prints exactly n numbers → total n² prints → O(n²).
Code
C++ Programs
Three complete programs: fixed n = 5, cin input, and a compact n = 3 demo. Use View Output to reveal sample results.
Example 1 — Fixed n = 5
Hard-coded width — sequence loop then fill loop on each descending row.
C++
#include <iostream>
using namespace std;
int main()
{
int n = 5;
int i, j;
for (i = n; i >= 1; i--)
{
for (j = i; j <= n; j++)
cout << j << " ";
for (j = 1; j < i; j++)
cout << n << " ";
cout << "\n";
}
return 0;
}
Output
5 5 5 5 5
4 5 5 5 5
3 4 5 5 5
2 3 4 5 5
1 2 3 4 5
How It Works
1. Outer descends.i starts at n (all fill) and ends at 1 (full sequence).
2. Sequence then pad. Print j from i to n, then print n exactly i - 1 times.
3. Newline once. Call cout << "\n" only after both inner loops finish.
Example 2 — User Input Width
Read n with cin, validate, then use the same sequence + fill core.
C++
#include <iostream>
using namespace std;
int main()
{
int n;
int i, j;
cout << "Enter the triangle width: ";
if (!(cin >> n) || n <= 0)
{
cout << "Please enter a positive integer.\n";
return 1;
}
for (i = n; i >= 1; i--)
{
for (j = i; j <= n; j++)
cout << j << " ";
for (j = 1; j < i; j++)
cout << n << " ";
cout << "\n";
}
return 0;
}
1. Prompt and validate. Reject failed reads and non-positive values before printing.
2. Same core. Sequence + fill matches Example 1 — only n comes from the user.
3. Safer input tip. Cap demos for readable output:
Safer input
if (!(cin >> n) || n < 1 || n > 9)
{
cout << "Enter a whole number from 1 to 9.\n";
return 1;
}
Example 3 — Compact n = 3
Same structure with width three — easy to confirm sequence vs fill counts.
C++
#include <iostream>
using namespace std;
int main()
{
int n = 3;
int i, j;
for (i = n; i >= 1; i--)
{
for (j = i; j <= n; j++)
cout << j << " ";
for (j = 1; j < i; j++)
cout << n << " ";
cout << "\n";
}
return 0;
}
Output
3 3 3
2 3 3
1 2 3
How It Works
1. Three rows.i = 3 → 3 3 3; i = 2 → 2 3 3; i = 1 → 1 2 3.
2. Count the pads. Fill runs i - 1 times — zero pads when i = 1.
3. Scale up next. Once the small demo is clear, use Examples 1–2 for n = 5 or user input.
Edge Cases & Pitfalls
Check these before calling the solution done.
no fill
Uneven row widths
Skipping the second loop leaves short top rows. Always pad i - 1 times so every row has width n.
i++
Wrong outer direction
Ascending i from 1 to n flips the pattern unless you rewrite both loops. Keep i descending for this shape.
\n inside
Broken rows
If cout << "\n" sits inside either inner loop, each number lands on its own line. Call it only after both loops finish.
n = 1
Single number
Output is just 1 — a good sanity check for input validation.
no space
Digits glued together
Use cout << j << " " so multi-digit values stay readable as n grows.
cin
Check the stream
Validate cin >> n before looping — a failed read leaves n unset or stale.
Analysis
Time and Space Complexity
Program
Time
Extra space
Fixed / input (Examples 1–2)
O(n²)
O(1)
Compact n = 3 (Example 3)
O(n²)
O(1)
Each of n rows prints exactly n numbers → n² prints → O(n²) time. Only a few loop variables are needed.
Remember
Key Takeaways
Descend first: outer i from n down to 1 builds the top-all-fill shape.
Two inner loops: sequence i..n, then pad i - 1 times with n.
Break the row:cout << j << " " in the inners; cout << "\n" after both.
Complexity:O(n²) time from n² prints; O(1) extra space.
One line: for each i from n down to 1, print i..n, pad with n, then newline.
Frequently Asked Questions
A fill-with-n rectangle of width n: for n=5 you get 5 5 5 5 5 / 4 5 5 5 5 / 3 4 5 5 5 / 2 3 4 5 5 / 1 2 3 4 5 — each row is a sequence then padded with n.
When i = 5, the sequence loop prints j = 5 once, then the fill loop runs 4 times — all values are 5, so the row is 5 5 5 5 5.
The first prints the increasing sequence i..n. The second fills remaining positions with n so every row has the same width.
When i = 1, the sequence loop prints j = 1 to 5 and the fill loop runs zero times — no padding needed.
cout << j << " " stays on the same line with a space. cout << "\n" ends the current line. Numbers use cout; the row break uses cout << "\n" after both inner loops.
Use a separate fill variable in the second loop only, while the sequence still runs up to n.
Rows will have different widths — the top row may be short while the bottom row is full length.
O(n²) for width n because each of n rows prints n numbers.
🤔
Did you know?
Each row prints an ascending sequence i..n, then pads with n so every row has width n. The second inner loop runs i - 1 times — still O(n²) total prints.
