A shrinking repeating number pattern prints digit i with a falling repeat count as i counts up — 11111, 2222, 333, 44, 5.
Remember
Rule: for i from 1 to rows
repeat print i exactly (rows - i + 1) times
11111
2222
333
44
5 ← rows = 5
Follows the inverted repeating triangle in Program 11; next is the alternating zigzag triangle in Program 13.
Approach
How to Solve It
Outer loop picks digit i (counting up). Inner loop runs from i to rows and prints i each time — that is rows - i + 1 repeats.
Method
Idea
Best for
Ascending outer + i..rows inner
Outer i = 1..rows; inner j = i..rows; print i
Learning, interviews, exams
Counted repeats
Inner j = 1..(rows - i + 1); print i
Same shape, clearer shrink formula
Pseudocode
Pseudocode
for i from 1 to rows:
for j from i to rows:
print i
print newline
Cheat sheet
Goal
Pattern
Pick each digit
for (i = 1; i <= rows; i++)
Repeat digit
for (j = i; j <= rows; j++) cout << i;
End of row
cout << "\n";
Repeat count
rows - i + 1 (shrinks as i grows)
vs Program 9
9: j = 1..i (grows); 12: j = i..rows (shrinks)
Printing Numbers vs Starting a New Line
API
Effect
Use for
cout << i
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 shrinking repeating pattern updates instantly — capped at 9 for readable single-digit 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
44
5
Trace
Worked Walkthrough — rows = 5
Trace how the digit rises while the repeat count falls.
i
Inner runs
Prints
1
5 times
11111
2
4 times
2222
3
3 times
333
5
1 time
5
Total digits = n(n+1)/2 → O(n²). Always print i, never j.
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 — outer picks digit i; inner runs from i to rows.
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++)
cout << i;
cout << "\n";
}
return 0;
}
Output
11111
2222
333
44
5
How It Works
1. Outer picks the digit.i starts at 1 and increases to rows.
2. Inner sets repeats.j runs from i to rows — that is rows - i + 1 prints of i.
3. Newline. Call cout << "\n" only after the inner loop finishes the row.
Example 2 — User Input Rows
Read rows with cin, validate, then run the same nested loops.
C++
#include <iostream>
using namespace std;
int main()
{
int rows;
int i, j;
cout << "Enter the number of rows: ";
if (!(cin >> rows) || rows <= 0)
{
cout << "Please enter a positive integer.\n";
return 1;
}
for (i = 1; i <= rows; i++)
{
for (j = i; j <= rows; j++)
cout << i;
cout << "\n";
}
return 0;
}
Output (when user enters 4)
Enter the number of rows: 4
1111
222
33
4
How It Works
1. Prompt and validate. Reject failed reads and non-positive values before printing.
2. Same core. Outer ascending / inner i..rows matches Example 1 — only rows comes from the user.
3. Safer input tip. Cap demos for readable single-digit 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 digit vs shrink count.
C++
#include <iostream>
using namespace std;
int main()
{
int rows = 3;
int i, j;
for (i = 1; i <= rows; i++)
{
for (j = i; j <= rows; j++)
cout << i;
cout << "\n";
}
return 0;
}
Output
111
22
3
How It Works
1. Three rows.i = 1 → 111; i = 2 → 22; i = 3 → 3.
2. Trace on paper. If you use j = 1..i instead, you get Program 9 (1, 22, 333).
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.
j = 1..i
Got Program 9 instead
Using for (j = 1; j <= i; j++) grows the row (1, 22, 333…). Keep j = i..rows to shrink.
i = rows..1
Got Program 11 instead
A descending outer loop with j = 1..i yields 55555, 4444… Keep outer 1..rows here.
print j
Wrong values on the row
If you print j instead of i, you get changing digits per position. Always print the outer variable.
\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
Count up: outer i = 1..rows picks a rising digit.
Shrinking repeats: inner j = i..rows runs rows - i + 1 times and always prints i.
Break the row:cout << i 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 i from 1 to rows, print i exactly rows - i + 1 times, then newline.
Frequently Asked Questions
A shrinking repeating pattern: for rows=5 you get 11111, 2222, 333, 44, 5 — digit rises each row while the repeat count falls.
The inner loop runs from j = i to rows. As i increases, the inner loop runs fewer times, so each row prints fewer copies of the row digit.
On the first row, i = 1 and the inner loop runs from 1 to rows, so it prints 1 exactly rows times.
Program 11 counts down and repeats digit i exactly i times (55555, 4444…). Program 12 counts up and uses for (j = i; j <= rows; j++) so digit i repeats rows - i + 1 times.
Program 9 grows: row i repeats i times (1, 22, 333). Program 12 shrinks: row i repeats rows - i + 1 times (11111, 2222, 333).
cout << i stays on the same line. cout << "\n" ends the current line. Repeated digits use cout << i; the row break uses cout << "\n" after the inner loop.
After prompting, use if (!(cin >> rows) || rows <= 0) to reject bad input before printing.
O(n²) for n rows because total digit prints are n+(n-1)+…+1 = n(n+1)/2.
🤔
Did you know?
Row digit i repeats rows − i + 1 times. The outer loop counts up from 1, while the inner loop runs from i to rows, so each row shrinks — still O(n²) total prints.
