Follows the decreasing continuous triangle in Program 38; next is the alternating 1 and 0 pattern in Program 40.
Approach
How to Solve It
One outer loop and two inner loops that rotate the digit sequence — tight digits with end="", then bare print().
Method
Idea
Best for
Forward + wrap
Print i..rows, then i-1..1
Learning, interviews, exams
User-input rows
Same logic with a variable width
Practice / demos
Pseudocode
Pseudocode
for i from 1 to rows:
for j from i to rows:
print j
for k from i-1 down to 1:
print k
print newline
Cheat sheet
Goal
Pattern
Next rotation
for i in range(1, rows + 1):
Forward segment
for j in range(i, rows + 1): print(j, end="")
Wrap segment
for k in range(i - 1, 0, -1): print(k, end="")
End the row
print()
Printing Numbers vs Starting a New Line
API
Effect
Use for
print(j, end="") / print(k, end="")
Stays on the same line
Each digit (no spaces)
print()
Ends the current line
After both inner loops
Print digits without a newline, then end the row once.
Try it
Live Preview
Change the row count and the rotating pattern updates instantly — each row stays the same width.
Whole numbers from 1 to 9 (keeps digits single-width). Tap a chip or type a value — the preview redraws as you go.
Live result5 rows · 5 digits/row
12345
23451
34521
45321
54321
Trace
Worked Walkthrough — rows = 4
Trace the forward climb and the wrap-around. Every row has exactly 4 digits.
i
Forward / wrap
Printed row
1
1234 / (none)
1234
2
234 / 1
2341
3
34 / 21
3421
4
4 / 321
4321
Forward length is rows - i + 1; wrap length is i - 1. Sum = rows.
Code
Python Programs
Three complete programs: fixed rows = 5, input() variant, and a compact rows = 3 demo. Use View Output to reveal sample results.
Example 1 — Fixed rows = 5
Two inner loops per row: forward i..rows, then wrap i-1..1.
Python
rows = 5
for i in range(1, rows + 1):
for j in range(i, rows + 1):
print(j, end="")
for k in range(i - 1, 0, -1):
print(k, end="")
print()
Output
12345
23451
34521
45321
54321
How It Works
1. Outer loop.i grows from 1 to 5 — each row starts one digit higher.
2. Forward segment. Print j from i to rows (the climb to the top).
3. Wrap segment. Print k from i - 1 down to 1 (skipped when i = 1).
Example 2 — User Input Rows
Read rows with input(), validate, then apply the same forward-and-wrap logic.
Python
try:
rows = int(input("Enter rows: "))
except ValueError:
print("Please enter a positive whole number.")
raise SystemExit(1)
if rows < 1:
print("Please enter a positive whole number.")
raise SystemExit(1)
for i in range(1, rows + 1):
for j in range(i, rows + 1):
print(j, end="")
for k in range(i - 1, 0, -1):
print(k, end="")
print()
Output (when user enters 3)
Enter rows: 3
123
231
321
How It Works
1. Prompt and validate. Catch ValueError; require rows >= 1 before printing.
2. Same core. Forward + wrap match Example 1 — only rows comes from the user.
3. Safer input tip. Cap demos for readable single-digit output:
Safer input tip
if rows < 1 or rows > 9:
print("Enter a whole number from 1 to 9.")
raise SystemExit(1)
Example 3 — Compact rows = 3
Same structure with only three rows — easy to confirm wrap length is i - 1.
Python
rows = 3
for i in range(1, rows + 1):
for j in range(i, rows + 1):
print(j, end="")
for k in range(i - 1, 0, -1):
print(k, end="")
print()
Output
123
231
321
How It Works
1. Same rules. Forward prints i..rows; wrap prints i-1..1.
2. Quick check. Every row has exactly 3 digits — 123, 231, 321.
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.
wrong wrap
Wrap ascending instead of descending
Using range(1, i) gives a pure cycle (23451 → 34512). This pattern wraps descending (34521).
newline early
print() between the two halves
That splits the rotation onto two lines. Call bare print() only after both loops finish.
extra spaces
Print a space after each digit
This pattern is tight digits only. Use print(j, end="") with no trailing space.
rows = 1
Single digit
The wrap loop does not run; only the forward loop prints 1.
rows > 9
Multi-digit cells
Values past 9 print as two characters and break the visual rotation. Cap demos at 9.
input()
Catch ValueError
Bare int(input()) crashes on non-numeric text — wrap it in try/except ValueError.
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)
Each of n rows prints exactly n digits → n² prints → O(n²) time. Only a few loop variables are needed.
Remember
Key Takeaways
Rule: print i..rows, then i-1..1 — each row has exactly rows digits.
Wrap: use range(i - 1, 0, -1) so the leftover digits descend to 1.
end="" vs print(): digits stay on the line; bare print() advances after both halves.
Complexity:O(n²) time; O(1) extra space.
One line: for each row i, print i..rows then wrap with i-1..1.
Frequently Asked Questions
For 5 rows: 12345, 23451, 34521, 45321, 54321 — each row starts at the row number and wraps back to 1.
The first loop prints i..rows (forward segment). The second loop prints i-1 down to 1 (wrap segment). Together they always produce rows digits.
After printing 2 3 4 5, the wrap loop prints i-1..1 — for i=2 that prints 1.
Exactly rows digits every time — (rows - i + 1) forward plus (i - 1) wrap = rows.
Program 37 builds palindrome rows (i..2 then 1..i). Program 39 rotates: i..rows then i-1..1.
Program 38 prints a shrinking-width continuous sequence. Program 39 keeps fixed width and rotates the digits.
Use try/except ValueError around int(input()) and require rows >= 1 — see Example 2.
O(n²) for n rows because each row prints n digits.
🤔
Did you know?
Each row starts at i, prints i..rows, then wraps with i-1..1. Row i always prints exactly rows digits — total digits = n².
