A repeating number triangle prints the row index i exactly i times on each row — 1, 22, 333, 4444, and so on.
Rule: for i from 1 to rows
for j from 1 to i: print i (no space)
1
22
333
4444
55555 ← rows = 5Follows the reverse left-growing triangle in Program 8; next is the descending repeating triangle in Program 10.
Outer loop picks the digit i. Inner loop runs i times and prints i each time — not j.
| Method | Idea | Best for |
|---|---|---|
Print i in the inner loop | Outer i = 1..rows; inner j = 1..i; print i | Learning, interviews, exams |
| Spaced digits | print(str(i) + " ", end="") instead of print(i, end="") | Easier reading on each row |
for i from 1 to rows:
for j from 1 to i:
print i
print newline| Goal | Pattern |
|---|---|
| Pick each row | for i in range(1, rows + 1): |
| Repeat digit | for j in range(1, i + 1): print(i, end="") |
| End of row | print() |
| Spaced form | print(str(i) + " ", end="") |
| vs Program 5 | 5 prints j; 9 prints i — same loop bounds |
| API | Effect | Use for |
|---|---|---|
print(i, end="") | Stays on the same line | Each repeated digit on the row |
print() | Ends the current line | After the inner loop finishes |
Digits use end=""; a bare print() ends the row once.
Change the row count and the repeating triangle updates instantly — capped at 9 for readable single-digit demos.
rows = 5Trace how the inner loop count equals the digit being printed.
i | Inner runs | Prints |
|---|---|---|
1 | 1 time | 1 |
2 | 2 times | 22 |
3 | 3 times | 333 |
5 | 5 times | 55555 |
Total digits = n(n+1)/2 → O(n²). The key mistake to avoid: printing j instead of i.
Three complete programs: fixed rows = 5, user-input rows, and a compact rows = 3 demo. Use View Output to reveal sample results.
rows = 5Hard-coded height — print i inside the inner loop, not j.
rows = 5
for i in range(1, rows + 1):
for j in range(1, i + 1):
print(i, end="")
print()1. Outer picks the digit. i is both the row index and the value to repeat.
2. Inner repeats it. j only controls how many times — it is never printed.
3. Newline. Call bare print() only after the inner loop finishes the row.
Read rows safely, validate, then run the same nested loops.
try:
rows = int(input("Enter the number of rows: "))
except ValueError:
print("Please enter a positive integer.")
raise SystemExit
if rows <= 0:
print("Please enter a positive integer.")
raise SystemExit
for i in range(1, rows + 1):
for j in range(1, i + 1):
print(i, end="")
print()1. Prompt and validate. Catch ValueError; reject non-positive values before looping.
2. Same core. Outer i / inner print-i matches Example 1 — only rows comes from the user.
3. Safer input tip. Cap demos for readable single-digit output:
try:
rows = int(input("Enter the number of rows: "))
except ValueError:
print("Enter a whole number from 1 to 9.")
raise SystemExit
if rows < 1 or rows > 9:
print("Enter a whole number from 1 to 9.")
raise SystemExitrows = 3Same structure with only three rows — easy to confirm you print i, not j.
rows = 3
for i in range(1, rows + 1):
for j in range(1, i + 1):
print(i, end="")
print()1. Three rows. i = 1 → 1; i = 2 → 22; i = 3 → 333.
2. Trace on paper. If you accidentally print j, row 3 becomes 123 (Program 5) — not 333.
3. Scale up next. Once the small demo is clear, use Examples 1–2 for five rows or user input.
Check these before calling the solution done.
If you print j instead of i, you get 1, 12, 123… Always print the outer variable.
Counting the outer loop down gives Program 10’s shape (5, 44, 333…). Keep outer 1..rows here.
If bare print() sits inside the inner loop, each digit lands on its own line. Call it only after the row finishes.
Output is just 1 — a good sanity check for input validation.
Digits 10+ make tight rows hard to read. Cap demos at 9 or switch to spaced print(str(i) + " ", end="").
Use try/except ValueError and require rows > 0 before looping — bad input should not crash.
| 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 = 1 + 2 + … + n = n(n+1)/2 → O(n²) time. Only a few loop variables are needed.
j = 1..i, but the printed value is i.print(i, end="") in the inner loop; bare print() after it.O(n²) time from n(n+1)/2 digits; O(1) extra space.One line: for each i, print i exactly i times, then newline.
Each row repeats the row number i exactly i times — outer i runs 1..rows, inner j prints i on every iteration — producing 1, 22, 333, and so on. Total prints grow as O(n²).
Continue with rows that repeat a descending digit (5, 44, 333).
12 people found this page helpful