A reverse-start alphabet triangle keeps the right edge fixed at a top letter while each new row begins one letter earlier — letters along the row still run forward.
Rule: start moves E→A; print start..top forward
E
DE
CDE
BCDE
ABCDE ← 5 rows, every row ends at ECompare Program 1 (always starts at A) and Program 2 (letters descend along the row). Program 4 reverses order inside each row instead.
Outer loop picks the first letter (top down to A). Inner loop prints forward from that letter up to top.
| Method | Idea | Best for |
|---|---|---|
Code points (ord/chr) | range(code, top + 1) prints start..top | Classic nested-loop drills |
| Spaced cells | print(chr(j) + " ", end="") | Easier-to-scan columns |
top = 'E' // or 'A' + rows - 1
for code from top down to 'A': // start letter
for j from code to top: // forward run
print j
print newline| Goal | Pattern |
|---|---|
| Fixed top | top = ord("E") |
| Rows → top | top = ord("A") + rows - 1 |
| Walk starts | for code in range(top, ord("A") - 1, -1): |
| Print forward | for j in range(code, top + 1): print(chr(j), end="") |
| End the row | print() |
| vs Program 1 | Here the start moves; Program 1 grows the end |
| vs Program 2 | Here letters increase; Program 2 decreases along the row |
| API | Effect | Use for |
|---|---|---|
print(chr(j), end="") | Stays on the same row | Each letter in the forward run |
print() | Ends the current row | After the inner loop |
Print letters without a newline, then end the row once.
Change the row count and the reverse-start triangle updates instantly — including the fixed top letter on every row.
rows = 3 (top = C)Outer start moves C → B → A. Inner loop always ends at C.
Start code | Inner run | Printed row |
|---|---|---|
C | C..C | C |
B | B..C | BC |
A | A..C | ABC |
Total letters are 1+2+…+n = n(n+1)/2 → O(n²).
Three complete programs: fixed top E, row-count input(), and spaced letters. Use View Output for sample results.
EOuter loop chooses the first letter; inner loop prints forward up to E.
top = ord("E")
for code in range(top, ord("A") - 1, -1):
for j in range(code, top + 1):
print(chr(j), end="")
print()1. Fix the right edge. top = ord("E") so every row ends at E.
2. Move the start. Outer code walks E → D → … → A.
3. Print forward. When start is C, the inner loop prints C, D, E → CDE.
Read how many rows to print and compute top = ord("A") + rows - 1. Cap at 26 so the top letter stays in A–Z.
raw = input("Enter the number of rows: ").strip()
try:
rows = int(raw)
except ValueError:
print("Please enter a whole number.")
raise SystemExit(1)
rows = max(1, min(rows, 26))
top = ord("A") + rows - 1
for code in range(top, ord("A") - 1, -1):
for j in range(code, top + 1):
print(chr(j), end="")
print()1. Validate rows. Parse an integer, then clamp to 1..26.
2. Derive top. For 4 rows, top becomes D, so every row ends at D.
Print a space after each letter so columns are easier to scan.
top = ord("E")
for code in range(top, ord("A") - 1, -1):
for j in range(code, top + 1):
print(chr(j) + " ", end="")
print()1. Same loops. Start still moves E→A; letters still run forward to E.
2. Extra space. chr(j) + " " separates letters without changing the pattern logic.
Check these before calling the solution done.
Printing top..code gives Program 2’s descending rows (E, ED, EDC). Keep range(code, top + 1).
range(top, ord("A"), -1)Stopping before A skips the full last row. Use ord("A") - 1 as the exclusive end.
print() inside the inner loopUse end="" for letters; call bare print() only after the forward run.
Cap rows at 26 so ord("A") + rows - 1 stays in A–Z.
Output is just A — a good sanity check.
Catch ValueError from int(...) before computing top.
| Program | Time | Extra space |
|---|---|---|
| Fixed / input (Examples 1–2) | O(n²) | O(1) |
| Spaced letters (Example 3) | O(n²) | O(1) |
There are n rows and row r prints r letters, so total work is n(n+1)/2.
One line: for each start from top down to A, print start..top forward, then newline.
This triangle changes only the starting letter of each row (E, D, C, …), while letters along the row still increase forward. In the 5-row example, every row ends at E, producing E, DE, CDE, BCDE, ABCDE.
Keep a growing left-to-right shape, but print each row’s letters in reverse (A, BA, CBA…).
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