A reverse alphabet triangle grows one letter per row, but each row starts at the current letter and counts down to A.
Rule: row i prints (A+i) .. A descending
A
BA
CBA
DCBA
EDCBA ← 5 rowsCompare Program 1 (ascends from A: A, AB, ABC) and Program 3 (start moves back, letters still run forward). Program 5 shrinks row width from ABCDE to A.
Outer loop picks the start letter; inner loop counts down to A with range(..., -1).
| Method | Idea | Best for |
|---|---|---|
Nested ord/chr | range(base + i, base - 1, -1) | Learning descending bounds |
| String slice | letters[i::-1] | Compact shortcut once the idea clicks |
base = 'A'
for i from 0 to rows - 1:
for code from (base + i) down to base: // inclusive A
print code
print newline| Goal | Pattern |
|---|---|
| Walk rows | for i in range(rows): |
| Count down to A | for code in range(base + i, base - 1, -1): |
| Print letter | print(chr(code), end="") |
| Include A | Stop at base - 1 (exclusive end) |
| End the row | print() |
| Slice shortcut | print(letters[i::-1]) |
| vs Program 1 | Here the inner loop goes down; Program 1 goes up |
| API | Effect | Use for |
|---|---|---|
print(chr(code), end="") | Stays on the same row | Each letter in the descending 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 triangle updates instantly — every row still ends at A.
rows = 3Start letter grows A → B → C; each inner run ends at A.
i | Inner run | Printed row |
|---|---|---|
0 | A..A | A |
1 | B..A | BA |
2 | C..A | CBA |
Total letters are 1+2+…+n = n(n+1)/2 → O(n²).
Three complete programs: fixed height, row-count input(), and a reverse-slice shortcut. Use View Output for sample results.
rows = 5Outer loop picks the start; inner loop counts down to A.
rows = 5
base = ord("A")
for i in range(rows):
for code in range(base + i, base - 1, -1):
print(chr(code), end="")
print()1. Pick the start. When i = 2, start is base + 2 (C).
2. Count down. range(base + i, base - 1, -1) prints C, B, A — the exclusive end base - 1 includes A.
3. Newline. Bare print() after the inner loop starts the next row.
Read how many rows to print. Cap at 26 so letters stay in A–Z.
raw = input("Enter the number of rows (max 26): ").strip()
try:
rows = int(raw)
except ValueError:
print("Please enter a whole number.")
raise SystemExit(1)
rows = max(1, min(rows, 26))
base = ord("A")
for i in range(rows):
for code in range(base + i, base - 1, -1):
print(chr(code), end="")
print()1. Validate rows. Parse an integer, then clamp to 1..26.
2. Same descending core. Four rows end at DCBA; every row still finishes on A.
letters[i::-1]Reverse-slice from index i to the start of the alphabet string.
rows = 5
letters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
for i in range(rows):
print(letters[i::-1])1. One slice. letters[i::-1] returns letters from index i down to 0.
2. Same shape. Identical output to Example 1 — keep the nested-loop version for exams that ask for descending bounds.
Check these before calling the solution done.
range(..., base, -1)range stops before the end value, so ending at base skips A. Use base - 1.
Ascending from A gives Program 1 (A, AB, ABC). Keep step -1 for this pattern.
print() inside the inner loopUse end="" for letters; call bare print() only after the descending run.
Cap rows at 26 so base + i stays in A–Z.
Output is just A — a good sanity check.
Catch ValueError from int(...) before the outer loop.
| Program | Time | Extra space |
|---|---|---|
| Nested loops (Examples 1–2) | O(n²) | O(1) |
| Slice shortcut (Example 3) | O(n²) | O(n) per row for the slice string |
There are n rows and row i prints i+1 letters, so total work is n(n+1)/2.
i prints (A+i)..A descending.base - 1, not base.One line: for each row i, print letters from A+i down to A.
Print ascending rows that shrink from ABCDE down to a single A.
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