DSA interview question

Pattern 5: Inverted Right-Angled Triangle of Stars

Easy · Algorithms · Topics: pattern building

The problem

After building a triangle that grows, the next exercise flips it: the first row has n stars and each following row has one fewer, down to a single star. It's the same shape as Pattern 2 upside down, and it teaches you to express a shrinking row length as a formula, n − i + 1 for row i. Getting that formula right, rather than guessing, is what stops the classic "one row too many or too few" bug.

Example for n = 4:

****
***
**
*

Function signature

vector<string> invertedTriangle(int n);  // row i (1-indexed) has n - i + 1 stars

Test cases

Input `n`	Output rows	Explanation
1	`"*"`	One row
3	`"*"`, `""`, `"*"`	Starts at n, shrinks by one
5	`"***"`, `""`, `""`, `""`, `""`	Last row has exactly one star

(Confirm: rows are left-aligned, stars have no separators, and there are exactly n rows.)

How interviewers phrase it

"Print an inverted right-angled triangle of stars."
"Now print the previous triangle upside down."

Clarifying questions to ask

Left-aligned or right-aligned?
Stars separated by spaces?
Print or return the rows?

Follow-ups to expect

"How did you get n − i + 1?"
"Is there another way to write it?"
"What's the complexity?"

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