Print a Right Pascal's Triangle Star Pattern

Easy10 min

Topics

BasicsPatternsLoops

Companies

TCSInfosysWipro

Given n, return 2n − 1 left-aligned rows of stars whose lengths go 1, 2, …, n and then back down to 1 — an arrow pointing to the right. (The name comes from its outline, not from Pascal's number triangle.) It is a Right Half Pyramid followed by its reverse, sharing the widest row. The first solution writes the two halves with two loops; the second uses the distance from the middle to give every row its width in a single loop.

Test Case 1:

Input:n = 3

Output:["*", "**", "***", "**", "*"]

Explanation:Rows grow from 1 star to n stars and then shrink back to 1, forming an arrow pointing right.

Test Case 2:

Input:n = 1

Output:["*"]

Explanation:Only the single widest row.

Constraints

◆1 ≤ n ≤ 20

Try the Dry Run

Mode

Approach

Approach & Solutions

Rising Loop, Then Falling LoopGood

Two plain loops: the first writes rows of 1, 2, …, n stars, the second writes rows of n − 1, …, 1 stars. The widest row is written only by the first loop, so it is not repeated. C uses a small helper that returns a row of a given number of stars.

TimeO(n²)

SpaceO(n²) for the output

Java

Try it out

class Solution {
    public String[] printRightPascalsTriangle(int n) {
        String[] result = new String[2 * n - 1];
        int idx = 0;
        for (int i = 1; i <= n; i++) {
            result[idx++] = "*".repeat(i);
        }
        for (int i = n - 1; i >= 1; i--) {
            result[idx++] = "*".repeat(i);
        }
        return result;
    }
}

One Loop, Width from the Distance to the MiddleOptimal

Use a single loop over all 2n − 1 rows. Row r is |n − 1 − r| rows away from the middle row, and each step away from the middle removes one star, so its width is n − |n − 1 − r|. One formula covers both the rising and the falling half.