Print a Full Star Pyramid

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Easy10 min
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Given an integer n, return the n rows of a centered, symmetric star pyramid — row i has (n − i) leading spaces, then (2i − 1) stars. Unlike the Right Half Pyramid (which grows by 1 star per row and is left-aligned), this pyramid grows by 2 stars per row — one on each side — which is exactly what keeps it centered and symmetric. The nested loopsNested LoopsAn outer loop over rows, one inner loop for the leading spaces, another for the (2i-1) stars. version builds each half explicitly; the built-in repetitionBuilt-in RepetitionBuilding the space-padding and the stars each with a repeat call, then concatenating the two pieces. version replaces both inner loops with two built-in calls.

Test Case 1:

Input:n = 4
Output:[" *", " ***", " *****", "*******"]
Explanation:Row i has (n − i) leading spaces, then (2i − 1) stars — a centered, symmetric pyramid.

Test Case 2:

Input:n = 1
Output:["*"]
Explanation:A single star, no spaces needed.

Constraints

  • ◆1 ≤ n ≤ 20

Try the Dry Run

Approach & Solutions

Nested Loops — Spaces, Then StarsGood

For row i, first append (n − i) spaces, then append (2i − 1) stars. The odd count 2i − 1 is what keeps the pyramid centered and symmetric — each row grows by exactly 2 stars (one on each side), unlike the Right Half Pyramid's one-star-per-row growth.

TimeO(n²)
SpaceO(n²) for the output
1class Solution { 2 public String[] printFullStarPyramid(int n) { 3 String[] result = new String[n]; 4 for (int i = 1; i <= n; i++) { 5 StringBuilder row = new StringBuilder(); 6 for (int s = 1; s <= n - i; s++) { 7 row.append(' '); 8 } 9 for (int j = 1; j <= 2 * i - 1; j++) { 10 row.append('*'); 11 } 12 result[i - 1] = row.toString(); 13 } 14 return result; 15 } 16}
Built-in Repetition for Spaces and StarsOptimal

Build each half of the row with a built-in repeat: (n − i) spaces concatenated with (2i − 1) stars. C has no repeat built-in, so it keeps both loops.

TimeO(n²)
SpaceO(n²)
1class Solution { 2 public String[] printFullStarPyramid(int n) { 3 String[] result = new String[n]; 4 for (int i = 1; i <= n; i++) { 5 result[i - 1] = " ".repeat(n - i) + "*".repeat(2 * i - 1); 6 } 7 return result; 8 } 9}

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