Problem
The first row is [1]. Each later row starts and ends with 1. Every interior entry is the sum of the two entries immediately above it. Return the first rows rows, beginning with the single 1.
Implement pascals_triangle(rows).
Input
rows: How many rows to generate.
Output
Return a list of rows, starting at the top; each row is a list of integers.
Constraints
0 ≤ rows ≤ 60. Every entry fits a signed 64-bit integer. Zero rows returns an empty list.
Examples
Input:
rows = 4
Expected output:
expected = [
[1],
[1, 1],
[1, 2, 1],
[1, 3, 3, 1]
]Input:
long long rows = 4;
Expected output:
std::vector<std::vector<long long>> expected = {
{1},
{1, 1},
{1, 2, 1},
{1, 3, 3, 1}
};Input:
long rows = 4;
Expected output:
long[][] expected = new long[][] {
new long[] {1},
new long[] {1, 1},
new long[] {1, 2, 1},
new long[] {1, 3, 3, 1}
};Explanation
The 2 in row 3 is 1 + 1. The two interior 3s in row 4 are 1 + 2 and 2 + 1.
Input:
rows = 0
Expected output:
expected = []Input:
long long rows = 0;
Expected output:
std::vector<std::vector<long long>> expected = {};Input:
long rows = 0;
Expected output:
long[][] expected = new long[][] {};Explanation
No rows are requested, so return an empty list.
How to approach it
Try on your own first, then open one hint at a time.
Hint 1
Each row starts and ends with 1, and interior values come from the two values above them.
Hint 2
Build rows one at a time from the previous row rather than recomputing from scratch.
Hint 3
Handle the zero-row case by returning an empty list immediately.
Expected complexity
O(rows^2) time, O(rows^2) extra space
References
This challenge is adapted for CodeUtility. Follow the statement and constraints above.