Same signature as v2/v3, but now n may be negative. Real head -n -K prints all lines except the last K. So when n < 0 return all lines except the last |n|; when n ≥ 0 behave exactly like v3 (return the first n, capped at the input length). If |n| is at least the number of lines, nothing is left, so return the empty string.
Function signature:
def head(lines: List[String], n: Int): String
Example (negative n):
Input: lines = ["a", "b", "c", "d", "e"], n = -2 Output: "a\nb\nc" When printed: a b c
Compute the cut-off with a single if/else expression: when
n < 0, the end index is math.max(0, lines.size + n); otherwise it's math.min(n, lines.size). Then take(end) and mkString("\n").
val end = if (n < 0) math.max(0, lines.size + n) else math.min(n, lines.size)
lines.take(end).mkString("\n")
The full solution:
class Solution {
def head(lines: List[String], n: Int): String = {
val end = if (n < 0) math.max(0, lines.size + n) else math.min(n, lines.size)
lines.take(end).mkString("\n")
}
}
Free plan: 30 tutor messages. Premium is unlimited.
Saved
Ctrl + Enter to check
