[
greedy
two pointers
]
BinarySearch 0107 Rocketship Rescue
Problem statement
https://binarysearch.com/problems/Rocketship-Rescue/
Solution
Equal to Leetcode 0881. Boats to Save People.
Complexity
It is O(n log n)
for time and O(n)
for space.
Code
class Solution:
def solve(self, people, limit):
people.sort()
beg, end, ans = 0, len(people) - 1, 0
while beg <= end:
if people[beg] + people[end] <= limit:
beg += 1
ans += 1
end -= 1
return ans