Description

The original description of the problem is clear enough:

You are given an array people where people[i] is the weight of the ith person, and an infinite number of boats where each boat can carry a maximum weight of limit. Each boat carries at most two people at the same time, provided the sum of the weight of those people is at most limit.

Return the minimum number of boats to carry every given person.

A few examples:

# input
people = [1, 2]
limit = 3

# output
1

# explanation
# 
# There is only one boat in this case:
#
# - [1, 2]

# input
people = [3, 2, 2, 1]
limit = 3

# output
3

# explanation
# 
# We need 3 boats:
#
# - [3]
# - [2, 1]
# - [2]

Heart of The Problem

For me, the most challenging issue is to… read the problem carefully. I understood the concepts of boats and limit, but somehow missed that:

Each boat carries at most two people at the same time, provided the sum of the weight of those people is at most limit.

It was almost like a suggestion for us to solve the problem with the technique “two pointer”. The steps for the algorithm is:

  • Sort people
  • Set a pointer at the beginning and another at the end of people
  • Try to put the two persons into a boat. There are only two cases:
    • Only the person at the end “fits”
    • The two persons fits
  • We can be sure that the process is optimal, since:
    • There can only be two people on a boat
    • One’s weight is at most limit, which means a the “heavy” person of the right side always occupies a boat, and it depends if we can fit the “light” person of the left side. We can always move the two cursors until they move past each other

For example, let us go through this case:

people = [1, 2, 2, 3, 4, 4, 5]
limit = 5

Let us simulate the process:

Iteration 1:

   v           v
  [1 2 2 3 4 4 5]

  In this iteration, we can only fits the last person on a boat.

  [5]

Iteration 2:

   v         v
  [1 2 2 3 4 4 5]

  Moving the right cursor left, we find out that `1` and `4` can be fit on a
  boat.

  [5] [1 4]

Iteration 3:

     v     v
  [1 2 2 3 4 4 5]

  [5] [1 4] [4]

Iteration 4:

     v   v  
  [1 2 2 3 4 4 5]

  [5] [1 4] [4] [2 3]

Iteration 5:

       v    
  [1 2 2 3 4 4 5]

  Now the two cursors are joined and we stop iterating.

  [5] [1 4] [4] [2 3] [2]

Implementation 1: Recursive Function

def recurse(people: List[int], limit: int) -> int:
    if len(people) <= 1:
        return len(people)
    elif people[-1] + people[0] <= limit:
        return 1 + recurse(people[1:-1], limit)
    else:
        return 1 + recurse(people[:-1], limit)


class Solution:
    def numRescueBoats(self, people: List[int], limit: int) -> int:
        return recurse(people=sorted(people), limit=limit)

In the this version, I implemented a naive recursive function:

  • Return 1 if there is only one person
  • If the last person and the first person can be fit on a boat, then the result is 1 plus a call to itself where there the two is excluded.
  • If the first person cannot be fit on a boat with the last, then we exclude the last one only.

Using the same input from above:

people = [1, 2, 2, 3, 4, 4, 5]
limit = 5

Our function calls are:

>>> recurse(people=[1, 2, 2, 3, 4, 4, 5], limit=5)
>>> 1 + recurse(people=[1, 2, 2, 3, 4, 4], limit=5)
>>> 1 + 1 + recurse(people=[2, 2, 3, 4], limit=5)
>>> 1 + 1 + 1 + recurse(people=[2, 2, 3], limit=5)
>>> 1 + 1 + 1 + 1 + recurse(people=[2], limit=5)
>>> 1 + 1 + 1 + 1 + 1
5

Before you are wondering, it TLE-ed.

Implementation 2: Variable Swapping

The idea of reducing people into a smaller array can be implemented using iteration.

class Solution:
    def numRescueBoats(self, people: List[int], limit: int) -> int:
        result = 0
        people = sorted(people)
        while len(people) > 1:
            if people[-1] + people[0] <= limit:
                result += 1
                people = people[1:-1]
            else:
                result += 1
                people = people[:-1]

        result += len(people)
        return result

Before you are questioning again, yes, this also TLE-ed. The problem lies in these assignments:

people = people[...]

Which create a new list that includes shallow copies of the old list’s elements. Using deque here should works, too.

Implementation 3: Two Pointers

Here comes the “intended” implementation:

class Solution:
    def numRescueBoats(self, people: List[int], limit: int) -> int:
        result = 0
        people = sorted(people)
        left_index = 0
        right_index = len(people) - 1
        while left_index <= right_index:
            if people[left_index] + people[right_index] <= limit:
                result += 1
                left_index += 1
                right_index -= 1
            else:
                result += 1
                right_index -= 1
        return result

Conclusion

I think this is a nice “medium” problem. The description’s wording and examples are not too cryptic nor deceiving in anyway. “Two pointer” is a “natural” technique that people can reasonably come up with in time even if they have not seen it before.