881. Boats to Save People

881. Boats to Save People (Medium)

The i-th person has weight people[i], and each boat can carry a maximum weight of limit.

Each boat carries at most 2 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.  (It is guaranteed each person can be carried by a boat.)

 

Example 1:

Input: people = [1,2], limit = 3
Output: 1
Explanation: 1 boat (1, 2)

Example 2:

Input: people = [3,2,2,1], limit = 3
Output: 3
Explanation: 3 boats (1, 2), (2) and (3)

Example 3:

Input: people = [3,5,3,4], limit = 5
Output: 4
Explanation: 4 boats (3), (3), (4), (5)

Note:

• 1 <= people.length <= 50000
• 1 <= people[i] <= limit <= 30000

Companies:
Google

Related Topics:
Two Pointers, Greedy

Solution 1. Greedy

// OJ: https://leetcode.com/problems/boats-to-save-people/
// Author: github.com/lzl124631x
// Time: O(NlogD) where N is to total number of people
//                and D is the distinct count of weights.
// Space: O(D)
class Solution {
public:
    int numRescueBoats(vector<int>& people, int limit) {
        map<int, int, greater<int>> m;
        for (int w : people) m[w]++;
        int ans = 0;
        while (m.size()) {
            int w = limit, cnt = 0;
            while (w > 0 && cnt < 2) {
                auto it = m.lower_bound(w);
                if (it == m.end()) break;
                w -= it->first;
                if (--it->second == 0) m.erase(it->first);
                ++cnt;
            }
            ++ans;
        }
        return ans;
    }
};

Solution 2. Greedy + Two Pointers

For the lightest person a, if a can pair with the heaviest person b, let a do so.

If not, then the heaviest person b can't pair with any one, let b go alone.

// OJ: https://leetcode.com/problems/boats-to-save-people/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int numRescueBoats(vector<int>& A, int limit) {
        sort(begin(A), end(A));
        int i = 0, j = A.size() - 1, ans = 0;
        while (i <= j) {
            ++ans;
            if (A[i] + A[j] <= limit) ++i;
            --j;
        }
        return ans;
    }
};