881. Boats to Save People

Difficulty: Medium

Topics: Array, Two Pointers, Greedy, Sorting

You are given an array people wherepeople[i] is the weight of the i^th 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.

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)

Constraints:

• 1 <= people.length <= 5 * 104
• 1 <= people[i] <= limit <= 3 * 104

Solution:

We can use a two-pointer greedy strategy combined with sorting. Here's the detailed approach:

1. Sort the Array:

   • First, sort the people array. Sorting helps us to easily pair the lightest and heaviest person together in one boat, if possible.

2. Two Pointer Strategy:

   • Use two pointers: one starting from the lightest person (left), and the other starting from the heaviest person (right).
   • Try to pair the heaviest person (right) with the lightest person (left). If the sum of their weights is less than or equal to the limit, they can share the same boat. Move both pointers (left++ and right--).
   • If they cannot be paired, send the heaviest person alone on a boat and move only the right pointer (right--).

3. Count Boats:

   • Each time we process a person (or pair), we count it as one boat.
   • Continue until all people are assigned to boats.

Let's implement this solution in PHP: 881. Boats to Save People

<?php
function numRescueBoats($people, $limit) {
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Test cases
echo numRescueBoats([1, 2], 3); // Output: 1
echo "\n";
echo numRescueBoats([3, 2, 2, 1], 3); // Output: 3
echo "\n";
echo numRescueBoats([3, 5, 3, 4], 5); // Output: 4
?>

Explanation:

1. Sorting:

   • For the example [3, 2, 2, 1], after sorting, it becomes [1, 2, 2, 3].

2. Two-pointer logic:

   • Start with left = 0 (lightest person, weight 1) and right = 3 (heaviest person, weight 3).
   • Check if people[0] + people[3] (1 + 3) is less than or equal to the limit (3). It is not, so send the heaviest person alone and decrement right to 2.
   • Now check people[0] + people[2] (1 + 2), which fits the limit. Pair them together and move both pointers (left = 1, right = 1).
   • The remaining person (weight 2) takes their own boat.

3. Boat Counting:

   • In each iteration, a boat is used whether we pair two people or send one person alone. This guarantees we use the minimum number of boats.

Time Complexity:

• Sorting the array: Sorting takes O(n log n) where n is the number of people.
• Two-pointer traversal: The two-pointer approach runs in O(n) because each pointer moves at most n times.

Thus, the overall time complexity is O(n log n) due to sorting.

Space Complexity:

• The space complexity is O(1) because no extra space is used beyond a few pointers and variables.

Example Walkthrough:

For the input:

$people = [3,5,3,4]; $limit = 5;

1. After sorting: [3, 3, 4, 5].
2. Initial pointers: left = 0, right = 3 (pointing at 5).
3. Check if people[0] + people[3] (3 + 5) ≤ 5 → False. The heaviest person (5) goes alone. right--.
4. Check if people[0] + people[2] (3 + 4) ≤ 5 → False. The next heaviest person (4) goes alone. right--.
5. Check if people[0] + people[1] (3 + 3) ≤ 5 → False. Each person (3) goes alone.

Thus, the final output is 4 boats.

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