What is the brute force approach and complexity for Boats to Save People?

Repeatedly find the heaviest person, then scan everyone for a partner who fits with time complexity O(N²) and space complexity O(N).

What is the optimal approach and complexity for Boats to Save People?

Sort; each step the heaviest leaves, taking the lightest along if they fit with optimal time complexity O(N log N) and space complexity O(1).

How do you identify the pattern for Boats to Save People?

Look for key signals in the prompt: "at most 2 people per boat" indicates Every boat is a pair (or a single) → a pairing problem, which suggests two pointers on a sorted array.; "minimum number of boats" indicates Make as many good pairs as possible. A greedy choice is needed for who pairs with whom.; "weight limit" indicates After sorting, the sum of the two ends tells us if the hardest person can be paired at all..

Boats to Save People

Medium·LeetCode #881 Pro OnlyInterview-Level

LeetCode

Problem Statement

people[i] is the weight of person i. You have as many boats as you need. Every boat can carry at most 2 people, and the total weight in a boat must be at most limit.

Return the smallest number of boats needed to carry everyone.

No single person is heavier than limit, so everyone can at least go alone.

Input

An integer array people and an integer limit.

Output

One integer: the minimum number of boats.

Example 1

Input: people = [1, 2], limit = 3

Output: 1

1 + 2 = 3 fits, so both share one boat.

Example 2

Input: people = [3, 2, 2, 1], limit = 3

Output: 3

Boats: [1, 2], [2], [3]. The 3 can't share with anyone, and the two 2s can't share with each other.

Example 3

Input: people = [3, 5, 3, 4], limit = 5

Output: 4

Every pair weighs more than 5, so everyone goes alone.

Constraints

• 1 ≤ people.length ≤ 5 · 10^4
• 1 ≤ people[i] ≤ limit ≤ 3 · 10^4

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“at most 2 people per boat”

→ Every boat is a pair (or a single) → a pairing problem, which suggests two pointers on a sorted array.

“minimum number of boats”

→ Make as many good pairs as possible. A greedy choice is needed for who pairs with whom.

“weight limit”

→ After sorting, the sum of the two ends tells us if the hardest person can be paired at all.

Approach Complexity Ladder

Brute Force — heaviest first, scan for a partnerO(N²)O(N)

Repeatedly find the heaviest person, then scan everyone for a partner who fits

Optimal — sort, then pair heaviest with lightestO(N log N)O(1)

Sort; each step the heaviest leaves, taking the lightest along if they fit

Brute Force Intuition

Keep a used array. Each round, scan for the heaviest person still on shore and put them in a new boat. Then scan everyone else and add the heaviest partner who still fits, so the boat is as full as possible. Repeat until everyone is on a boat.

Recognition rule for this sub-pattern:

Start at both ends of a sorted (or symmetric) structure. Each step, move the side that cannot do better.

Remember this · Two Pointers

Trigger

Sorted array, pairs/triplets with a target, comparing from both ends, compacting an array in place.

Think

If I compare the elements at two positions, can I safely discard one side?

Key question

Which pointer moves, and why can the skipped elements never be the answer?

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