LeetCode #881 — MEDIUM

Boats to Save People

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an array people where people[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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Two Pointers · Greedy

Example 1

[1,2]
3

Example 2

[3,2,2,1]
3

Example 3

[3,5,3,4]
5

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #881: Boats to Save People
class Solution {
    public int numRescueBoats(int[] people, int limit) {
        Arrays.sort(people);
        int ans = 0;
        for (int i = 0, j = people.length - 1; i <= j; --j) {
            if (people[i] + people[j] <= limit) {
                ++i;
            }
            ++ans;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #881: Boats to Save People
func numRescueBoats(people []int, limit int) int {
	sort.Ints(people)
	ans := 0
	for i, j := 0, len(people)-1; i <= j; j-- {
		if people[i]+people[j] <= limit {
			i++
		}
		ans++
	}
	return ans
}

# Accepted solution for LeetCode #881: Boats to Save People
class Solution:
    def numRescueBoats(self, people: List[int], limit: int) -> int:
        people.sort()
        ans = 0
        i, j = 0, len(people) - 1
        while i <= j:
            if people[i] + people[j] <= limit:
                i += 1
            j -= 1
            ans += 1
        return ans

// Accepted solution for LeetCode #881: Boats to Save People
struct Solution;

impl Solution {
    fn num_rescue_boats(mut people: Vec<i32>, limit: i32) -> i32 {
        let n = people.len();
        people.sort_unstable();
        people.reverse();
        let mut i = 0;
        let mut j = n - 1;
        let mut res = 0;
        while i <= j {
            res += 1;
            if people[i] + people[j] <= limit {
                j -= 1;
            }
            i += 1;
        }
        res
    }
}

#[test]
fn test() {
    let people = vec![1, 2];
    let limit = 3;
    let res = 1;
    assert_eq!(Solution::num_rescue_boats(people, limit), res);
    let people = vec![3, 2, 2, 1];
    let limit = 3;
    let res = 3;
    assert_eq!(Solution::num_rescue_boats(people, limit), res);
    let people = vec![3, 5, 3, 4];
    let limit = 5;
    let res = 4;
    assert_eq!(Solution::num_rescue_boats(people, limit), res);
}

// Accepted solution for LeetCode #881: Boats to Save People
function numRescueBoats(people: number[], limit: number): number {
    people.sort((a, b) => a - b);
    let ans = 0;
    for (let i = 0, j = people.length - 1; i <= j; --j) {
        if (people[i] + people[j] <= limit) {
            ++i;
        }
        ++ans;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(log n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.