LeetCode #1011 — MEDIUM

Capacity To Ship Packages Within D Days

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

The Problem

Problem Statement

A conveyor belt has packages that must be shipped from one port to another within days days.

The i^th package on the conveyor belt has a weight of weights[i]. Each day, we load the ship with packages on the conveyor belt (in the order given by weights). We may not load more weight than the maximum weight capacity of the ship.

Return the least weight capacity of the ship that will result in all the packages on the conveyor belt being shipped within days days.

Example 1:

Input: weights = [1,2,3,4,5,6,7,8,9,10], days = 5
Output: 15
Explanation: A ship capacity of 15 is the minimum to ship all the packages in 5 days like this:
1st day: 1, 2, 3, 4, 5
2nd day: 6, 7
3rd day: 8
4th day: 9
5th day: 10

Note that the cargo must be shipped in the order given, so using a ship of capacity 14 and splitting the packages into parts like (2, 3, 4, 5), (1, 6, 7), (8), (9), (10) is not allowed.

Example 2:

Input: weights = [3,2,2,4,1,4], days = 3
Output: 6
Explanation: A ship capacity of 6 is the minimum to ship all the packages in 3 days like this:
1st day: 3, 2
2nd day: 2, 4
3rd day: 1, 4

Example 3:

Input: weights = [1,2,3,1,1], days = 4
Output: 3
Explanation:
1st day: 1
2nd day: 2
3rd day: 3
4th day: 1, 1

Constraints:

1 <= days <= weights.length <= 5 * 10⁴
1 <= weights[i] <= 500

Step 01

Brute Force Baseline

Problem summary: A conveyor belt has packages that must be shipped from one port to another within days days. The ith package on the conveyor belt has a weight of weights[i]. Each day, we load the ship with packages on the conveyor belt (in the order given by weights). We may not load more weight than the maximum weight capacity of the ship. Return the least weight capacity of the ship that will result in all the packages on the conveyor belt being shipped within days days.

Baseline thinking

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

Pattern signal: Array · Binary Search

Example 1

[1,2,3,4,5,6,7,8,9,10]
5

Example 2

[3,2,2,4,1,4]
3

Example 3

[1,2,3,1,1]
4

Core Insight

What unlocks the optimal approach

Binary search on the answer. We need a function possible(capacity) which returns true if and only if we can do the task in D days.

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 #1011: Capacity To Ship Packages Within D Days
class Solution {
    public int shipWithinDays(int[] weights, int days) {
        int left = 0, right = 0;
        for (int w : weights) {
            left = Math.max(left, w);
            right += w;
        }
        while (left < right) {
            int mid = (left + right) >> 1;
            if (check(mid, weights, days)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }

    private boolean check(int mx, int[] weights, int days) {
        int ws = 0, cnt = 1;
        for (int w : weights) {
            ws += w;
            if (ws > mx) {
                ws = w;
                ++cnt;
            }
        }
        return cnt <= days;
    }
}

// Accepted solution for LeetCode #1011: Capacity To Ship Packages Within D Days
func shipWithinDays(weights []int, days int) int {
	var left, right int
	for _, w := range weights {
		if left < w {
			left = w
		}
		right += w
	}
	return left + sort.Search(right, func(mx int) bool {
		mx += left
		ws, cnt := 0, 1
		for _, w := range weights {
			ws += w
			if ws > mx {
				ws = w
				cnt++
			}
		}
		return cnt <= days
	})
}

# Accepted solution for LeetCode #1011: Capacity To Ship Packages Within D Days
class Solution:
    def shipWithinDays(self, weights: List[int], days: int) -> int:
        def check(mx):
            ws, cnt = 0, 1
            for w in weights:
                ws += w
                if ws > mx:
                    cnt += 1
                    ws = w
            return cnt <= days

        left, right = max(weights), sum(weights) + 1
        return left + bisect_left(range(left, right), True, key=check)

// Accepted solution for LeetCode #1011: Capacity To Ship Packages Within D Days
struct Solution;

impl Solution {
    fn ship_within_days(weights: Vec<i32>, d: i32) -> i32 {
        let mut sum = 0;
        let mut max = 0;
        for &w in &weights {
            sum += w;
            max = max.max(w);
        }
        let mut l = max;
        let mut h = sum;
        while l < h {
            let m = l + (h - l) / 2;
            if Self::days(m, &weights) <= d {
                h = m;
            } else {
                l = m + 1;
            }
        }
        l
    }

    fn days(capacity: i32, weights: &[i32]) -> i32 {
        let mut cur = 0;
        let mut res = 1;
        for &w in weights {
            cur += w;
            if cur > capacity {
                res += 1;
                cur = w;
            }
        }
        res
    }
}

#[test]
fn test() {
    let weights = vec![1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
    let d = 5;
    let res = 15;
    assert_eq!(Solution::ship_within_days(weights, d), res);
    let weights = vec![3, 2, 2, 4, 1, 4];
    let d = 3;
    let res = 6;
    assert_eq!(Solution::ship_within_days(weights, d), res);
    let weights = vec![1, 2, 3, 1, 1];
    let d = 4;
    let res = 3;
    assert_eq!(Solution::ship_within_days(weights, d), res);
}

// Accepted solution for LeetCode #1011: Capacity To Ship Packages Within D Days
function shipWithinDays(weights: number[], days: number): number {
    let left = 0;
    let right = 0;
    for (const w of weights) {
        left = Math.max(left, w);
        right += w;
    }
    const check = (mx: number) => {
        let ws = 0;
        let cnt = 1;
        for (const w of weights) {
            ws += w;
            if (ws > mx) {
                ws = w;
                ++cnt;
            }
        }
        return cnt <= days;
    };
    while (left < right) {
        const mid = (left + right) >> 1;
        if (check(mid)) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}

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(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.