LeetCode #3638 — MEDIUM

Maximum Balanced Shipments

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

Solve on LeetCode

The Problem

Problem Statement

You are given an integer array weight of length n, representing the weights of n parcels arranged in a straight line. A shipment is defined as a contiguous subarray of parcels. A shipment is considered balanced if the weight of the last parcel is strictly less than the maximum weight among all parcels in that shipment.

Select a set of non-overlapping, contiguous, balanced shipments such that each parcel appears in at most one shipment (parcels may remain unshipped).

Return the maximum possible number of balanced shipments that can be formed.

Example 1:

Input: weight = [2,5,1,4,3]

Output: 2

Explanation:

We can form the maximum of two balanced shipments as follows:

Shipment 1: [2, 5, 1]
- Maximum parcel weight = 5
- Last parcel weight = 1, which is strictly less than 5. Thus, it's balanced.
Shipment 2: [4, 3]
- Maximum parcel weight = 4
- Last parcel weight = 3, which is strictly less than 4. Thus, it's balanced.

It is impossible to partition the parcels to achieve more than two balanced shipments, so the answer is 2.

Example 2:

Input: weight = [4,4]

Output: 0

Explanation:

No balanced shipment can be formed in this case:

A shipment [4, 4] has maximum weight 4 and the last parcel's weight is also 4, which is not strictly less. Thus, it's not balanced.
Single-parcel shipments [4] have the last parcel weight equal to the maximum parcel weight, thus not balanced.

As there is no way to form even one balanced shipment, the answer is 0.

Constraints:

2 <= n <= 10⁵
1 <= weight[i] <= 10⁹

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: You are given an integer array weight of length n, representing the weights of n parcels arranged in a straight line. A shipment is defined as a contiguous subarray of parcels. A shipment is considered balanced if the weight of the last parcel is strictly less than the maximum weight among all parcels in that shipment. Select a set of non-overlapping, contiguous, balanced shipments such that each parcel appears in at most one shipment (parcels may remain unshipped). Return the maximum possible number of balanced shipments that can be formed.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming · Stack · Greedy

Example 1

[2,5,1,4,3]

Example 2

[4,4]

Step 02

Core Insight

What unlocks the optimal approach

Use a monotonic stack to find for each end index <code>i</code> the nearest previous index <code>j</code> with <code>weight[j] > weight[i]</code> (set <code>j = -1</code> if none).
Then set <code>dp[i] = best[j] + 1</code> (using <code>best[-1] = 0</code>), and update <code>best[i] = max(best[i-1], dp[i])</code>; the result is <code>best[n-1]</code>.

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 #3638: Maximum Balanced Shipments
class Solution {
    public int maxBalancedShipments(int[] weight) {
        int ans = 0;
        int mx = 0;
        for (int x : weight) {
            mx = Math.max(mx, x);
            if (x < mx) {
                ++ans;
                mx = 0;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3638: Maximum Balanced Shipments
func maxBalancedShipments(weight []int) (ans int) {
	mx := 0
	for _, x := range weight {
		mx = max(mx, x)
		if x < mx {
			ans++
			mx = 0
		}
	}
	return
}

# Accepted solution for LeetCode #3638: Maximum Balanced Shipments
class Solution:
    def maxBalancedShipments(self, weight: List[int]) -> int:
        ans = mx = 0
        for x in weight:
            mx = max(mx, x)
            if x < mx:
                ans += 1
                mx = 0
        return ans

// Accepted solution for LeetCode #3638: Maximum Balanced Shipments
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3638: Maximum Balanced Shipments
// class Solution {
//     public int maxBalancedShipments(int[] weight) {
//         int ans = 0;
//         int mx = 0;
//         for (int x : weight) {
//             mx = Math.max(mx, x);
//             if (x < mx) {
//                 ++ans;
//                 mx = 0;
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3638: Maximum Balanced Shipments
function maxBalancedShipments(weight: number[]): number {
    let [ans, mx] = [0, 0];
    for (const x of weight) {
        mx = Math.max(mx, x);
        if (x < mx) {
            ans++;
            mx = 0;
        }
    }
    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 × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.

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.