How do you solve Maximum Balanced Shipments on LeetCode?

Maximum Balanced Shipments (LeetCode #3638) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Use of monotonic stack for efficient index tracking

What is the time complexity of Maximum Balanced Shipments?

The optimal solution for Maximum Balanced Shipments runs in O(n) time and uses O(n) space. We traverse the array once, and each index is pushed and popped from the stack at most once.

What is the brute force solution for Maximum Balanced Shipments?

The brute force approach for Maximum Balanced Shipments is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every possible subarray to see if it forms a balanced shipment. This approach is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Balanced Shipments?

Maximum Balanced Shipments uses the following concepts: Array, Dynamic Programming, Stack, Greedy, Monotonic Stack. The recommended patterns to study are: Dynamic Programming, Monotonic Stack.

What are the interview tips for Maximum Balanced Shipments?

Explain your thought process clearly Practice coding the monotonic stack pattern Be prepared to discuss time and space complexities

What are common mistakes when solving Maximum Balanced Shipments?

Not using a stack correctly for nearest larger element Confusing balanced shipment conditions

#3638

Maximum Balanced Shipments

Medium

ArrayDynamic ProgrammingStackGreedyMonotonic StackDynamic ProgrammingMonotonic Stack

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Use a monotonic stack to efficiently track the nearest larger parcel weight, allowing us to calculate balanced shipments in linear time.

⚙️

Algorithm

3 steps

1Step 1: Initialize a stack to track indices of parcels and a dp array to store the maximum shipments up to each index.
2Step 2: For each parcel, use the stack to find the nearest previous index with a larger weight.
3Step 3: Update dp based on the found index and maintain a best array to track the maximum shipments.

solution.py13 lines

1def maxBalancedShipments(weight):
2    n = len(weight)
3    dp = [0] * n
4    best = [0] * n
5    stack = []
6    for i in range(n):
7        while stack and weight[stack[-1]] <= weight[i]:
8            stack.pop()
9        j = stack[-1] if stack else -1
10        dp[i] = (dp[j] + 1) if j != -1 else 1
11        best[i] = max(best[i-1], dp[i])
12        stack.append(i)
13    return best[-1]

ℹ

Complexity note: We traverse the array once, and each index is pushed and popped from the stack at most once.

1Use of monotonic stack for efficient index tracking
2Dynamic programming to store optimal solutions

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.