How do you solve Maximum Capacity Within Budget on LeetCode?

Maximum Capacity Within Budget (LeetCode #3814) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting helps to efficiently pair machines within budget.

What is the time complexity of Maximum Capacity Within Budget?

The optimal solution for Maximum Capacity Within Budget runs in O(n log n) time and uses O(n) space. Sorting the machines takes O(n log n), and the two-pointer search is O(n).

What is the brute force solution for Maximum Capacity Within Budget?

The brute force approach for Maximum Capacity Within Budget is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every pair of machines to see if their combined cost is within budget. This guarantees finding the maximum capacity but is inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Capacity Within Budget?

Maximum Capacity Within Budget uses the following concepts: Array, Two Pointers, Binary Search, Sorting. The recommended patterns to study are: Two Pointers, Sorting.

What are the interview tips for Maximum Capacity Within Budget?

Explain your thought process clearly. Discuss edge cases like empty arrays. Practice similar problems to build confidence.

What are common mistakes when solving Maximum Capacity Within Budget?

Not considering the case of selecting only one machine. Overlooking the strict budget condition.

#3814

Maximum Capacity Within Budget

Medium

ArrayTwo PointersBinary SearchSortingTwo PointersSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log n)Space O(n)

Sort machines by cost and use a two-pointer technique to find the best combination of machines within budget efficiently.

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Algorithm

3 steps

1Step 1: Pair costs with capacities and sort them by costs.
2Step 2: Create a prefix array to store the maximum capacity up to each index.
3Step 3: Use two pointers to find the best pair of machines that fit within the budget.

solution.py15 lines

1def maxCapacity(costs, capacity, budget):
2    machines = sorted(zip(costs, capacity))
3    n = len(machines)
4    prefMax = [0] * n
5    prefMax[0] = machines[0][1]
6    for i in range(1, n):
7        prefMax[i] = max(prefMax[i-1], machines[i][1])
8    max_capacity = 0
9    for i in range(n):
10        j = n - 1
11        while j > i and machines[i][0] + machines[j][0] >= budget:
12            j -= 1
13        if j > i:
14            max_capacity = max(max_capacity, machines[i][1] + prefMax[j])
15    return max_capacity

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Complexity note: Sorting the machines takes O(n log n), and the two-pointer search is O(n).

1Sorting helps to efficiently pair machines within budget.
2Using a prefix max array allows quick access to the best capacity.

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