LeetCode #3814 — MEDIUM

Maximum Capacity Within Budget

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

The Problem

Problem Statement

You are given two integer arrays costs and capacity, both of length n, where costs[i] represents the purchase cost of the i^th machine and capacity[i] represents its performance capacity.

You are also given an integer budget.

You may select at most two distinct machines such that the total cost of the selected machines is strictly less than budget.

Return the maximum achievable total capacity of the selected machines.

Example 1:

Input: costs = [4,8,5,3], capacity = [1,5,2,7], budget = 8

Output: 8

Explanation:

Choose two machines with costs[0] = 4 and costs[3] = 3.
The total cost is 4 + 3 = 7, which is strictly less than budget = 8.
The maximum total capacity is capacity[0] + capacity[3] = 1 + 7 = 8.

Example 2:

Input: costs = [3,5,7,4], capacity = [2,4,3,6], budget = 7

Output: 6

Explanation:

Choose one machine with costs[3] = 4.
The total cost is 4, which is strictly less than budget = 7.
The maximum total capacity is capacity[3] = 6.

Example 3:

Input: costs = [2,2,2], capacity = [3,5,4], budget = 5

Output: 9

Explanation:

Choose two machines with costs[1] = 2 and costs[2] = 2.
The total cost is 2 + 2 = 4, which is strictly less than budget = 5.
The maximum total capacity is capacity[1] + capacity[2] = 5 + 4 = 9.

Constraints:

1 <= n == costs.length == capacity.length <= 10⁵
1 <= costs[i], capacity[i] <= 10⁵
1 <= budget <= 2 * 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given two integer arrays costs and capacity, both of length n, where costs[i] represents the purchase cost of the ith machine and capacity[i] represents its performance capacity. You are also given an integer budget. You may select at most two distinct machines such that the total cost of the selected machines is strictly less than budget. Return the maximum achievable total capacity of the selected machines.

Baseline thinking

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

Pattern signal: Array · Two Pointers · Binary Search

Example 1

[4,8,5,3]
[1,5,2,7]
8

Example 2

[3,5,7,4]
[2,4,3,6]
7

Example 3

[2,2,2]
[3,5,4]
5

Step 02

Core Insight

What unlocks the optimal approach

Sort machines by increasing <code>costs</code>, keeping <code>capacity</code> aligned.
Build a prefix array where <code>prefMax[i]</code> stores the maximum <code>capacity</code> among machines with index <code><= i</code>.
For selecting one machine, take the maximum <code>capacity</code> among all machines with <code>cost < budget</code>.
For selecting two machines, fix machine <code>i</code> and binary search the largest index <code>j < i</code> with <code>costs[j] < budget - costs[i]</code>.
Update the answer with <code>capacity[i] + prefMax[j]</code> whenever such <code>j</code> exists.

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 #3814: Maximum Capacity Within Budget
class Solution {
    public int maxCapacity(int[] costs, int[] capacity, int budget) {
        List<int[]> arr = new ArrayList<>();
        for (int k = 0; k < costs.length; k++) {
            int a = costs[k], b = capacity[k];
            if (a < budget) {
                arr.add(new int[] {a, b});
            }
        }
        if (arr.isEmpty()) {
            return 0;
        }
        arr.sort(Comparator.comparingInt(o -> o[0]));
        TreeSet<int[]> remain = new TreeSet<>((x, y) -> {
            if (x[0] != y[0]) {
                return x[0] - y[0];
            }
            return x[1] - y[1];
        });
        for (int i = 0; i < arr.size(); i++) {
            remain.add(new int[] {arr.get(i)[1], i});
        }
        int i = 0, j = arr.size() - 1;
        int ans = remain.last()[0];
        while (i < j) {
            remain.remove(new int[] {arr.get(i)[1], i});
            while (i < j && arr.get(i)[0] + arr.get(j)[0] >= budget) {
                remain.remove(new int[] {arr.get(j)[1], j});
                j--;
            }
            if (!remain.isEmpty()) {
                ans = Math.max(ans, arr.get(i)[1] + remain.last()[0]);
            }
            i++;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3814: Maximum Capacity Within Budget
type Node struct {
	b int
	i int
}

type MaxHeap []Node

func (h MaxHeap) Len() int { return len(h) }
func (h MaxHeap) Less(i, j int) bool {
	if h[i].b != h[j].b {
		return h[i].b > h[j].b
	}
	return h[i].i > h[j].i
}
func (h MaxHeap) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *MaxHeap) Push(x interface{}) {
	*h = append(*h, x.(Node))
}
func (h *MaxHeap) Pop() interface{} {
	old := *h
	n := len(old)
	x := old[n-1]
	*h = old[:n-1]
	return x
}

func maxCapacity(costs []int, capacity []int, budget int) int {
	arr := make([][2]int, 0)
	for k := 0; k < len(costs); k++ {
		a, b := costs[k], capacity[k]
		if a < budget {
			arr = append(arr, [2]int{a, b})
		}
	}
	if len(arr) == 0 {
		return 0
	}
	sort.Slice(arr, func(i, j int) bool {
		if arr[i][0] != arr[j][0] {
			return arr[i][0] < arr[j][0]
		}
		return arr[i][1] < arr[j][1]
	})
	alive := make([]bool, len(arr))
	h := &MaxHeap{}
	for i := 0; i < len(arr); i++ {
		alive[i] = true
		heap.Push(h, Node{arr[i][1], i})
	}
	i, j := 0, len(arr)-1
	for h.Len() > 0 && !alive[(*h)[0].i] {
		heap.Pop(h)
	}
	ans := (*h)[0].b
	for i < j {
		alive[i] = false
		for i < j && arr[i][0]+arr[j][0] >= budget {
			alive[j] = false
			j--
		}
		for h.Len() > 0 && !alive[(*h)[0].i] {
			heap.Pop(h)
		}
		if h.Len() > 0 {
			if arr[i][1]+(*h)[0].b > ans {
				ans = arr[i][1] + (*h)[0].b
			}
		}
		i++
	}
	return ans
}

# Accepted solution for LeetCode #3814: Maximum Capacity Within Budget
class Solution:
    def maxCapacity(self, costs: List[int], capacity: List[int], budget: int) -> int:
        arr = []
        for a, b in zip(costs, capacity):
            if a < budget:
                arr.append((a, b))
        if not arr:
            return 0
        arr.sort()
        remain = SortedList()
        for i, (_, b) in enumerate(arr):
            remain.add((b, i))
        i, j = 0, len(arr) - 1
        ans = remain[-1][0]
        while i < j:
            remain.discard((arr[i][1], i))
            while i < j and arr[i][0] + arr[j][0] >= budget:
                remain.discard((arr[j][1], j))
                j -= 1
            if remain:
                ans = max(ans, arr[i][1] + remain[-1][0])
            i += 1
        return ans

// Accepted solution for LeetCode #3814: Maximum Capacity Within Budget
// 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 #3814: Maximum Capacity Within Budget
// class Solution {
//     public int maxCapacity(int[] costs, int[] capacity, int budget) {
//         List<int[]> arr = new ArrayList<>();
//         for (int k = 0; k < costs.length; k++) {
//             int a = costs[k], b = capacity[k];
//             if (a < budget) {
//                 arr.add(new int[] {a, b});
//             }
//         }
//         if (arr.isEmpty()) {
//             return 0;
//         }
//         arr.sort(Comparator.comparingInt(o -> o[0]));
//         TreeSet<int[]> remain = new TreeSet<>((x, y) -> {
//             if (x[0] != y[0]) {
//                 return x[0] - y[0];
//             }
//             return x[1] - y[1];
//         });
//         for (int i = 0; i < arr.size(); i++) {
//             remain.add(new int[] {arr.get(i)[1], i});
//         }
//         int i = 0, j = arr.size() - 1;
//         int ans = remain.last()[0];
//         while (i < j) {
//             remain.remove(new int[] {arr.get(i)[1], i});
//             while (i < j && arr.get(i)[0] + arr.get(j)[0] >= budget) {
//                 remain.remove(new int[] {arr.get(j)[1], j});
//                 j--;
//             }
//             if (!remain.isEmpty()) {
//                 ans = Math.max(ans, arr.get(i)[1] + remain.last()[0]);
//             }
//             i++;
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3814: Maximum Capacity Within Budget
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #3814: Maximum Capacity Within Budget
// class Solution {
//     public int maxCapacity(int[] costs, int[] capacity, int budget) {
//         List<int[]> arr = new ArrayList<>();
//         for (int k = 0; k < costs.length; k++) {
//             int a = costs[k], b = capacity[k];
//             if (a < budget) {
//                 arr.add(new int[] {a, b});
//             }
//         }
//         if (arr.isEmpty()) {
//             return 0;
//         }
//         arr.sort(Comparator.comparingInt(o -> o[0]));
//         TreeSet<int[]> remain = new TreeSet<>((x, y) -> {
//             if (x[0] != y[0]) {
//                 return x[0] - y[0];
//             }
//             return x[1] - y[1];
//         });
//         for (int i = 0; i < arr.size(); i++) {
//             remain.add(new int[] {arr.get(i)[1], i});
//         }
//         int i = 0, j = arr.size() - 1;
//         int ans = remain.last()[0];
//         while (i < j) {
//             remain.remove(new int[] {arr.get(i)[1], i});
//             while (i < j && arr.get(i)[0] + arr.get(j)[0] >= budget) {
//                 remain.remove(new int[] {arr.get(j)[1], j});
//                 j--;
//             }
//             if (!remain.isEmpty()) {
//                 ans = Math.max(ans, arr.get(i)[1] + remain.last()[0]);
//             }
//             i++;
//         }
//         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(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.

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.