LeetCode #2861 — MEDIUM

Maximum Number of Alloys

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

The Problem

Problem Statement

You are the owner of a company that creates alloys using various types of metals. There are n different types of metals available, and you have access to k machines that can be used to create alloys. Each machine requires a specific amount of each metal type to create an alloy.

For the i^th machine to create an alloy, it needs composition[i][j] units of metal of type j. Initially, you have stock[i] units of metal type i, and purchasing one unit of metal type i costs cost[i] coins.

Given integers n, k, budget, a 1-indexed 2D array composition, and 1-indexed arrays stock and cost, your goal is to maximize the number of alloys the company can create while staying within the budget of budget coins.

All alloys must be created with the same machine.

Return the maximum number of alloys that the company can create.

Example 1:

Input: n = 3, k = 2, budget = 15, composition = [[1,1,1],[1,1,10]], stock = [0,0,0], cost = [1,2,3]
Output: 2
Explanation: It is optimal to use the 1^st machine to create alloys.
To create 2 alloys we need to buy the:
- 2 units of metal of the 1^st type.
- 2 units of metal of the 2^nd type.
- 2 units of metal of the 3^rd type.
In total, we need 2 * 1 + 2 * 2 + 2 * 3 = 12 coins, which is smaller than or equal to budget = 15.
Notice that we have 0 units of metal of each type and we have to buy all the required units of metal.
It can be proven that we can create at most 2 alloys.

Example 2:

Input: n = 3, k = 2, budget = 15, composition = [[1,1,1],[1,1,10]], stock = [0,0,100], cost = [1,2,3]
Output: 5
Explanation: It is optimal to use the 2^nd machine to create alloys.
To create 5 alloys we need to buy:
- 5 units of metal of the 1^st type.
- 5 units of metal of the 2^nd type.
- 0 units of metal of the 3^rd type.
In total, we need 5 * 1 + 5 * 2 + 0 * 3 = 15 coins, which is smaller than or equal to budget = 15.
It can be proven that we can create at most 5 alloys.

Example 3:

Input: n = 2, k = 3, budget = 10, composition = [[2,1],[1,2],[1,1]], stock = [1,1], cost = [5,5]
Output: 2
Explanation: It is optimal to use the 3^rd machine to create alloys.
To create 2 alloys we need to buy the:
- 1 unit of metal of the 1^st type.
- 1 unit of metal of the 2^nd type.
In total, we need 1 * 5 + 1 * 5 = 10 coins, which is smaller than or equal to budget = 10.
It can be proven that we can create at most 2 alloys.

Constraints:

1 <= n, k <= 100
0 <= budget <= 10⁸
composition.length == k
composition[i].length == n
1 <= composition[i][j] <= 100
stock.length == cost.length == n
0 <= stock[i] <= 10⁸
1 <= cost[i] <= 100

Step 01

Brute Force Baseline

Problem summary: You are the owner of a company that creates alloys using various types of metals. There are n different types of metals available, and you have access to k machines that can be used to create alloys. Each machine requires a specific amount of each metal type to create an alloy. For the ith machine to create an alloy, it needs composition[i][j] units of metal of type j. Initially, you have stock[i] units of metal type i, and purchasing one unit of metal type i costs cost[i] coins. Given integers n, k, budget, a 1-indexed 2D array composition, and 1-indexed arrays stock and cost, your goal is to maximize the number of alloys the company can create while staying within the budget of budget coins. All alloys must be created with the same machine. Return the maximum number of alloys that the company can create.

Baseline thinking

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

Pattern signal: Array · Binary Search

Example 1

3
2
15
[[1,1,1],[1,1,10]]
[0,0,0]
[1,2,3]

Example 2

3
2
15
[[1,1,1],[1,1,10]]
[0,0,100]
[1,2,3]

Example 3

2
3
10
[[2,1],[1,2],[1,1]]
[1,1]
[5,5]

Step 02

Core Insight

What unlocks the optimal approach

Use binary search to find the answer.

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 #2861: Maximum Number of Alloys
class Solution {
    int n;
    int budget;
    List<List<Integer>> composition;
    List<Integer> stock;
    List<Integer> cost;

    boolean isValid(long target) {
        for (List<Integer> currMachine : composition) {
            long remain = budget;
            for (int j = 0; j < n && remain >= 0; j++) {
                long need = Math.max(0, currMachine.get(j) * target - stock.get(j));
                remain -= need * cost.get(j);
            }
            if (remain >= 0) {
                return true;
            }
        }
        return false;
    }

    public int maxNumberOfAlloys(int n, int k, int budget, List<List<Integer>> composition,
        List<Integer> stock, List<Integer> cost) {
        this.n = n;
        this.budget = budget;
        this.composition = composition;
        this.stock = stock;
        this.cost = cost;
        int l = -1;
        int r = budget / cost.get(0) + stock.get(0);
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            if (isValid(mid)) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        return l;
    }
}

// Accepted solution for LeetCode #2861: Maximum Number of Alloys
func maxNumberOfAlloys(n int, k int, budget int, composition [][]int, stock []int, cost []int) int {
	isValid := func(target int) bool {
		for _, currMachine := range composition {
			remain := budget
			for i, x := range currMachine {
				need := max(0, x*target-stock[i])
				remain -= need * cost[i]
			}
			if remain >= 0 {
				return true
			}
		}
		return false
	}

	l, r := 0, budget+stock[0]
	for l < r {
		mid := (l + r + 1) >> 1
		if isValid(mid) {
			l = mid
		} else {
			r = mid - 1
		}
	}
	return l
}

# Accepted solution for LeetCode #2861: Maximum Number of Alloys
class Solution:
    def maxNumberOfAlloys(
        self,
        n: int,
        k: int,
        budget: int,
        composition: List[List[int]],
        stock: List[int],
        cost: List[int],
    ) -> int:
        ans = 0
        for c in composition:
            l, r = 0, budget + stock[0]
            while l < r:
                mid = (l + r + 1) >> 1
                s = sum(max(0, mid * x - y) * z for x, y, z in zip(c, stock, cost))
                if s <= budget:
                    l = mid
                else:
                    r = mid - 1
            ans = max(ans, l)
        return ans

// Accepted solution for LeetCode #2861: Maximum Number of Alloys
/**
 * [2861] Maximum Number of Alloys
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    fn check(
        n: usize,
        count: i32,
        budget: i32,
        composition: &Vec<i32>,
        stock: &Vec<i32>,
        cost: &Vec<i32>,
    ) -> bool {
        let mut need = 0i64;

        for i in 0..n {
            let number = composition[i] * count - stock[i];

            if number > 0 {
                need += number as i64 * cost[i] as i64;
            }
        }

        need <= budget as i64
    }

    pub fn max_number_of_alloys(
        n: i32,
        k: i32,
        budget: i32,
        composition: Vec<Vec<i32>>,
        stock: Vec<i32>,
        cost: Vec<i32>,
    ) -> i32 {
        let mut result = 0;
        let (n, k) = (n as usize, k as usize);

        for m in 0..k {
            let mut left = 0;
            let mut right = vec![0; n];

            for i in 0..n {
                right[i] = (stock[i] + budget / cost[i]) / composition[m][i];
            }

            let mut right = *right.iter().max().unwrap();

            while left < right {
                let mid = (left + right) / 2;

                if Solution::check(n, mid, budget, &composition[m], &stock, &cost) {
                    left = mid + 1;
                } else {
                    right = mid;
                }

                dbg!(left, right);
            }

            if !Solution::check(n, left, budget, &composition[m], &stock, &cost) {
                left = left - 1;
            }

            result = result.max(left);
        }
        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2861() {
        assert_eq!(
            Solution::max_number_of_alloys(
                3,
                2,
                15,
                vec![vec![1, 1, 1], vec![1, 1, 10]],
                vec![0, 0, 0],
                vec![1, 2, 3]
            ),
            2
        );
    }
}

// Accepted solution for LeetCode #2861: Maximum Number of Alloys
function maxNumberOfAlloys(
    n: number,
    k: number,
    budget: number,
    composition: number[][],
    stock: number[],
    cost: number[],
): number {
    let l = 0;
    let r = budget + stock[0];
    const isValid = (target: number): boolean => {
        for (const currMachine of composition) {
            let remain = budget;
            for (let i = 0; i < n; ++i) {
                let need = Math.max(0, target * currMachine[i] - stock[i]);
                remain -= need * cost[i];
            }
            if (remain >= 0) {
                return true;
            }
        }
        return false;
    };
    while (l < r) {
        const mid = (l + r + 1) >> 1;
        if (isValid(mid)) {
            l = mid;
        } else {
            r = mid - 1;
        }
    }
    return l;
}

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.