LeetCode #3789 — MEDIUM

Minimum Cost to Acquire Required Items

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

The Problem

Problem Statement

You are given five integers cost1, cost2, costBoth, need1, and need2.

There are three types of items available:

An item of type 1 costs cost1 and contributes 1 unit to the type 1 requirement only.
An item of type 2 costs cost2 and contributes 1 unit to the type 2 requirement only.
An item of type 3 costs costBoth and contributes 1 unit to both type 1 and type 2 requirements.

You must collect enough items so that the total contribution toward type 1 is at least need1 and the total contribution toward type 2 is at least need2.

Return an integer representing the minimum possible total cost to achieve these requirements.

Example 1:

Input: cost1 = 3, cost2 = 2, costBoth = 1, need1 = 3, need2 = 2

Output: 3

Explanation:

After buying three type 3 items, which cost 3 * 1 = 3, the total contribution to type 1 is 3 (>= need1 = 3) and to type 2 is 3 (>= need2 = 2).
Any other valid combination would cost more, so the minimum total cost is 3.

Example 2:

Input: cost1 = 5, cost2 = 4, costBoth = 15, need1 = 2, need2 = 3

Output: 22

Explanation:

We buy need1 = 2 items of type 1 and need2 = 3 items of type 2: 2 * 5 + 3 * 4 = 10 + 12 = 22.
Any other valid combination would cost more, so the minimum total cost is 22.

Example 3:

Input: cost1 = 5, cost2 = 4, costBoth = 15, need1 = 0, need2 = 0

Output: 0

Explanation:

Since no items are required (need1 = need2 = 0), we buy nothing and pay 0.

Constraints:

1 <= cost1, cost2, costBoth <= 10⁶
0 <= need1, need2 <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given five integers cost1, cost2, costBoth, need1, and need2. There are three types of items available: An item of type 1 costs cost1 and contributes 1 unit to the type 1 requirement only. An item of type 2 costs cost2 and contributes 1 unit to the type 2 requirement only. An item of type 3 costs costBoth and contributes 1 unit to both type 1 and type 2 requirements. You must collect enough items so that the total contribution toward type 1 is at least need1 and the total contribution toward type 2 is at least need2. Return an integer representing the minimum possible total cost to achieve these requirements.

Baseline thinking

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

Pattern signal: Math · Greedy

Example 1

Example 2

Example 3

Step 02

Core Insight

What unlocks the optimal approach

First, use <code>min(costBoth, cost1 + cost2)</code> for <code>min(need1, need2)</code>.
For the remaining type 1 requirement <code>rem1</code>, use cost <code>min(costBoth, cost1)</code> for the rest.
Do the same for type 2, using cost <code>min(costBoth, cost2)</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 #3789: Minimum Cost to Acquire Required Items
class Solution {
    public long minimumCost(int cost1, int cost2, int costBoth, int need1, int need2) {
        long a = (long) need1 * cost1 + (long) need2 * cost2;
        long b = (long) costBoth * Math.max(need1, need2);
        int mn = Math.min(need1, need2);
        long c = (long) costBoth * mn + (long) (need1 - mn) * cost1 + (long) (need2 - mn) * cost2;
        return Math.min(a, Math.min(b, c));
    }
}

// Accepted solution for LeetCode #3789: Minimum Cost to Acquire Required Items
func minimumCost(cost1 int, cost2 int, costBoth int, need1 int, need2 int) int64 {
	a := int64(need1)*int64(cost1) + int64(need2)*int64(cost2)
	b := int64(costBoth) * int64(max(need1, need2))
	mn := min(need1, need2)
	c := int64(costBoth)*int64(mn) +
		int64(need1-mn)*int64(cost1) +
		int64(need2-mn)*int64(cost2)
	return min(a, min(b, c))
}

# Accepted solution for LeetCode #3789: Minimum Cost to Acquire Required Items
class Solution:
    def minimumCost(
        self, cost1: int, cost2: int, costBoth: int, need1: int, need2: int
    ) -> int:
        a = need1 * cost1 + need2 * cost2
        b = costBoth * max(need1, need2)
        mn = min(need1, need2)
        c = costBoth * mn + (need1 - mn) * cost1 + (need2 - mn) * cost2
        return min(a, b, c)

// Accepted solution for LeetCode #3789: Minimum Cost to Acquire Required Items
// 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 #3789: Minimum Cost to Acquire Required Items
// class Solution {
//     public long minimumCost(int cost1, int cost2, int costBoth, int need1, int need2) {
//         long a = (long) need1 * cost1 + (long) need2 * cost2;
//         long b = (long) costBoth * Math.max(need1, need2);
//         int mn = Math.min(need1, need2);
//         long c = (long) costBoth * mn + (long) (need1 - mn) * cost1 + (long) (need2 - mn) * cost2;
//         return Math.min(a, Math.min(b, c));
//     }
// }

// Accepted solution for LeetCode #3789: Minimum Cost to Acquire Required Items
function minimumCost(
    cost1: number,
    cost2: number,
    costBoth: number,
    need1: number,
    need2: number,
): number {
    const a = need1 * cost1 + need2 * cost2;
    const b = costBoth * Math.max(need1, need2);
    const mn = Math.min(need1, need2);
    const c = costBoth * mn + (need1 - mn) * cost1 + (need2 - mn) * cost2;
    return Math.min(a, b, c);
}

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(1)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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.