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3789. Minimum Cost to Acquire Required Items

Description

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

Solutions

Solution 1: Case Analysis

We can divide the purchasing strategy into three cases:

  1. Only buy Type 1 and Type 2 items. The total cost is \(a = \textit{need1} \times \textit{cost1} + \textit{need2} \times \textit{cost2}\).
  2. Only buy Type 3 items. The total cost is \(b = \textit{costBoth} \times \max(\textit{need1}, \textit{need2})\).
  3. Buy some Type 3 items, and purchase Type 1 and Type 2 items separately for the remaining needs. Let \(\textit{mn} = \min(\textit{need1}, \textit{need2})\), then the total cost is \(c = \textit{costBoth} \times \textit{mn} + (\textit{need1} - \textit{mn}) \times \textit{cost1} + (\textit{need2} - \textit{mn}) \times \textit{cost2}\).

Finally, we return the minimum value among the three cases, \(\min(a, b, c)\).

The time complexity is \(O(1)\), and the space complexity is \(O(1)\).

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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)

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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));
    }
}

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class Solution {
public:
    long long minimumCost(int cost1, int cost2, int costBoth, int need1, int need2) {
        long long a = 1LL * need1 * cost1 + 1LL * need2 * cost2;
        long long b = 1LL * costBoth * max(need1, need2);
        int mn = min(need1, need2);
        long long c = 1LL * costBoth * mn
            + 1LL * (need1 - mn) * cost1
            + 1LL * (need2 - mn) * cost2;
        return min({a, b, c});
    }
};

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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))
}

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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);
}

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