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2335. Minimum Amount of Time to Fill Cups

Description

You have a water dispenser that can dispense cold, warm, and hot water. Every second, you can either fill up 2 cups with different types of water, or 1 cup of any type of water.

You are given a 0-indexed integer array amount of length 3 where amount[0], amount[1], and amount[2] denote the number of cold, warm, and hot water cups you need to fill respectively. Return the minimum number of seconds needed to fill up all the cups.

 

Example 1:

Input: amount = [1,4,2]
Output: 4
Explanation: One way to fill up the cups is:
Second 1: Fill up a cold cup and a warm cup.
Second 2: Fill up a warm cup and a hot cup.
Second 3: Fill up a warm cup and a hot cup.
Second 4: Fill up a warm cup.
It can be proven that 4 is the minimum number of seconds needed.

Example 2:

Input: amount = [5,4,4]
Output: 7
Explanation: One way to fill up the cups is:
Second 1: Fill up a cold cup, and a hot cup.
Second 2: Fill up a cold cup, and a warm cup.
Second 3: Fill up a cold cup, and a warm cup.
Second 4: Fill up a warm cup, and a hot cup.
Second 5: Fill up a cold cup, and a hot cup.
Second 6: Fill up a cold cup, and a warm cup.
Second 7: Fill up a hot cup.

Example 3:

Input: amount = [5,0,0]
Output: 5
Explanation: Every second, we fill up a cold cup.

 

Constraints:

  • amount.length == 3
  • 0 <= amount[i] <= 100

Solutions

Solution 1

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class Solution:
    def fillCups(self, amount: List[int]) -> int:
        ans = 0
        while sum(amount):
            amount.sort()
            ans += 1
            amount[2] -= 1
            amount[1] = max(0, amount[1] - 1)
        return ans

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class Solution {
    public int fillCups(int[] amount) {
        int ans = 0;
        while (amount[0] + amount[1] + amount[2] > 0) {
            Arrays.sort(amount);
            ++ans;
            amount[2]--;
            amount[1] = Math.max(0, amount[1] - 1);
        }
        return ans;
    }
}

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class Solution {
public:
    int fillCups(vector<int>& amount) {
        int ans = 0;
        while (amount[0] + amount[1] + amount[2]) {
            sort(amount.begin(), amount.end());
            ++ans;
            amount[2]--;
            amount[1] = max(0, amount[1] - 1);
        }
        return ans;
    }
};

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func fillCups(amount []int) int {
    ans := 0
    for amount[0]+amount[1]+amount[2] > 0 {
        sort.Ints(amount)
        ans++
        amount[2]--
        if amount[1] > 0 {
            amount[1]--
        }
    }
    return ans
}

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function fillCups(amount: number[]): number {
    amount.sort((a, b) => a - b);
    let [a, b, c] = amount;
    let diff = a + b - c;
    if (diff <= 0) return c;
    else return Math.floor((diff + 1) / 2) + c;
}

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impl Solution {
    pub fn fill_cups(mut amount: Vec<i32>) -> i32 {
        amount.sort();
        let dif = amount[0] + amount[1] - amount[2];
        if dif <= 0 {
            return amount[2];
        }
        (dif + 1) / 2 + amount[2]
    }
}

Solution 2

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class Solution:
    def fillCups(self, amount: List[int]) -> int:
        amount.sort()
        if amount[0] + amount[1] <= amount[2]:
            return amount[2]
        return (sum(amount) + 1) // 2

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class Solution {
    public int fillCups(int[] amount) {
        Arrays.sort(amount);
        if (amount[0] + amount[1] <= amount[2]) {
            return amount[2];
        }
        return (amount[0] + amount[1] + amount[2] + 1) / 2;
    }
}

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class Solution {
public:
    int fillCups(vector<int>& amount) {
        sort(amount.begin(), amount.end());
        if (amount[0] + amount[1] <= amount[2]) {
            return amount[2];
        }
        return (amount[0] + amount[1] + amount[2] + 1) / 2;
    }
};

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func fillCups(amount []int) int {
    sort.Ints(amount)
    if amount[0]+amount[1] <= amount[2] {
        return amount[2]
    }
    return (amount[0] + amount[1] + amount[2] + 1) / 2
}

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