2335

2335. Minimum Amount of Time to Fill Cups (Easy)

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

Related Topics:
Array, Greedy, Sorting, Heap (Priority Queue)

Similar Questions:

• Construct Target Array With Multiple Sums (Hard)
• Maximum Score From Removing Stones (Medium)
• Maximum Running Time of N Computers (Hard)
• Minimum Cost to Make Array Equal (Hard)

Solution 1.

// OJ: https://leetcode.com/problems/minimum-amount-of-time-to-fill-cups
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int fillCups(vector<int>& A) {
        int ans = 0;
        priority_queue<int> pq;
        for (int n : A) {
            if (n) pq.push(n);
        }
        while (pq.size() >= 2) {
            int a = pq.top();
            pq.pop();
            int b = pq.top();
            pq.pop();
            if (a - 1 > 0) pq.push(a - 1);
            if (b - 1 > 0) pq.push(b - 1);
            ++ans;
        }
        if (pq.size()) ans += pq.top();
        return ans;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/minimum-amount-of-time-to-fill-cups
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
// Ref: https://leetcode.com/problems/minimum-amount-of-time-to-fill-cups/solutions/2261394/java-c-python-max-max-a-sum-a-1-2/
class Solution {
public:
    int fillCups(vector<int>& A) {
        int mx = 0, sum = 0;
        for (int n : A) {
            mx = max(mx, n);
            sum += n;
        }
        return max(mx, (sum + 1) / 2);
    }
};