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Leetcode 2335. Minimum Amount of Time to Fill Cups

Problem Statement

You are given three integers representing the number of cups of water to be filled with three different types of water dispensers. Assume that:

• You can fill exactly one cup at a time utilizing a single dispenser.
• You may use all three dispensers simultaneously.

The task is to find the minimum amount of time required to fill all the cups.

Clarifying Questions

1. Input Range: What is the range of the integers representing the water cups for each dispenser?
   • Answer: The provided integers can be any non-negative value.
2. Output: Should the returned value just be the minimum amount of time in integer form?
   • Answer: Yes, the function should return a single integer representing the minimum time required.
3. Simultaneous Filling: Can we assume it is always beneficial to try and use multiple dispensers at the same time when possible?
   • Answer: Yes, using multiple dispensers simultaneously should help minimize the time.

Strategy

To minimize the time to fill all cups using the dispensers:

1. Sort the Cups: Start by sorting the three values to always deal with them in a non-decreasing order.
2. Simultaneous Filling: Utilize the fact that multiple dispensers can be used at the same time to always try and fill the highest remaining cups.

A clear approach involves these steps:

1. Sort the array of cup counts.
2. Pair the largest two values to be filled at the same time whenever their sum is greater than or equal to the third value.
3. If at some point the highest value of the cup count is greater than the sum of the other two, then the highest value is the minimum time needed because the other two will eventually be filled up simultaneously before the last cup is completely filled.

Given the input array [a, b, c] and sorting such that a <= b <= c, the minimum time can be efficiently derived.

Code

#include <algorithm> // For std::sort
#include <vector>

int fillCups(std::vector<int>& amount) {
    // Sort the vector to simplify the logic
    std::sort(amount.begin(), amount.end());
    
    // if largest number of cups to fill (amount[2]) is less than or equal 
    // to sum of the other two (amount[0] + amount[1]), it will take at 
    // least (sum of all) / 2 rounded up, otherwise it will take amount[2]
    if (amount[2] <= amount[0] + amount[1]) {
        return (amount[0] + amount[1] + amount[2] + 1) / 2;
    } else {
        return amount[2];
    }
}

Time Complexity

The time complexity analysis is as follows:

• Sorting: The primary sorting operation will take (O(3 \log 3)). Since the size of the array is fixed at 3, sorting is essentially a constant time operation, (O(1)).
• Comparison and calculation: Following the sort, the operations are constant time, (O(1)).

Thus, the overall time complexity is (O(1)). The algorithm operates in constant time due to the fixed number of elements.

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