Minimum Amount Of Time To Fill Cups - Leetcode Solution

Problem:

You have a number of goals to reach in your list. Each goal consists of several cups that need to be filled with different amounts of water. Your task is to find the minimum amount of time needed to fill all the cups such that each cup is filled with the exact amount of water required.

You are given:

• An array of n integers, goals, where each goals[i] is the number of cups in the i-th goal.
• An array of n integers, batchSize, where each batchSize[i] is the number of cups you can fill at once from the i-th faucet.
• An array of n arrays, the j-th subarray of which, jugs[i], contains the amounts of water required to fill each cup in the i-th goal.

You can fill the cups in any order, and you can reuse a faucet for multiple cups at once if sufficient water is available.

Return the minimum amount of time required to fill all the cups in the order given by goals.

Solution:

To solve the problem, we need to find a way to distribute the water from the faucets to the cups such that all the cups are filled with the exact amount of water required. The problem can be solved by using a graph-based approach using the method of Maximum Bipartite Matching. This approach can be broken down into the following steps:

1. Constructing the Graph:

Construct a bipartite graph with two sets of nodes - the Faucet Set and Cup Set. Each node in the Faucet Set represents a faucet, which can fill a certain number of cups at once, and each node in the Cup Set represents a cup that needs to be filled. An edge is drawn from each faucet node to each cup node if the faucet can fill the cup.

2. Computing Maximum Bipartite Matching:

We can find the maximum flow of water that can be transferred from the faucets to the cups using the Ford-Fulkerson algorithm. Finally, the number of times we need to repeat the process to fill all the cups completely is the minimum amount of time required to fill all the cups.

3. Time Complexity:

The time complexity of this approach is O(m * n * log (m * n)), where m is the maximum number of cups that can be filled from a single faucet at once and n is the total number of cups.

4. Space Complexity:

The space complexity of this approach is O(m * n), where m is the maximum number of cups that can be filled from a single faucet at once and n is the total number of cups.

1