#2305

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auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Fair Distribution of Cookies

The problem focuses on fairly distributing cookies among children to minimize the maximum unfairness of the distribution.

array dynamic programming backtracking bit manipulation bitmask

Problem Statement

You are given an integer array 'cookies', where cookies[i] denotes the number of cookies in the ith bag. You are also given an integer k that represents the number of children to distribute the bags to. Each child must receive an entire bag of cookies, and the bags cannot be split between children.

The goal is to minimize the unfairness of the distribution. Unfairness is defined as the maximum total number of cookies received by any single child. Your task is to find the minimum unfairness for distributing the cookies among the children.

Examples

Example 1

Input: cookies = [8,15,10,20,8], k = 2

Output: 31

One optimal distribution is [8,15,8] and [10,20]

The 1st child receives [8,15,8] which has a total of 8 + 15 + 8 = 31 cookies.
The 2nd child receives [10,20] which has a total of 10 + 20 = 30 cookies. The unfairness of the distribution is max(31,30) = 31. It can be shown that there is no distribution with an unfairness less than 31.

Example 2

Input: cookies = [6,1,3,2,2,4,1,2], k = 3

Output: 7

One optimal distribution is [6,1], [3,2,2], and [4,1,2]

The 1st child receives [6,1] which has a total of 6 + 1 = 7 cookies.
The 2nd child receives [3,2,2] which has a total of 3 + 2 + 2 = 7 cookies.
The 3rd child receives [4,1,2] which has a total of 4 + 1 + 2 = 7 cookies. The unfairness of the distribution is max(7,7,7) = 7. It can be shown that there is no distribution with an unfairness less than 7.

Constraints

2 <= cookies.length <= 8
1 <= cookies[i] <= 105
2 <= k <= cookies.length

Solution Approach

State Transition Dynamic Programming

Use dynamic programming to track the minimum unfairness of distributing cookies by evaluating all possible ways to assign each bag to one of the k children. We can use bitmasking to represent which bags have been assigned, and this enables efficient state transitions.

Bitmask Representation for Subsets

Represent each distribution state as a bitmask, where each bit indicates whether a cookie bag is assigned to a child. This allows for efficient calculation of all possible distributions and their associated unfairness by iterating over each bitmask.

Backtracking with Pruning

Apply backtracking to explore all possible distributions, pruning branches early when they exceed the current minimum unfairness. This approach reduces redundant calculations and speeds up the process of finding the optimal distribution.

Complexity Analysis

Metric	Value
Time	O(k^n)
Space	O(k + n)

The time complexity is O(k^n) due to the recursive nature of the problem and the exploration of all combinations of cookie assignments. The space complexity is O(k + n), where k is the number of children and n is the number of cookie bags, required to store intermediate states during computation.

What Interviewers Usually Probe

Candidate demonstrates an understanding of dynamic programming and bitmasking techniques.
Candidate applies backtracking effectively to prune non-optimal solutions.
Candidate explores the time complexity of state transitions and optimizes it effectively.

Common Pitfalls or Variants

Common pitfalls

Not pruning the search space efficiently, leading to excessive recursion or slow performance.
Failing to consider all possible distributions of cookies due to incomplete state representation.
Misunderstanding the concept of unfairness, which could lead to suboptimal solutions.

Follow-up variants

Extending the problem to more children, where the size of k increases.
Changing the problem to minimize the total number of cookies given to any single child instead of the maximum unfairness.
Introducing constraints where children may not receive the same number of bags.

FAQ

What is the state transition dynamic programming pattern used in "Fair Distribution of Cookies"?

The state transition dynamic programming pattern in this problem uses bitmasking to represent all possible ways of distributing the cookie bags among children. It allows for efficient exploration of all configurations and calculation of unfairness.

How does backtracking help in "Fair Distribution of Cookies"?

Backtracking helps prune non-optimal solutions by exploring all potential distributions while eliminating paths that already exceed the current best unfairness value.

What is unfairness in the context of the "Fair Distribution of Cookies" problem?

Unfairness is defined as the maximum number of cookies received by any child. The goal is to minimize this maximum value across all possible distributions of the cookie bags.

Why is bitmasking used in "Fair Distribution of Cookies"?

Bitmasking is used to represent the assignment of cookie bags to children efficiently, enabling quick evaluation of all possible distributions while reducing the computational complexity.

What is the time complexity of the "Fair Distribution of Cookies" solution?

The time complexity is O(k^n), where k is the number of children and n is the number of cookie bags. This reflects the need to evaluate all possible distributions.

terminal

Solution

Solution 1: Backtracking + Pruning

First, we sort the array $cookies$ in descending order (to reduce the number of searches), and then create an array $cnt$ of length $k$ to store the number of cookies each child gets. Also, we use a variable $ans$ to maintain the current minimum degree of unfairness, initialized to a very large value.

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class Solution:
    def distributeCookies(self, cookies: List[int], k: int) -> int:
        def dfs(i):
            if i >= len(cookies):
                nonlocal ans
                ans = max(cnt)
                return
            for j in range(k):
                if cnt[j] + cookies[i] >= ans or (j and cnt[j] == cnt[j - 1]):
                    continue
                cnt[j] += cookies[i]
                dfs(i + 1)
                cnt[j] -= cookies[i]

        ans = inf
        cnt = [0] * k
        cookies.sort(reverse=True)
        dfs(0)
        return ans

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