Problem Description

You have an integer array cookies where cookies[i] represents the number of cookies in the i-th bag. You need to distribute all these bags to k children with the following constraints:

• Each bag must be given to exactly one child (you cannot split a bag between multiple children)
• All cookies in a bag stay together

The unfairness of a distribution is defined as the maximum total number of cookies that any single child receives.

Your task is to find the minimum possible unfairness across all valid distributions.

For example, if you have cookies = [8, 15, 10, 20, 8] and k = 2 children:

• One possible distribution: Child 1 gets bags with [8, 15, 8] (total = 31 cookies), Child 2 gets bags with [10, 20] (total = 30 cookies). The unfairness is max(31, 30) = 31.
• Another distribution: Child 1 gets bags with [20, 8] (total = 28 cookies), Child 2 gets bags with [8, 15, 10] (total = 33 cookies). The unfairness is max(28, 33) = 33.

The goal is to find the distribution strategy that minimizes this maximum value (the unfairness).

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• No: The problem doesn't involve nodes and edges or graph traversal. We're dealing with distributing bags of cookies to children.

Need to solve for kth smallest/largest?

• No: We're not looking for a kth smallest or largest element. We need to find the minimum possible unfairness value.

Involves Linked Lists?

• No: The problem uses an array of cookies, not linked lists.

Does the problem have small constraints?

• Yes: Looking at the problem, we need to distribute bags to k children, and typically these distribution problems have relatively small constraints (usually k ≤ 10 and array length ≤ 15-20). We need to explore all possible ways to distribute the bags.

Brute force / Backtracking?

• Yes: Since we have small constraints and need to explore all possible distributions to find the optimal one, backtracking is the appropriate approach. We need to:
  • Try assigning each bag to each possible child
  • Track the current distribution's unfairness
  • Backtrack when we've assigned all bags or when a distribution exceeds our current best
  • Explore all possibilities to find the minimum unfairness

Conclusion: The flowchart correctly leads us to use a Backtracking approach. This makes sense because we need to exhaustively search through all possible ways to distribute the cookie bags among the children, keeping track of the minimum unfairness found, and pruning branches that cannot lead to better solutions.

Intuition

When we need to distribute items among multiple recipients and find the optimal distribution, we're essentially dealing with a decision tree where at each step we decide which recipient gets which item. Since we want to minimize the maximum cookies any child receives, we need to explore different distribution strategies.

The key insight is that this is a classic assignment problem with a twist - we care about minimizing the maximum load rather than the total. Think of it like distributing work among employees where you want to ensure no one is overworked. We can't use a greedy approach (like always giving to the child with the least cookies) because local optimal choices don't guarantee a global optimum.

For example, if we have cookies [20, 15, 10] and 2 children, greedily giving the largest bag first might not lead to the best solution. We need to try different combinations to find the truly optimal distribution.

Since we must try all possible distributions to guarantee finding the minimum unfairness, backtracking becomes our natural choice. We systematically:

1. Take a bag of cookies
2. Try giving it to each child
3. Recursively distribute the remaining bags
4. Track the maximum cookies any child gets in each complete distribution
5. Backtrack to try other possibilities

To make this efficient, we can add pruning strategies:

• Sort cookies in descending order to fail fast (larger values are more likely to exceed our current best answer early)
• Skip distributions where a child already has more cookies than our current best answer
• Skip giving cookies to a child if the previous child has the same amount (avoids duplicate distributions)

This exhaustive search with intelligent pruning ensures we find the minimum possible unfairness while avoiding unnecessary computation.

Learn more about Dynamic Programming, Backtracking and Bitmask patterns.

Solution Approach

The implementation uses a depth-first search (DFS) with backtracking to explore all possible distributions of cookie bags to children.

Data Structures:

• cnt: An array of size k that tracks the current number of cookies each child has
• ans: A variable to store the minimum unfairness found so far, initialized to infinity

Algorithm Steps:

1. Preprocessing: Sort the cookies array in descending order. This optimization helps with pruning - when we assign larger cookie bags first, we can more quickly identify and skip branches that will exceed our current best answer.

2. DFS Function: The recursive dfs(i) function handles distributing the i-th bag of cookies:

   • Base Case: When i >= len(cookies), all bags have been distributed. Calculate the current unfairness as max(cnt) and update ans if this distribution is better.

   • Recursive Case: For the current bag i, try giving it to each child j:

     Pruning conditions (skip if):

     • cnt[j] + cookies[i] >= ans: Adding this bag to child j would already meet or exceed our current best unfairness, so no point exploring this branch
     • j > 0 and cnt[j] == cnt[j-1]: If the current child has the same number of cookies as the previous child, giving the bag to either produces equivalent distributions (avoids duplicates)

     Backtracking steps:

     • Add cookies[i] to child j's count: cnt[j] += cookies[i]
     • Recursively distribute remaining bags: dfs(i + 1)
     • Backtrack by removing the cookies: cnt[j] -= cookies[i]

3. Main Execution:

   • Initialize cnt with zeros and ans with infinity
   • Start the DFS from bag 0
   • Return the minimum unfairness found

The sorting optimization combined with the two pruning conditions significantly reduces the search space. The first pruning condition acts as an upper bound check, while the second eliminates symmetric distributions, making the algorithm efficient enough for the given constraints.

Example Walkthrough

Let's trace through a small example with cookies = [6, 4, 5] and k = 2 children.

Step 1: Preprocessing

• Sort cookies in descending order: [6, 5, 4]
• Initialize cnt = [0, 0] (tracking cookies for each child)
• Initialize ans = infinity

Step 2: DFS Exploration

Starting with dfs(0) - distributing bag with 6 cookies:

Branch 1: Give bag 0 (6 cookies) to child 0

• cnt = [6, 0]

• Call dfs(1) - distributing bag with 5 cookies

  Branch 1.1: Give bag 1 (5 cookies) to child 0

  • cnt = [11, 0]
  • Call dfs(2) - distributing bag with 4 cookies
    • Must give to child 1 (child 0 already has 11)
    • cnt = [11, 4]
    • All bags distributed: unfairness = max(11, 4) = 11
    • Update ans = 11
    • Backtrack: cnt = [11, 0]
  • Backtrack: cnt = [6, 0]

  Branch 1.2: Give bag 1 (5 cookies) to child 1

  • cnt = [6, 5]

  • Call dfs(2) - distributing bag with 4 cookies

    Branch 1.2.1: Give to child 0

    • cnt = [10, 5]
    • All bags distributed: unfairness = max(10, 5) = 10
    • Update ans = 10 (better than 11)
    • Backtrack: cnt = [6, 5]

    Branch 1.2.2: Give to child 1

    • cnt = [6, 9]
    • All bags distributed: unfairness = max(6, 9) = 9
    • Update ans = 9 (better than 10)
    • Backtrack: cnt = [6, 5]

  • Backtrack: cnt = [6, 0]

• Backtrack: cnt = [0, 0]

Branch 2: Give bag 0 (6 cookies) to child 1

• cnt = [0, 6]

• Call dfs(1) - distributing bag with 5 cookies

  Branch 2.1: Give bag 1 (5 cookies) to child 0

  • cnt = [5, 6]

  • Call dfs(2) - distributing bag with 4 cookies

    Branch 2.1.1: Give to child 0

    • cnt = [9, 6]
    • Pruning: 9 >= ans (9), skip this branch
    • Backtrack: cnt = [5, 6]

    Branch 2.1.2: Give to child 1

    • cnt = [5, 10]
    • Pruning: 10 >= ans (9), skip this branch
    • Backtrack: cnt = [5, 6]

  • Backtrack: cnt = [0, 6]

  Branch 2.2: Give bag 1 (5 cookies) to child 1

  • cnt = [0, 11]
  • Pruning: 11 >= ans (9), skip entire subtree
  • Backtrack: cnt = [0, 6]

• Backtrack: cnt = [0, 0]

Result: The minimum unfairness is 9, achieved when:

• Child 0 gets bag with 6 cookies
• Child 1 gets bags with 5 and 4 cookies (total = 9)

The pruning conditions helped us skip several branches:

• When a child's current total would exceed our best answer
• This reduced our search from exploring all 8 possible distributions to examining only the necessary ones

Solution Implementation

Time and Space Complexity

Time Complexity: O(k^n) where n is the number of cookies and k is the number of children.

The algorithm uses backtracking to try all possible distributions of cookies among k children. For each cookie (n total cookies), we have at most k choices of which child to give it to. This creates a recursion tree with:

• Depth: n (one level for each cookie)
• Branching factor: at most k at each level

Although there are pruning conditions (cnt[j] + cookies[i] >= ans and cnt[j] == cnt[j - 1]) that reduce the actual number of branches explored, the worst-case scenario still requires exploring k^n possibilities.

Space Complexity: O(n + k)

The space complexity consists of:

• Recursion call stack: O(n) - The maximum depth of recursion is n (the number of cookies)
• Array cnt: O(k) - Stores the current distribution of cookies for k children
• Sorting the cookies array is done in-place, so no additional space for that
• Variable ans: O(1)

Therefore, the total space complexity is O(n + k).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Missing Early Termination for Empty Children

The Pitfall: The current pruning condition if child > 0 and children_cookies[child] == children_cookies[child - 1] only avoids duplicate states between consecutive children with the same cookie count. However, when multiple children have 0 cookies (which is common early in the recursion), the algorithm still explores giving the current bag to each of these empty children separately, even though these distributions are equivalent.

For example, with 3 children all having 0 cookies, the algorithm tries:

• Give bag to child 0 (explores full subtree)
• Give bag to child 1 (explores same subtree again)
• Give bag to child 2 (explores same subtree yet again)

The Solution: Add an early termination when encountering an empty child. Once we've explored giving a bag to one empty child, we don't need to try other empty children:

for child in range(k):
    # Pruning condition 1: Skip if exceeds current best
    if children_cookies[child] + cookies[cookie_index] >= self.min_unfairness:
        continue
  
    # Pruning condition 2: Skip duplicate states
    if child > 0 and children_cookies[child] == children_cookies[child - 1]:
        continue
  
    # Give cookie to current child
    children_cookies[child] += cookies[cookie_index]
  
    # Recursively distribute remaining cookies
    backtrack(cookie_index + 1)
  
    # Backtrack
    children_cookies[child] -= cookies[cookie_index]
  
    # IMPORTANT: If this child had 0 cookies, no need to try other empty children
    if children_cookies[child] == 0:
        break

2. Integer Overflow with Large Cookie Values

The Pitfall: While Python handles arbitrary precision integers automatically, in other languages or when optimizing with NumPy arrays, the sum of cookies could overflow standard integer types if cookie values are large.

The Solution: Ensure appropriate data types are used for tracking cookie counts, or add validation for input constraints:

# Add input validation if needed
if any(c > 10**9 for c in cookies) or len(cookies) * max(cookies) > 10**18:
    # Handle potential overflow scenarios
    pass

3. Not Handling Edge Case: k > len(cookies)

The Pitfall: When there are more children than cookie bags, some children will inevitably get 0 cookies. The current solution handles this correctly, but it's worth noting that the minimum unfairness in this case is always 0 (since at least one child gets nothing, and we can give all bags to one child).

The Solution: Add an optimization for this edge case:

def distributeCookies(self, cookies: List[int], k: int) -> int:
    # Edge case: more children than bags
    if k > len(cookies):
        return max(cookies)  # Best we can do is give one bag to one child
  
    # Rest of the implementation...