LeetCode #2305 — MEDIUM

Fair Distribution of Cookies

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an integer array cookies, where cookies[i] denotes the number of cookies in the i^th bag. You are also given an integer k that denotes the number of children to distribute all the bags of cookies to. All the cookies in the same bag must go to the same child and cannot be split up.

The unfairness of a distribution is defined as the maximum total cookies obtained by a single child in the distribution.

Return the minimum unfairness of all distributions.

Example 1:

Input: cookies = [8,15,10,20,8], k = 2
Output: 31
Explanation: One optimal distribution is [8,15,8] and [10,20]
- The 1^st child receives [8,15,8] which has a total of 8 + 15 + 8 = 31 cookies.
- The 2^nd 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
Explanation: One optimal distribution is [6,1], [3,2,2], and [4,1,2]
- The 1^st child receives [6,1] which has a total of 6 + 1 = 7 cookies.
- The 2^nd child receives [3,2,2] which has a total of 3 + 2 + 2 = 7 cookies.
- The 3^rd 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

Step 01

Brute Force Baseline

Problem summary: 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 denotes the number of children to distribute all the bags of cookies to. All the cookies in the same bag must go to the same child and cannot be split up. The unfairness of a distribution is defined as the maximum total cookies obtained by a single child in the distribution. Return the minimum unfairness of all distributions.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Backtracking · Bit Manipulation

Example 1

[8,15,10,20,8]
2

Example 2

[6,1,3,2,2,4,1,2]
3

Core Insight

What unlocks the optimal approach

We have to give each bag to one of the children. How can we enumerate all of the possibilities?
Use recursion and keep track of the current number of cookies each child has. Once all the bags have been distributed, find the child with the most cookies.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2305: Fair Distribution of Cookies
class Solution {
    private int[] cookies;
    private int[] cnt;
    private int k;
    private int n;
    private int ans = 1 << 30;

    public int distributeCookies(int[] cookies, int k) {
        n = cookies.length;
        cnt = new int[k];
        // 升序排列
        Arrays.sort(cookies);
        this.cookies = cookies;
        this.k = k;
        // 这里搜索顺序是 n-1, n-2,...0
        dfs(n - 1);
        return ans;
    }

    private void dfs(int i) {
        if (i < 0) {
            // ans = Arrays.stream(cnt).max().getAsInt();
            ans = 0;
            for (int v : cnt) {
                ans = Math.max(ans, v);
            }
            return;
        }
        for (int j = 0; j < k; ++j) {
            if (cnt[j] + cookies[i] >= ans || (j > 0 && cnt[j] == cnt[j - 1])) {
                continue;
            }
            cnt[j] += cookies[i];
            dfs(i - 1);
            cnt[j] -= cookies[i];
        }
    }
}

// Accepted solution for LeetCode #2305: Fair Distribution of Cookies
func distributeCookies(cookies []int, k int) int {
	sort.Sort(sort.Reverse(sort.IntSlice(cookies)))
	cnt := make([]int, k)
	ans := 1 << 30
	var dfs func(int)
	dfs = func(i int) {
		if i >= len(cookies) {
			ans = slices.Max(cnt)
			return
		}
		for j := 0; j < k; j++ {
			if cnt[j]+cookies[i] >= ans || (j > 0 && cnt[j] == cnt[j-1]) {
				continue
			}
			cnt[j] += cookies[i]
			dfs(i + 1)
			cnt[j] -= cookies[i]
		}
	}
	dfs(0)
	return ans
}

# Accepted solution for LeetCode #2305: Fair Distribution of Cookies
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

// Accepted solution for LeetCode #2305: Fair Distribution of Cookies
/**
 * [2305] Fair Distribution of Cookies
 *
 * You are given an integer array cookies, where cookies[i] denotes the number of cookies in the i^th bag. You are also given an integer k that denotes the number of children to distribute all the bags of cookies to. All the cookies in the same bag must go to the same child and cannot be split up.
 * The unfairness of a distribution is defined as the maximum total cookies obtained by a single child in the distribution.
 * Return the minimum unfairness of all distributions.
 *  
 * Example 1:
 *
 * Input: cookies = [8,15,10,20,8], k = 2
 * Output: 31
 * Explanation: One optimal distribution is [8,15,8] and [10,20]
 * - The 1^st child receives [8,15,8] which has a total of 8 + 15 + 8 = 31 cookies.
 * - The 2^nd 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
 * Explanation: One optimal distribution is [6,1], [3,2,2], and [4,1,2]
 * - The 1^st child receives [6,1] which has a total of 6 + 1 = 7 cookies.
 * - The 2^nd child receives [3,2,2] which has a total of 3 + 2 + 2 = 7 cookies.
 * - The 3^rd 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] <= 10^5
 * 	2 <= k <= cookies.length
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/fair-distribution-of-cookies/
// discuss: https://leetcode.com/problems/fair-distribution-of-cookies/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn distribute_cookies(cookies: Vec<i32>, k: i32) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2305_example_1() {
        let cookies = vec![8, 15, 10, 20, 8];
        let k = 2;

        let result = 31;

        assert_eq!(Solution::distribute_cookies(cookies, k), result);
    }

    #[test]
    #[ignore]
    fn test_2305_example_2() {
        let cookies = vec![6, 1, 3, 2, 2, 4, 1, 2];
        let k = 3;

        let result = 7;

        assert_eq!(Solution::distribute_cookies(cookies, k), result);
    }
}

// Accepted solution for LeetCode #2305: Fair Distribution of Cookies
function distributeCookies(cookies: number[], k: number): number {
    const cnt = new Array(k).fill(0);
    let ans = 1 << 30;
    const dfs = (i: number) => {
        if (i >= cookies.length) {
            ans = Math.max(...cnt);
            return;
        }
        for (let j = 0; j < k; ++j) {
            if (cnt[j] + cookies[i] >= ans || (j && cnt[j] == cnt[j - 1])) {
                continue;
            }
            cnt[j] += cookies[i];
            dfs(i + 1);
            cnt[j] -= cookies[i];
        }
    };
    dfs(0);
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.