Distribute Money to Maximum Children - Solution & Explanation
Problem Statement
You are given an integer money denoting the amount of money (in dollars) that you have and another integer children denoting the number of children that you must distribute the money to.
You have to distribute the money according to the following rules:
- All money must be distributed.
- Everyone must receive at least
1dollar. - Nobody receives
4dollars.
Return the maximum number of children who may receive exactly 8 dollars if you distribute the money according to the aforementioned rules. If there is no way to distribute the money, return -1.
Example 1:
Input: money = 20, children = 3 Output: 1 Explanation: The maximum number of children with 8 dollars will be 1. One of the ways to distribute the money is: - 8 dollars to the first child. - 9 dollars to the second child. - 3 dollars to the third child. It can be proven that no distribution exists such that number of children getting 8 dollars is greater than 1.
Example 2:
Input: money = 16, children = 2 Output: 2 Explanation: Each child can be given 8 dollars.
Constraints:
1 <= money <= 2002 <= children <= 30
Approach Overview
Problem Overview: You are given money dollars and children. Each child must receive at least $1. The goal is to maximize the number of children who receive exactly $8, while ensuring no child ends up with exactly $4. If distributing at least $1 to every child is impossible, return -1.
Approach 1: Brute Force Simulation (O(children) time, O(1) space)
Try every possible number of children that could receive exactly $8. For a candidate value k, compute the required money: 8 * k plus at least 1 dollar for each of the remaining children. Check whether the remaining money creates an invalid case where one child ends up with $4. Iterate from the maximum possible k down to 0 and return the first valid configuration. This approach directly models the constraints and is useful for reasoning about edge cases, but it performs multiple checks across the range of children.
Approach 2: Greedy Distribution (O(1) time, O(1) space)
The optimal solution relies on a greedy observation combined with simple math. First give every child $1. If money < children, the distribution is impossible and the answer is -1. After this base allocation, every child needs an additional $7 to reach $8. Compute how many such upgrades are possible using money / 7, capped by the number of children.
Two edge cases require adjustment. If every child receives $8 but leftover money remains, at least one child would exceed $8, so reduce the count by one. Another special case occurs when exactly one child does not receive $8 and the leftover money equals $3. That child would end up with $4 (1 + 3), which is forbidden. Reduce the count by one to avoid that configuration.
This greedy reasoning works because maximizing $8 allocations always consumes the remaining budget in chunks of $7 after the initial distribution. No additional data structures are required, making it a constant-time calculation driven purely by arithmetic constraints.
Recommended for interviews: The greedy solution is what interviewers expect. It shows you can convert problem constraints into arithmetic reasoning and handle tricky edge cases. Explaining the brute force approach first demonstrates understanding of the constraints, while the greedy optimization highlights problem-solving maturity.
Approach 1: Greedy Approach
The idea is to determine the maximum number of children that can receive exactly 8 dollars. Since each child must receive at least 1 dollar, we start with giving each child 1 dollar. This leaves us with money - children dollars. We need to maximize the number of children receiving exactly 8 dollars by ensuring that the remaining children post this allocation do not violate the rules.
In this solution, we first check if it's possible to distribute at least 1 dollar to each child. If not, return -1. Then, we calculate how much more money can be distributed after giving each child 1 dollar. By dividing this money by the additional needed to reach 8 dollars (7 more dollars needed), we determine the maximum children who can receive 8 dollars. We also ensure not exceeding the number of children to prevent robustness issues.
Complexity
The time complexity is O(1) since the operations are straightforward arithmetic calculations. The space complexity is also O(1) as no additional space is used beyond simple variables.
Approach 2: Brute Force Approach
This approach evaluates all possible distributions across children to determine one that maximizes the number getting exactly 8 dollars. While inefficient for large inputs, it ensures an understanding of distribution constraints by considering every possible partition systematically.
Using a nested loop pattern, this C solution tries every combination of children and checks if their total expenditure matches the initial amount provided without breaching any specified rules. Once determined, it uses resultant combinations to find the maximum possible eights allocation. However, this approach iteratively checks all potential distributions exhaustively.
Complexity
The time complexity is O(children) as the dominant operations loop across possible combinations of children receiving 8. Space complexity is O(1), as aside from loop management variables, no additional persistent data structures are maintained.
Approach 3: Case analysis
If money \lt children, then there must be a child who did not receive money, return -1.
If money \gt 8 times children, then there are children-1 children who received 8 dollars, and the remaining child received money - 8 times (children-1) dollars, return children-1.
If money = 8 times children - 4, then there are children-2 children who received 8 dollars, and the remaining two children shared the remaining 12 dollars (as long as it is not 4, 8 dollars is fine), return children-2.
If we assume that there are x children who received 8 dollars, then the remaining money is money- 8 times x, as long as it is greater than or equal to the number of remaining children children-x, it can meet the requirements. Therefore, we only need to find the maximum value of x, which is the answer.
Time complexity O(1), space complexity O(1).
Complexity Comparison
| Approach | Complexity |
|---|---|
| Greedy Approach | The time complexity is O(1) since the operations are straightforward arithmetic calculations. The space complexity is also O(1) as no additional space is used beyond simple variables. |
| Brute Force Approach | The time complexity is O(children) as the dominant operations loop across possible combinations of children receiving 8. Space complexity is O(1), as aside from loop management variables, no additional persistent data structures are maintained. |
| Case analysis | — |
Detailed Complexity Analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Simulation | O(children) | O(1) | Useful for reasoning about constraints or validating edge cases during initial problem solving |
| Greedy Distribution | O(1) | O(1) | Best approach for interviews and production due to constant-time math-based reasoning |
Video Solution
Distribute Money to Maximum Children || Greedy Algorithm || Leetcode-2591 • Aryan Mittal • 1,973 views views
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