#2591

Easy

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Distribute Money to Maximum Children

Determine the maximum number of children who can each receive exactly 8 dollars using a greedy approach with validation checks.

math greedy

Problem Statement

You have a total amount of money represented by an integer and a specific number of children to distribute it to. Each child must receive at least 1 dollar, and the total money must be exactly allocated among all children. Your goal is to maximize how many children receive exactly 8 dollars.

Return the largest number of children who can receive 8 dollars under these rules. If it is impossible to distribute the money according to these constraints, return -1.

Examples

Example 1

Input: money = 20, children = 3

Output: 1

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

Each child can be given 8 dollars.

Constraints

1 <= money <= 200
2 <= children <= 30

Solution Approach

Greedy assignment of 8 dollars

Start by giving 8 dollars to as many children as possible without exceeding the total money. This ensures the maximum number of children receive 8 dollars, aligning with the primary greedy pattern.

Check remaining money against constraints

After assigning 8 dollars, verify if the leftover money is enough to give at least 1 dollar to each remaining child. This invariant validation prevents illegal distributions and guarantees correctness.

Adjust for impossible distributions

If the remaining money cannot satisfy all children with at least 1 dollar, reduce the number of children assigned 8 dollars by one and recheck. Repeat until a valid distribution is found or return -1 if none exists.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(1) because the maximum number of children is limited to 30, making simple arithmetic sufficient. Space complexity is O(1) as no extra structures are required beyond variable storage.

What Interviewers Usually Probe

You may be asked to explain why a simple greedy approach works for maximizing 8-dollar allocations.
Expect clarifying questions on validating the remaining money after initial assignments.
Interviewers might test edge cases like distributing all money evenly or having a remainder less than the number of remaining children.

Common Pitfalls or Variants

Common pitfalls

Assuming you can always assign 8 dollars to floor(money/8) children without checking leftover constraints.
Ignoring the rule that each child must get at least 1 dollar in validation steps.
Failing to decrement the 8-dollar assignments when leftover money is insufficient, leading to incorrect -1 results.

Follow-up variants

Maximize children receiving a fixed amount other than 8 dollars, requiring similar greedy validation.
Distribute money with additional constraints such as maximum cap per child, testing more complex invariants.
Calculate minimum money required to guarantee at least k children receive 8 dollars.

FAQ

What is the main strategy for Distribute Money to Maximum Children?

Use a greedy approach by giving 8 dollars to as many children as possible, then validate the remaining money for feasibility.

Can this problem be solved without iterative checking?

Yes, due to small constraints, arithmetic checks with conditional adjustment suffice, avoiding full iteration.

Why must we validate leftover money?

Validation ensures every child receives at least 1 dollar and the total distribution matches the available money.

What happens if leftover money is too small for remaining children?

Reduce the number of 8-dollar allocations until leftover money can satisfy all remaining children or return -1.

Does this problem always require exactly 8 dollars per greedy choice?

Yes, the 8-dollar target is the key to the greedy pattern, and adjusting allocations revolves around it.

terminal

Solution

Solution 1: Case analysis

If $money \lt children$, then there must be a child who did not receive money, return $-1$.

1

2

3

4

5

6

7

8

9

10

class Solution:
    def distMoney(self, money: int, children: int) -> int:
        if money < children:
            return -1
        if money > 8 * children:
            return children - 1
        if money == 8 * children - 4:
            return children - 2
        # money-8x >= children-x, x <= (money-children)/7
        return (money - children) // 7

Continue Practicing

#2598 smallest missing non negative integer after operations #2600 k items with the maximum sum #2601 prime subtraction operation