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Distribute Money to Maximum Children

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Greedy Algorithms

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 1 dollar.
Nobody receives 4 dollars.

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 <= 200
2 <= children <= 30

Try coding it in LeetCode

Distribute Money to Maximum Children

Easy

Asked by:

73 views

Topics:

Greedy Algorithms

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 1 dollar.
Nobody receives 4 dollars.

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 <= 200
2 <= children <= 30

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the values of 'money' and the length of the 'children' array? Can they be zero, negative, or very large?
Is it possible for 'money' to be less than the number of children? If so, what should I return?
If there's leftover money after giving $1 to all children except one, is that considered a valid distribution for that child to get the remaining money that isn't $4?
Should I prioritize distributing to as many children as possible, even if some children get less than $1?
If there are multiple ways to distribute the money to the maximum number of children, is any valid distribution acceptable, or is there a specific distribution I should aim for?

Brute Force Solution

Approach

The brute force method involves checking every single way you could possibly distribute the money among the children. We aim to find the maximum number of children that can receive money while adhering to the specific rules.

Here's how the algorithm would work step-by-step:

Start by giving money to only one child and see if it's possible while following the rules (each child gets exactly 8 unless we don't have enough money left for that).
Then, consider giving money to two children and check every possible amount given to each child.
Continue by checking the possibility of giving money to three children, then four, and so on, always verifying if the rules are met for each combination.
For each possible scenario, make sure the money given out doesn't exceed the total amount of money available.
Remember the rule that each child, except potentially the last one, has to get exactly 8 dollars. The last child could get less than 8 dollars.
Count how many children receive money in each of the scenarios that satisfy all the rules.
Finally, find the scenario where the most children receive money and return that number.

Code Implementation

def distribute_money_to_maximum_children_brute_force(money, children):
    max_children_receiving_money = -1

    # Iterate through all possible numbers of children receiving money
    for num_children_to_give_money_to in range(1, children + 1):
        
        if num_children_to_give_money_to * 1 > money:
            continue

        # Check all combinations of money distribution
        if num_children_to_give_money_to == 1:
            if money >= 1:
                max_children_receiving_money = max(max_children_receiving_money, 1)

        else:
            # Trying to give exactly 8 to each child but the last one
            money_needed_for_first_children = 8 * (num_children_to_give_money_to - 1)

            if money_needed_for_first_children <= money:
                money_left_for_last_child = money - money_needed_for_first_children

                # Ensure the last child receives at least 1 dollar.
                if money_left_for_last_child >= 1:
                    max_children_receiving_money = max(max_children_receiving_money, num_children_to_give_money_to)

    # Check if every child receives exactly 8, and the last child does not receive 4
    if children > 0 and money >= 8 * children:
        if money - 8 * children != 4:
            max_children_receiving_money = children

    return max_children_receiving_money

Big(O) Analysis

Time Complexity

O(8^n) – The brute force method iterates through each possible number of children from 1 to n, where n is the number of children available. For each number of children k, it explores all possible distributions of money. Since each child (except potentially the last) receives 8 dollars, the possible distributions grow exponentially with the number of children. The algorithm is essentially trying all combinations and the number of combinations scales with 8^n. Thus the time complexity is O(8^n).

Space Complexity

O(1) – The described brute force approach checks different combinations of money distribution among children. This involves a few variables to keep track of the number of children considered and the money distributed in each scenario. The algorithm does not create any auxiliary data structures like lists or hash maps whose size depends on the number of children (N, which is the number of children given in the input) or the amount of money. Therefore, the auxiliary space remains constant, independent of the input size, resulting in O(1) space complexity.

Optimal Solution

Approach

The goal is to give money to as many children as possible following specific rules. We want to prioritize giving the ideal amount to as many kids as we can, and then handle any leftover money carefully to avoid breaking a rule.

Here's how the algorithm would work step-by-step:

First, see if you have enough money to give every child the ideal amount, which is a set amount of money per child. If so, give the ideal amount to each child and you are done.
If you don't have enough money for everyone to get the ideal amount, try to give as many kids as possible the ideal amount without going over the total amount of money. Keep track of how many kids got the ideal amount and how much money is left over.
Now, with the leftover money and the remaining children, try to give the remaining children any amount of money less than the ideal amount. The goal is to give at least some money to each child possible.
There's a rule that one specific amount of money cannot be given to a child. If you accidentally gave a child that amount, or you are about to give a child that amount, then take back some money from a child who got the ideal amount earlier and give it to that problem child, but be sure they do not end up with the specific disallowed amount.
After considering all of these constraints, determine the maximum number of children that could be given some amount of money.

Code Implementation

def distribute_money_to_maximum_children(money, number_of_children):
    ideal_amount = 8

    if money < number_of_children:
        return 0

    children_getting_ideal_amount = min(money // ideal_amount, number_of_children)
    money_remaining = money - children_getting_ideal_amount * ideal_amount

    # If everyone gets ideal, and there's remainder, one less gets ideal.
    if children_getting_ideal_amount == number_of_children and money_remaining > 0:
        children_getting_ideal_amount -= 1
        
    remaining_children = number_of_children - children_getting_ideal_amount

    # Need to redistribute to avoid giving exactly 4.
    if remaining_children == 1 and money_remaining == 4:
        if children_getting_ideal_amount > 0:
            children_getting_ideal_amount -= 1
        else:
            return number_of_children - 1
            
        #Adjust counts accordingly after redistribution
        remaining_children = number_of_children - children_getting_ideal_amount

    # If there's still money left, and kids to give to, increment.
    if money_remaining > 0 and remaining_children > 0:
        children_getting_some_money = min(remaining_children, money_remaining) 
        return children_getting_ideal_amount + remaining_children

    #Return when all kids have ideal amount
    return children_getting_ideal_amount

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the list of children once to distribute the ideal amount of money and calculate the remainder. It then might iterate through the remaining children and previously funded children a constant number of times to adjust the amounts based on the disallowed amount. Since the number of iterations is directly proportional to the number of children (n) and there are no nested loops dependent on n, the time complexity is O(n).

Space Complexity

O(1) – The algorithm primarily uses a few integer variables to track the number of children, the amount of money distributed, and potentially some counters. No auxiliary data structures like arrays or hash maps are used that would scale with the input (number of children or total money). Therefore, the space required remains constant irrespective of the input size N (number of children or amount of money), resulting in O(1) space complexity.

Edge Cases

Case	How to Handle
money is negative	Return 0 because it's impossible to distribute negative money.
children is negative or zero	Return 0, as you can't distribute money to a non-positive number of children.
money is 0 and children is positive	Return 0, since no child can receive $1 in this scenario.
money is less than children	Return 0, as each child needs at least $1.
money is exactly children	Return the number of children as each gets exactly $1.
money is just enough to give each child $1 and no more	The standard distribution algorithm works correctly in this case; ensure it doesn't over-allocate.
money is much larger than the number of children	The algorithm should still execute efficiently, distributing to as many children as possible then handling the remainder correctly.
Only one child and sufficient money for them to get $4 or more, after attempting to give $1 to each child initially	The condition where a child gets exactly 4 should be handled correctly to prevent over-allocation, especially when there is only one child.

money is negative

How to Handle:

Return 0 because it's impossible to distribute negative money.

children is negative or zero

How to Handle:

Return 0, as you can't distribute money to a non-positive number of children.

money is 0 and children is positive

How to Handle:

Return 0, since no child can receive $1 in this scenario.

money is less than children

How to Handle:

Return 0, as each child needs at least $1.

money is exactly children

How to Handle:

Return the number of children as each gets exactly $1.

money is just enough to give each child $1 and no more

How to Handle:

The standard distribution algorithm works correctly in this case; ensure it doesn't over-allocate.

money is much larger than the number of children

How to Handle:

The algorithm should still execute efficiently, distributing to as many children as possible then handling the remainder correctly.

Only one child and sufficient money for them to get $4 or more, after attempting to give $1 to each child initially

How to Handle:

The condition where a child gets exactly 4 should be handled correctly to prevent over-allocation, especially when there is only one child.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the values of 'money' and the length of the 'children' array? Can they be zero, negative, or very large?
Is it possible for 'money' to be less than the number of children? If so, what should I return?
If there's leftover money after giving $1 to all children except one, is that considered a valid distribution for that child to get the remaining money that isn't $4?
Should I prioritize distributing to as many children as possible, even if some children get less than $1?
If there are multiple ways to distribute the money to the maximum number of children, is any valid distribution acceptable, or is there a specific distribution I should aim for?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start by giving money to only one child and see if it's possible while following the rules (each child gets exactly 8 unless we don't have enough money left for that).
Then, consider giving money to two children and check every possible amount given to each child.
Continue by checking the possibility of giving money to three children, then four, and so on, always verifying if the rules are met for each combination.
For each possible scenario, make sure the money given out doesn't exceed the total amount of money available.
Remember the rule that each child, except potentially the last one, has to get exactly 8 dollars. The last child could get less than 8 dollars.
Count how many children receive money in each of the scenarios that satisfy all the rules.
Finally, find the scenario where the most children receive money and return that number.

Code Implementation

def distribute_money_to_maximum_children_brute_force(money, children):
    max_children_receiving_money = -1

    # Iterate through all possible numbers of children receiving money
    for num_children_to_give_money_to in range(1, children + 1):
        
        if num_children_to_give_money_to * 1 > money:
            continue

        # Check all combinations of money distribution
        if num_children_to_give_money_to == 1:
            if money >= 1:
                max_children_receiving_money = max(max_children_receiving_money, 1)

        else:
            # Trying to give exactly 8 to each child but the last one
            money_needed_for_first_children = 8 * (num_children_to_give_money_to - 1)

            if money_needed_for_first_children <= money:
                money_left_for_last_child = money - money_needed_for_first_children

                # Ensure the last child receives at least 1 dollar.
                if money_left_for_last_child >= 1:
                    max_children_receiving_money = max(max_children_receiving_money, num_children_to_give_money_to)

    # Check if every child receives exactly 8, and the last child does not receive 4
    if children > 0 and money >= 8 * children:
        if money - 8 * children != 4:
            max_children_receiving_money = children

    return max_children_receiving_money

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

First, see if you have enough money to give every child the ideal amount, which is a set amount of money per child. If so, give the ideal amount to each child and you are done.
If you don't have enough money for everyone to get the ideal amount, try to give as many kids as possible the ideal amount without going over the total amount of money. Keep track of how many kids got the ideal amount and how much money is left over.
Now, with the leftover money and the remaining children, try to give the remaining children any amount of money less than the ideal amount. The goal is to give at least some money to each child possible.
There's a rule that one specific amount of money cannot be given to a child. If you accidentally gave a child that amount, or you are about to give a child that amount, then take back some money from a child who got the ideal amount earlier and give it to that problem child, but be sure they do not end up with the specific disallowed amount.
After considering all of these constraints, determine the maximum number of children that could be given some amount of money.

Code Implementation

def distribute_money_to_maximum_children(money, number_of_children):
    ideal_amount = 8

    if money < number_of_children:
        return 0

    children_getting_ideal_amount = min(money // ideal_amount, number_of_children)
    money_remaining = money - children_getting_ideal_amount * ideal_amount

    # If everyone gets ideal, and there's remainder, one less gets ideal.
    if children_getting_ideal_amount == number_of_children and money_remaining > 0:
        children_getting_ideal_amount -= 1
        
    remaining_children = number_of_children - children_getting_ideal_amount

    # Need to redistribute to avoid giving exactly 4.
    if remaining_children == 1 and money_remaining == 4:
        if children_getting_ideal_amount > 0:
            children_getting_ideal_amount -= 1
        else:
            return number_of_children - 1
            
        #Adjust counts accordingly after redistribution
        remaining_children = number_of_children - children_getting_ideal_amount

    # If there's still money left, and kids to give to, increment.
    if money_remaining > 0 and remaining_children > 0:
        children_getting_some_money = min(remaining_children, money_remaining) 
        return children_getting_ideal_amount + remaining_children

    #Return when all kids have ideal amount
    return children_getting_ideal_amount

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
money is negative	Return 0 because it's impossible to distribute negative money.
children is negative or zero	Return 0, as you can't distribute money to a non-positive number of children.
money is 0 and children is positive	Return 0, since no child can receive $1 in this scenario.
money is less than children	Return 0, as each child needs at least $1.
money is exactly children	Return the number of children as each gets exactly $1.
money is just enough to give each child $1 and no more	The standard distribution algorithm works correctly in this case; ensure it doesn't over-allocate.
money is much larger than the number of children	The algorithm should still execute efficiently, distributing to as many children as possible then handling the remainder correctly.
Only one child and sufficient money for them to get $4 or more, after attempting to give $1 to each child initially	The condition where a child gets exactly 4 should be handled correctly to prevent over-allocation, especially when there is only one child.

money is negative

How to Handle:

Return 0 because it's impossible to distribute negative money.

children is negative or zero

How to Handle:

Return 0, as you can't distribute money to a non-positive number of children.

money is 0 and children is positive

How to Handle:

Return 0, since no child can receive $1 in this scenario.

money is less than children

How to Handle:

Return 0, as each child needs at least $1.

money is exactly children

How to Handle:

Return the number of children as each gets exactly $1.

money is just enough to give each child $1 and no more

How to Handle:

The standard distribution algorithm works correctly in this case; ensure it doesn't over-allocate.

money is much larger than the number of children

How to Handle:

The algorithm should still execute efficiently, distributing to as many children as possible then handling the remainder correctly.

Only one child and sufficient money for them to get $4 or more, after attempting to give $1 to each child initially

How to Handle:

The condition where a child gets exactly 4 should be handled correctly to prevent over-allocation, especially when there is only one child.