Problem Description

You have a certain amount of money (in dollars) and need to distribute it among a group of children following specific rules:

1. All money must be distributed - You cannot keep any money; every dollar must be given out
2. Everyone gets at least $1 - Each child must receive a minimum of one dollar
3. Nobody can receive exactly $4 - This specific amount is forbidden for any child

Given money (total dollars available) and children (number of children), your goal is to maximize the number of children who receive exactly $8.

The problem asks you to find the maximum number of children who can get exactly $8 while satisfying all the rules above. If it's impossible to distribute the money according to these rules, return -1.

For example:

• If you have less money than children, it's impossible to give everyone at least $1, so return -1
• If you have exactly enough to give everyone $8, you need to be careful about edge cases (like when someone would end up with$4)
• The key is to maximize those getting $8 while ensuring the remaining money can be distributed legally among the rest

The solution involves analyzing different cases based on the relationship between money and children, considering scenarios where:

• Not enough money exists to give everyone at least $1
• Too much money exists (more than giving everyone $8)
• Special cases where the remaining money would force someone to receive exactly $4
• The general case where we calculate the maximum number of $8 distributions possible

Intuition

The key insight is to maximize the number of children receiving exactly $8 while ensuring the remaining money can be legally distributed among the rest.

Let's think about this step by step. If x children receive $8 each, then:

• Money used: 8 * x dollars
• Remaining money: money - 8 * x dollars
• Remaining children: children - x

For a valid distribution, the remaining money must be enough to give at least $1 to each remaining child, so we need: money - 8 * x >= children - x

Rearranging this inequality: money - 8 * x >= children - x money - children >= 8 * x - x money - children >= 7 * x x <= (money - children) / 7

This tells us the maximum possible value of x is (money - children) // 7.

However, we need to handle special edge cases:

1. Not enough money: If money < children, we can't give everyone at least $1, so return -1.

2. Too much money: If money > 8 * children, even after giving everyone $8, we'd have leftover money. Since all money must be distributed, we give children - 1 people exactly $8 and dump all the extra money on the last person.

3. **The forbidden $4 case**: If `money = 8 * children - 4`, following our formula would give us `children - 1` people with `$8and one person with$4` (which is forbidden). To avoid this, we can only give `children - 2` people exactly `$8, leaving $12` to distribute between the last two people (they could get `$5and$7`, or `$6and$6`, etc.).

After handling these edge cases, the general formula (money - children) // 7 gives us the maximum number of children who can receive exactly $8.

Learn more about Greedy and Math patterns.

Solution Approach

The solution uses case analysis to handle different scenarios systematically. Let's walk through each case in the implementation:

Case 1: Insufficient money

if money < children:
    return -1

If we have less money than the number of children, it's impossible to give everyone at least $1. Return -1 immediately.

Case 2: Excess money

if money > 8 * children:
    return children - 1

When we have more than 8 * children dollars, we can't give everyone exactly $8 because there would be leftover money. The optimal strategy is to:

• Give children - 1 people exactly $8
• Give the remaining person all the leftover money: money - 8 * (children - 1) dollars

This maximizes the count at children - 1.

Case 3: The forbidden $4 edge case

if money == 8 * children - 4:
    return children - 2

This special case occurs when following the general formula would leave exactly one person with $4. Since $4 is forbidden:

• We can only give children - 2 people exactly $8
• This uses 8 * (children - 2) = 8 * children - 16 dollars
• Remaining money: (8 * children - 4) - (8 * children - 16) = 12 dollars
• These $12 are distributed between the last two people (e.g., $5 and $7, or $6 and $6)

Case 4: General case

return (money - children) // 7

For all other scenarios, we use the formula derived from our inequality analysis. If x children get $8:

• Constraint: money - 8x >= children - x (remaining money must cover remaining children)
• Rearranging: x <= (money - children) / 7
• Maximum x: (money - children) // 7 (using integer division)

This formula ensures:

• Each of the x children gets exactly $8
• The remaining children - x children share money - 8x dollars
• Everyone gets at least $1 and nobody gets exactly $4

The algorithm runs in O(1) time and space complexity since it only performs constant-time arithmetic operations and comparisons.

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Example: money = 20, children = 3

We want to maximize how many children get exactly $8, while ensuring everyone gets at least$1 and nobody gets exactly $4.

Step 1: Check which case applies

• Is money < children? No, 20 ≥ 3, so we have enough money.
• Is money > 8 * children? No, 20 ≤ 24, so we don't have excess money.
• Is money == 8 * children - 4? Let's check: 8 * 3 - 4 = 20. Yes! This is the special $4 case.

Step 2: Apply Case 3 (The forbidden $4 edge case)

If we naively tried to maximize, we might think:

• Give 2 children $8 each =$16
• Remaining: 1 child gets $20 -$16 = $4 ❌ (This is forbidden!)

Instead, we return children - 2 = 3 - 2 = 1.

This means only 1 child gets exactly $8:

• 1 child gets $8
• Remaining money: $20 -$8 = $12
• These $12 are split between the other 2 children (e.g.,$5 and $7, or$6 and $6)

Let's verify with another example: money = 16, children = 3

Step 1: Check which case applies

• Is money < children? No, 16 ≥ 3
• Is money > 8 * children? No, 16 < 24
• Is money == 8 * children - 4? No, 16 ≠ 20

Step 2: Apply Case 4 (General case)

Calculate: (money - children) // 7 = (16 - 3) // 7 = 13 // 7 = 1

So 1 child gets exactly $8:

• 1 child gets $8
• Remaining: 2 children share $16 -$8 = $8
• They could get $1 and$7, $2 and$6, $3 and$5, etc. (avoiding $4)

This distribution satisfies all constraints and maximizes the number of children receiving exactly $8.

Solution Implementation

Time and Space Complexity

The time complexity of this solution is O(1) because the function performs a fixed number of arithmetic operations and comparisons regardless of the input size. All operations - the conditional checks (if statements), arithmetic calculations (money - children, money // 7, 8 * children, etc.), and the return statements - execute in constant time.

The space complexity is O(1) as the function only uses a constant amount of extra space. No additional data structures are created, and the space usage doesn't depend on the input values money or children. The function only performs calculations using the input parameters and returns a single integer value.

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

Common Pitfalls

1. Overlooking the Special Case of $4

One of the most common mistakes is forgetting to handle the case where the remaining distribution would leave exactly one child with $4. Many solutions correctly handle the general formula but miss this edge case.

Incorrect approach:

def distMoney(self, money: int, children: int) -> int:
    if money < children:
        return -1
    if money > 8 * children:
        return children - 1
    # Missing the special case check!
    return (money - children) // 7

Why it fails: When money = 8 * children - 4, using the general formula gives us children - 1 as the answer. However, this would mean giving children - 1 people $8 each, leaving exactly$4 for the last person, which violates the rules.

Solution: Always check for the special case before applying the general formula:

if money == 8 * children - 4:
    return children - 2

2. Incorrect Handling of Excess Money

Another pitfall is misunderstanding what happens when there's more money than needed to give everyone $8. Some might think all children can still get exactly$8.

Incorrect approach:

if money >= 8 * children:
    return children  # Wrong! Someone must take the excess

Why it fails: If money > 8 * children, there's leftover money that must be distributed. Since all money must be given out and nobody can keep exactly $8 if they're receiving extra, the maximum is children - 1.

Solution: Recognize that excess money must go to someone, reducing the count by 1:

if money > 8 * children:
    return children - 1

3. Using the Wrong Formula for the General Case

Some solutions attempt to use simpler division without considering the constraint properly.

Incorrect approach:

return money // 8  # Simply dividing total money by 8

Why it fails: This doesn't account for the requirement that every child must receive at least $1. It might suggest more children can get$8 than is actually possible.

Solution: Use the correct formula derived from the constraint money - 8x >= children - x:

return (money - children) // 7

4. Not Validating the Remaining Distribution

Even with the correct formula, failing to verify that the remaining money can be distributed validly (avoiding $4) can lead to subtle bugs in edge cases.

Best Practice: When implementing, mentally verify each case:

• For x children getting $8: they use 8x dollars
• Remaining children: children - x
• Remaining money: money - 8x
• Each remaining child gets: (money - 8x) / (children - x) on average
• Ensure this doesn't force anyone to get exactly $4