To maximize children receiving exactly $8 while ensuring all money is distributed, every child gets at least $1, and no child receives $4, the algorithm first gives each child $1, then uses integer division of remaining money by 7 to determine how many children can receive $8, with edge cases requiring adjustment when remaining money is $3 or when excess money prevents any child from having exactly $8.
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LeetCode#2591 Distribute Money to Maximum Children - PythonAdded:
Distribute money to maximum children.
All the money must be distributed.
Everyone must receive at least $1 and nobody [music] receives $4. Return the maximum number of children who may receive exactly $8. And so we can give [music] $8 to the first child. This leaves us with $12 left. We can give $8 to the second child. [music] This leaves us with $4. And so we would have to give all $4 to the last child. [music] And so in this case, notice the maximum number of children with $8 is [music] two. But notice this is an incorrect distribution because one child has $4 and it violates the rule. And so we would have to back up and redistribute [music] differently.
We could give the second child $9 and the third child $3. [music] In this distribution, all the rules are satisfied and it can be proved given these exact [music] parameters, we can only form one child with a maximum of $8. We will solve this in constant time and space. The idea is to first give each child exactly $1. This is to ensure that one rule is satisfied. [music] And so the remaining amount of money we have is $17. And so with the remaining amount of money, we want to determine how much children we could distribute seven more dollars to to get a total of eight. And to figure that out, you can take your money and [music] integer division it by seven. And so in this case, we can give seven more dollars to two children. [music] We can give seven more to child one and seven more to child two.
>> [music] >> And so now we have a total of three more dollars. And notice two of the children [music] have exactly $8. So we only need to worry about one more child. Now if we ever get into this scenario where the money is three and the children is one, then we know that one child must redistribute its money so we don't end up with a child of $4. So, in this scenario, we return the result of the integer division subtracted by one. And so, in this [music] case, we will have only one child with $8. Now, the only other edge case we need to consider is when all the children have eight, but we have excess money. [music] But after we gave both the children exactly eight candies, we had zero left. But notice, we have excess money that we need [music] to distribute. So, it is impossible to have any of the children have exactly [music] eight. So, in this case, one of the child will have to take the excess dollars. And so, notice in this scenario, we do the same procedure of subtracting one, and this will give us the number of children with exactly eight candies. And so, these were the two edge cases we [music] need to consider. In all the other scenarios, we would just return the result of money [music] integer division seven. And here's the code to do so.
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