Welcome to Subscribe On Youtube
Formatted question description: https://leetcode.ca/all/2591.html
2591. Distribute Money to Maximum Children
Description
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
Solutions
Solution 1: 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 $children1$ children who received $8$ dollars, and the remaining child received $money  8 \times (children1)$ dollars, return $children1$.
If $money = 8 \times children  4$, then there are $children2$ 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 $children2$.
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 $childrenx$, 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 $O(1)$.

class Solution { public int distMoney(int money, int children) { if (money < children) { return 1; } if (money > 8 * children) { return children  1; } if (money == 8 * children  4) { return children  2; } // money8x >= childrenx, x <= (moneychildren)/7 return (money  children) / 7; } }

class Solution { public: int distMoney(int money, int children) { if (money < children) { return 1; } if (money > 8 * children) { return children  1; } if (money == 8 * children  4) { return children  2; } // money8x >= childrenx, x <= (moneychildren)/7 return (money  children) / 7; } };

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 # money8x >= childrenx, x <= (moneychildren)/7 return (money  children) // 7

func distMoney(money int, children int) int { if money < children { return 1 } if money > 8*children { return children  1 } if money == 8*children4 { return children  2 } // money8x >= childrenx, x <= (moneychildren)/7 return (money  children) / 7 }

function distMoney(money: number, children: number): number { if (money < children) { return 1; } if (money > 8 * children) { return children  1; } if (money === 8 * children  4) { return children  2; } return Math.floor((money  children) / 7); }

impl Solution { pub fn dist_money(money: i32, children: i32) > i32 { if money < children { return 1 } if money > children * 8 { return children  1 } if money == children * 8  4 { return children  2 } (money  children) / 7 } }