Distribute Money to Maximum Children

1. If money < children, it's impossible to give each child at least $1. Return -1.
2. Iterate through the number of children who could receive $8, let's call it k, from the maximum possible (children) down to 0.
3. For each k, calculate the money that would be spent (k * 8) and the remaining money (rem_money) and remaining children (rem_children).
4. If the money spent exceeds the total money, this k is not feasible, so we continue to k-1.
5. Check if the rem_money can be validly distributed among rem_children:
   • If rem_children is 0, rem_money must also be 0. If so, we found a valid distribution for k children, and since we iterate downwards, this is the maximum. Return k.
   • If rem_children is greater than 0, we must ensure two conditions: (a) each remaining child can get at least $1 (rem_money >= rem_children), and (b) we don't run into the special case where we are forced to give the last child $4 (this happens only when rem_children == 1 and rem_money == 4).
6. If both conditions in step 5 are met, k is a valid answer. Return k.
7. If the loop completes, it implies no solution was found, which should only happen if the initial check money < children was true. A fallback return of -1 handles this.

The core idea is to iterate and check. We want to maximize the number of children, k, who get $8. We can simply try all possible values for k and see which ones are valid.

To be efficient, we should start our search from the highest possible value of k, which is children, and go down to 0. The first value of k that allows a valid distribution of the entire money will be our answer.

For a given k, we check its validity as follows:

1. Calculate the money given to these k children: spent = k * 8. If spent > money, this k is impossible.
2. Calculate the remaining money rem_money = money - spent and remaining children rem_children = children - k.
3. Now, we must distribute rem_money among rem_children.
   • If rem_children == 0, all children were given $8. This is only valid if rem_money == 0.
   • If rem_children > 0, every one of them must get at least $1, so rem_money >= rem_children. Additionally, we must not be forced to give a child exactly $4. This problematic scenario only occurs when there is exactly one child left (rem_children == 1) and the exact amount of money left is $4 (rem_money == 4).

If a value of k passes these checks, it's a valid solution. Since we are iterating downwards, the first one we find is the maximum.

1class Solution {2    public int distMoney(int money, int children) {3        if (money < children) {4            return -1;5        }6        // Iterate from the maximum possible children getting $8 downwards.7        for (int k = children; k >= 0; k--) {8            int spent_money = k * 8;9            if (spent_money > money) {10                continue;11            }12            int rem_money = money - spent_money;13            int rem_children = children - k;14 15            if (rem_children == 0) {16                if (rem_money == 0) {17                    return k; // Perfect distribution18                }19            } else {20                // Check if remaining money can be distributed21                if (rem_money >= rem_children) {22                    // Check for the forbidden $4 case23                    if (rem_children == 1 && rem_money == 4) {24                        continue; // This k is invalid25                    }26                    return k; // Found a valid distribution27                }28            }29        }30        return -1; // Should not be reached if money >= children31    }32}

1. Handle the base case: if money < children, return -1.
2. Conceptually, give every child $1 first to satisfy a core rule. The remaining money is money - children.
3. The problem now is to distribute this remaining amount to the children, where each needs an additional $7 to reach a total of $8.
4. Calculate the maximum number of children who can receive an additional $7: count = (money - children) / 7.
5. Calculate the leftover money after this distribution: remainder = (money - children) % 7.
6. Analyze the results based on a few distinct cases:
   • If count is greater than children, it means we have more than enough money to give everyone $8. We can't give everyone $8 because money would be left over. The answer is children - 1.
   • If count equals children, we can give everyone $8 if and only if the remainder is 0. If there is a remainder, one child must take it, so only children - 1 children get exactly $8.
   • If count is less than children, count is our candidate answer. However, we must check the $4 rule. This rule is violated if we are left with one child (children - count == 1) who has $1, and the remainder to be given is $3 (totaling $4). In this specific scenario, the answer is count - 1. Otherwise, the answer is count.

This approach solves the problem without any loops by performing a case analysis.

First, we handle the trivial invalid case: if money < children, it's impossible, so we return -1.

To simplify the problem, we can first satisfy the rule that everyone gets at least $1. We distribute $1 to each of the children, which uses up children dollars. The remaining money is money - children. Now, our goal is to distribute this remaining money to give as many children as possible an additional $7 to make their total $8.

The number of children we can give an additional $7 to is count = (money - children) / 7. The money left after this is remainder = (money - children) % 7.

We can now determine the final answer with a few checks:

1. Too much money (count > children): This happens if money was initially much larger than 8 * children. We can't give everyone $8 because money would be left over. The best we can do is give children - 1 kids $8 and the last kid all the rest. The answer is children - 1.

2. Just enough money (count == children): We have enough to give everyone an additional $7. If remainder == 0, everyone gets exactly $8. The answer is children. If remainder > 0, the extra money must go to one child, meaning only children - 1 children end up with exactly $8.

3. Not enough money (count < children): We can give count children $8. The remainder must be distributed among the children - count kids who still have only $1. The only problem arises if there's exactly one child left (children - count == 1) and the remainder is exactly 3, as this child would receive a total of $1 + $3 = $4. In this specific scenario, we must break up one of the $8 packages, reducing the number of children with $8 by one. So the answer is count - 1. In all other sub-cases, count is the correct answer.

1class Solution {2    public int distMoney(int money, int children) {3        // 1. Not enough money to give every child $1.4        if (money < children) {5            return -1;6        }7 8        // 2. Special case: too much money. We can't give everyone $8.9        // The max is children - 1. The last child gets the large remainder.10        if (money > 8 * children) {11            return children - 1;12        }13 14        // 3. Special case: giving k = children - 1 leaves the last child with $4.15        // This happens when money = 8 * (children - 1) + 4, which is 8 * children - 4.16        // In this case, the answer must be k = children - 2.17        if (money == 8 * children - 4) {18            return children - 2;19        }20 21        // 4. General case: The number of children getting $8 is limited by the need22        // to give the rest at least $1. money - 8k >= children - k => k <= (money - children) / 7.23        return (money - children) / 7;24    }25}