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Leetcode 2591. Distribute Money to Maximum Children

Problem Statement

You are given an integer money representing the total amount of money you have, and an integer children representing the number of children to distribute the money to. You need to distribute all the money according to the following rules:

  1. Each child must receive at least 1 dollar.
  2. Each child can receive at most 8 dollars.

Your goal is to maximize the number of children who receive exactly 8 dollars.

Return the maximum number of children who can receive exactly 8 dollars.

Clarifying Questions

Before we proceed, let’s clarify a few points to ensure the problem is well-understood:

Strategy

  1. Check if the money is less than children. If yes, return 0.
  2. Calculate the maximum number of children that can be given exactly 8 dollars by dividing money by 8 and then taking the minimum with children (min(money / 8, children)).
  3. Distribute money giving the maximum number of children 8 dollars each, then distribute the remaining money to ensure that every child gets at least 1 dollar.
  4. Adjust the counts if any child receiving 8 dollars needs to be lowered to accommodate the remaining money and conditions.

Code

Here is the C++ implementation of the solution:

#include <algorithm>
using namespace std;

class Solution {
public:
    int distMoney(int money, int children) {
        // Check if not enough money to give each child 1 dollar
        if (money < children) return 0;

        // Calculate initial number of children we can give exactly 8 dollars
        int maxChildrenWithEightDollars = min(money / 8, children);
        
        // Calculate remaining money after giving 8 dollars to maxChildrenWithEightDollars children
        int remainingMoney = money - (8 * maxChildrenWithEightDollars);
        int remainingChildren = children - maxChildrenWithEightDollars;

        // Adjust the counts if we have more children than remaining money
        while (remainingMoney < remainingChildren && maxChildrenWithEightDollars > 0) {
            // We need to adjust by reducing the number of children getting exactly 8 dollars
            maxChildrenWithEightDollars--;
            remainingMoney += 8;
            remainingChildren++;
        }

        // If there is still not enough remaining money for the remaining children, return 0
        if (remainingMoney < remainingChildren) return 0;

        return maxChildrenWithEightDollars;
    }
};

Time Complexity

The time complexity of this solution is (O(1)). The operations performed are simple arithmetic operations and conditional checks, which run in constant time.

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