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You are given an integer money and another integer children. Distribute the given amount of money to the maximum number of children such that:

  1. Each child gets at least 1 unit of money.
  2. No child gets exactly 4 units of money.

Return the maximum number of children who can receive money under the specified constraints. If it is not possible to satisfy the constraints, return -1.

Clarifying Questions

  1. Can money and children be negative?
    • No, both money and children are non-negative integers.
  2. Is it possible for the number of children to be zero while the money is non-zero?
    • Yes, and in that case, the function should return 0, since there are no children to distribute to.
  3. What if the money is zero, but the number of children is non-zero?
    • In this case, it’s impossible to distribute money while fulfilling the constraints, so the function should return -1.

Strategy

  1. Initial Allocation:
    • Each child should receive at least 1 unit of money. Therefore, check if the total children can get at least 1 unit each without exceeding the money. If not, return -1.
  2. Avoid Exact 4 Units:
    • Next, distribute the remaining money keeping in mind that no child should end up with exactly 4 units.
    • If the remaining money after the initial allocation is distributed in a way that forms exactly 4 units for any child, adjust the distribution to avoid it.
  3. Return Max Children:
    • If there are methods to distribute money within the constraints and cover all children, then return the number of children. If not, return as per constraints violation.

Code

def max_children_with_money(money, children):
    # Step 1: Ensure minimum money per child is 1 unit
    if money < children:
        return -1
    
    # Step 2: We initially assume each child gets 1 unit of money
    initial_allocation = children
    remaining_money = money - children
    
    # Step 3: Distribute the remaining money
    can_satisfy = (remaining_money // 3 + initial_allocation <= children) and \
                  (remaining_money % 3 != 1 or remaining_money > 3)
    
    if not can_satisfy:
        return -1
    
    # Step 4: Calculate maximum possible children
    return children

# Example Usage
print(max_children_with_money(20, 4))  # Expected output: 4
print(max_children_with_money(7, 2))  # Expected output: 2
print(max_children_with_money(10, 3))  # Expected output: -1

Time Complexity

The time complexity of this solution is O(1) since we are using basic arithmetic operations that are essentially constant time checks irrespective of the size of money and children. There are no loops or recursive calls involved in the solution.

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