Maximum Number Of Balls In A Box - Leetcode Solution

Problem Statement:

You are given an infinite number of boxes. Each box has a maximum capacity of numBalls. You have numBalls balls and you want to distribute the balls among the boxes such that the maximum number of balls in any box is minimum.

Example 1:

Input: numBalls = 7, numBoxes = 3 Output: 2 Explanation: We can distribute the 7 balls as follows:

• Box 1 contains 2 balls.
• Box 2 contains 2 balls.
• Box 3 contains 3 balls. The maximum number of balls in any box is 3.

Example 2:

Input: numBalls = 10, numBoxes = 3 Output: 4 Explanation: We can distribute the 10 balls as follows:

• Box 1 contains 4 balls.
• Box 2 contains 3 balls.
• Box 3 contains 3 balls. The maximum number of balls in any box is 4.

Approach:

The approach to solve this problem is to use binary search. We can apply binary search on the maximum number of balls that can be distributed in a box. As the number of balls that can be distributed in a box increases, the maximum number of balls in any box decreases. We can set the lower limit of the binary search to 1, as each box should contain at least 1 ball. We can set the upper limit of the binary search to numBalls, as all the balls can be put in a single box. For each mid value, we can check if it is possible to distribute the balls in the boxes such that no box contains more than mid balls.

To distribute the balls, we can start with the first box and keep adding balls until it reaches the mid value. We can then move to the next box and continue adding balls until it reaches the mid value. If we have more balls to distribute after all the boxes have been filled, we can start again with the first box. If we can distribute all the balls without any box having more than mid balls, we can try to increase the value of mid and continue the binary search. Otherwise, we have to decrease the value of mid and continue the binary search.

Solution:

Here is the Python solution for Maximum Number Of Balls In A Box problem on Leetcode:

class Solution: def getNumBoxes(self, numBalls: int, boxCapacity: int, maxBallsPerBox: int) -> int: low = 1 high = numBalls while low < high: mid = (low + high) // 2 if self.canDistribute(numBalls, boxCapacity, mid, maxBallsPerBox): high = mid else: low = mid + 1 return low

def canDistribute(self, numBalls: int, boxCapacity: int, mid: int, maxBallsPerBox: int) -> bool:
    numBoxes = (numBalls + mid - 1) // mid
    if numBoxes > boxCapacity:
        return False
    ballsPerBox = [mid] * numBoxes
    totalBalls = numBoxes * mid
    i = 0
    while totalBalls < numBalls:
        ballsToAdd = min(maxBallsPerBox - mid, numBalls - totalBalls)
        ballsPerBox[i] += ballsToAdd
        totalBalls += ballsToAdd
        i = (i + 1) % numBoxes
    for balls in ballsPerBox:
        if balls > maxBallsPerBox:
            return False
    return True

Explanation:

The solution is implemented in the class Solution. The getNumBoxes method takes numBalls (total number of balls), boxCapacity (maximum capacity of each box), and maxBallsPerBox (maximum number of balls that can be put into each box) as input and returns the maximum number of balls in any box.

Inside the getNumBoxes method, we set the lower limit of the binary search to 1 and the upper limit to numBalls. We then run a while loop until the lower limit is less than the upper limit. Inside the while loop, we find the mid value by taking the average of the lower and upper limits. We then check if we can distribute the balls such that no box contains more than mid balls using the canDistribute method. If we can distribute the balls, we try to decrease the value of mid by setting high to mid. Otherwise, we try to increase the value of mid by setting low to mid+1.

The canDistribute method takes numBalls, boxCapacity, mid, and maxBallsPerBox as input and returns True if we can distribute the balls such that no box contains more than mid balls. Inside the method, we calculate the number of boxes needed to distribute the balls using the formula (numBalls + mid - 1) // mid. We then check if the number of boxes is less than or equal to boxCapacity. If not, we return False.

We initialize ballsPerBox list to contain mid for each box. We then add balls to the boxes starting with the first box until we run out of balls or each box has reached the maximum number of balls specified by maxBallsPerBox. If we still have balls to distribute, we start again with the first box. We then check if any box has more than maxBallsPerBox balls. If yes, we return False. Otherwise, we return True.

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