Minimum Size of Bag

XXX. Minimum Size of Bag

Binary Search

Problem Statement:

Given an array nums representing the number of balls in each bag, and an integer maxOperations, determine the minimum possible size of the largest bag after performing at most maxOperations. An operation consists of splitting a bag into two bags.

Algorithm:

1. Use binary search to find the minimum possible size of the largest bag. The search range is between 1 and the maximum value in nums.
2. For each midpoint in the binary search, check if it is possible to achieve the maximum size mid using the helper function isPossible.
3. The isPossible function calculates the number of operations required to ensure no bag exceeds size limit. If the number of operations is less than or equal to maxOperations, update the answer and search for a smaller size.
4. If the current size is not feasible, adjust the search range to look for larger sizes.

Complexity:

Time Complexity: O(n log(max(nums))), where n is the length of nums and max(nums) is the maximum value in the array.
Space Complexity: O(1), as no additional data structures are used.

Java Implementation:

public int minimumSize(int[] nums, int maxOperations) {
    // Define the binary search range
    int left = 1, right = Arrays.stream(nums).max().getAsInt(), answer = right;

    // Binary search to find the minimum possible size
    while (left <= right) {
        int mid = left + (right - left) / 2; // Calculate mid-point

        // Check if it's possible to split the bags to achieve size 'mid'
        if (isPossible(mid, nums, maxOperations)) {
            answer = mid; // Update answer
            right = mid - 1; // Search for smaller sizes
        } else
            left = mid + 1; // Search for larger sizes
    }

    return answer; // Return the minimum size found
}

private boolean isPossible(int limit, int[] nums, int maxOperations) {
    int operations = 0; // Track the number of operations needed

    for (int balls : nums) // Iterate through each bag
        if (balls > limit) // Check if the bag needs splitting
            operations += (balls - 1) / limit; // Calculate required splits

    // If operations exceed maxOperations, return false
    if (operations > maxOperations) return false;

    return true; // Otherwise, it's possible to achieve the size
}

Explanation:

The problem is solved using binary search to minimize the largest bag size while ensuring the constraints on the number of operations are met:

• The minimumSize function defines the search range and uses binary search to find the smallest feasible size for the largest bag.
• The isPossible helper function calculates the number of operations needed to ensure no bag exceeds the size limit. The formula (balls - 1) / limit determines how many splits are required for a bag with balls balls.
• If the operations exceed maxOperations, the size is not feasible, and the binary search continues to explore larger sizes. Otherwise, the size is feasible, and the binary search explores smaller sizes.