1760. Minimum Limit of Balls in a Bag | Daily LeetCode Solution | Java #leetcodechallenge #coding

This problem seeks to minimize the largest possible ball count in any bag by dividing the balls into smaller bags with a maximum number of allowable operations.

Approach:Binary Search for Optimal Penalty:
Use binary search to find the smallest possible penalty 𝑃 such that the number of operations required does not exceed maxOperations.
The penalty 𝑃 represents the maximum number of balls allowed in any single bag.
Helper Functions:
getMax(nums): Finds the maximum number of balls in any bag, which serves as the upper bound for binary search.
canSplit(nums, maxOperations, penalty): Checks if the array can be split with a maximum ball count of penalty in each bag using at most maxOperations.
For each bag, the required operations to split balls into smaller bags is calculated as (num−1)/penalty.
Binary Search Logic:
Initialize left to 1 (minimum possible penalty) and right to the maximum number of balls in a bag.
At each iteration, calculate the middle penalty (mid).
If canSplit returns true for the current penalty, update the upper bound (right = mid). Otherwise, move the lower bound (left = mid + 1).
Return Result:
The final value of left is the minimum penalty 𝑃 that satisfies the conditions.

Code Explanation:
Input:
nums: Array of integers representing the number of balls in each bag.
maxOperations: Maximum number of operations allowed to split the bags.
Output:
The minimum possible penalty 𝑃.

Complexity:
Time: (𝑛⋅log(max)), where 𝑛 is the number of bags and max is the maximum number of balls in any bag.
Space: 𝑂(1).

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[AlgoXploration]

class Solution {
public int getMax(int[] nums) {
int max = 0;
for(int num : nums)
max = Math.max(max, num);

return max;
}

public boolean canSplit(int[] nums, int maxOperations, int penalty) {
int operations = 0;
for(int num : nums) {
operations += (num-1)/penalty;
}

return operations <= maxOperations;
}
public int minimumSize(int[] nums, int maxOperations) {
int left = 1;
int right = getMax(nums);
while(left < right) {
int mid = left + (right - left) / 2;
if(canSplit(nums, maxOperations, mid))
right = mid;
else
left = mid+1;

}

return left;
}
}