// Runtime: 26 ms (Top 86.75%) | Memory: 59.70 MB (Top 45.73%)

class Solution {

public int minimumSize(int[] nums, int maxOperations) {
//initiate the boundary for possible answers, here if you let min=1 it will still work for most cases except for some corner cases. We make max=100000000 because nums[i] <= 10^9. You can choose to sort the array and make the max= arr.max, at the price of time consumption.
//The answer should be the minimized max value.
    int min = 0;
    int max = 1000000000;
	//Compared with min<max or min <= max, min + 1 < max will avoid infinite loops e.g. when min = 2, max = 3
    while (min +1< max) {
        int mid = (max - min)/2 + min;
		//count indicates the operation times with atmost mid balls in bag.
        int count = 0;
        for (int a: nums) {
		//this is the same as Math. ceil(a/mid) - 1=> math.ceil(a/mid) gives the number of divided bags, we subtract the number by 1 to get the subdivision operation times.
            count+=(a-1)/mid;
        }
		//if count < maxOperations, max WOULD be further minimized and set to mid; 
		//if count = maxOperations, max still COULD be further minimized and set to mid. 
		//so we combine < and = cases together in one if condition
        if (count <= maxOperations) {
		//max = mid - 1 will not work in this case becasue mid could be the correct answer. 
		//To not miss the correct answer we set a relatively "loose" boundary for max and min.
            max = mid;
        } else{
            min = mid;
        }
    }
	//Now we find the minimized max value
    return max;
}
}