Minimum Limit of Balls in a Bag LeetCode solution in Java

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