Minimum Limit of Balls in a Bag LeetCode solution in C++

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class Solution {
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 public:
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  bool isPossible(vector<int> &nums, int mid_penalty, int maxOperations) {
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    int operations = 0;
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    for (int i = 0; i < nums.size(); i++) {
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      operations += (nums[i] / mid_penalty); // Operations Nedded to divide that element.
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      if (nums[i] % mid_penalty == 0)        // if it is completely divisible means
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                                             // less 1 less is needed for that nums.
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        operations--;
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    }
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    return operations <= maxOperations ? 1 : 0; // If operations are less than maxOperations it is
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                                                // one of our ans.
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  }
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  int minimumSize(vector<int> &nums, int maxOperations) {
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    int low_penalty = 1, high_penalty = *max_element(nums.begin(), nums.end());
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    int ans = high_penalty;
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    while (low_penalty <= high_penalty) {
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      int mid_penalty = low_penalty + (high_penalty - low_penalty) / 2;
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      if (mid_penalty == 0) // To avoid divison by zero.
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        break;
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      if (isPossible(nums, mid_penalty, maxOperations)) {
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        ans = mid_penalty;
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        high_penalty = mid_penalty - 1;
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      } else
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        low_penalty = mid_penalty + 1;
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    }
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    return ans;
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  }
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};