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1283. Find the Smallest Divisor Given a Threshold 👍

  • Time: $O(n\log\max(\texttt{nums}))$
  • Space: $O(1)$
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class Solution {
 public:
  int smallestDivisor(vector<int>& nums, int threshold) {
    int l = 1;
    int r = ranges::max(nums);

    while (l < r) {
      const int m = (l + r) / 2;
      if (sumDivision(nums, m) <= threshold)
        r = m;
      else
        l = m + 1;
    }

    return l;
  }

 private:
  int sumDivision(const vector<int>& nums, int m) {
    int sum = 0;
    for (const int num : nums)
      sum += (num - 1) / m + 1;
    return sum;
  }
};

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class Solution {
  public int smallestDivisor(int[] nums, int threshold) {
    int l = 1;
    int r = Arrays.stream(nums).max().getAsInt();

    while (l < r) {
      final int m = (l + r) / 2;
      if (sumDivision(nums, m) <= threshold)
        r = m;
      else
        l = m + 1;
    }

    return l;
  }

  private int sumDivision(int[] nums, int m) {
    int sum = 0;
    for (final int num : nums)
      sum += (num - 1) / m + 1;
    return sum;
  }
}

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class Solution:
  def smallestDivisor(self, nums: list[int], threshold: int) -> int:
    l = 1
    r = max(nums)

    while l < r:
      m = (l + r) // 2
      if sum((num - 1) // m + 1 for num in nums) <= threshold:
        r = m
      else:
        l = m + 1

    return l