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2137. Pour Water Between Buckets to Make Water Levels Equal 👍

  • Time: $O(n\log \frac{\max}{10^{-5}})$
  • Space: $O(1)$
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class Solution {
 public:
  double equalizeWater(vector<int>& buckets, int loss) {
    constexpr double kErr = 1e-5;
    const double kPercentage = (100 - loss) / (double)100;
    double l = 0.0;
    double r = ranges::max(buckets);

    while (r - l > kErr) {
      const double m = (l + r) / 2;
      if (canFill(buckets, m, kPercentage))
        l = m;
      else
        r = m;
    }

    return l;
  }

 private:
  bool canFill(const vector<int>& buckets, double target, double kPercentage) {
    double extra = 0.0;
    double need = 0.0;
    for (const int bucket : buckets)
      if (bucket > target)
        extra += bucket - target;
      else
        need += target - bucket;
    return extra * kPercentage >= need;
  }
};

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class Solution {
  public double equalizeWater(int[] buckets, int loss) {
    final double kErr = 1e-5;
    final double kPercentage = (100 - loss) / (double) 100;
    double l = 0.0;
    double r = Arrays.stream(buckets).max().getAsInt();

    while (r - l > kErr) {
      final double m = (l + r) / 2;
      if (canFill(buckets, m, kPercentage))
        l = m;
      else
        r = m;
    }

    return l;
  }

  private boolean canFill(int[] buckets, double target, double kPercentage) {
    double extra = 0.0;
    double need = 0.0;
    for (final int bucket : buckets)
      if (bucket > target)
        extra += bucket - target;
      else
        need += target - bucket;
    return extra * kPercentage >= need;
  };
}

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class Solution:
  def equalizeWater(self, buckets: list[int], loss: int) -> float:
    kErr = 1e-5
    kPercentage = (100 - loss) / 100
    l = 0.0
    r = max(buckets)

    def canFill(target: float) -> bool:
      extra = 0
      need = 0
      for bucket in buckets:
        if bucket > target:
          extra += bucket - target
        else:
          need += target - bucket
      return extra * kPercentage >= need

    while r - l > kErr:
      m = (l + r) / 2
      if canFill(m):
        l = m
      else:
        r = m

    return l