Question

1093. 大样本统计

English Version

题目描述

我们对 0 到 255 之间的整数进行采样，并将结果存储在数组 count 中：count[k] 就是整数 k 在样本中出现的次数。

计算以下统计数据:

• minimum ：样本中的最小元素。
• maximum ：样品中的最大元素。
• mean ：样本的平均值，计算为所有元素的总和除以元素总数。
• median ：
  • 如果样本的元素个数是奇数，那么一旦样本排序后，中位数 median 就是中间的元素。
  • 如果样本中有偶数个元素，那么中位数median 就是样本排序后中间两个元素的平均值。
• mode ：样本中出现次数最多的数字。保众数是 唯一 的。

以浮点数数组的形式返回样本的统计信息_ _[minimum, maximum, mean, median, mode] 。与真实答案误差在_ _10-5_ _内的答案都可以通过。

 

示例 1：

输入：count = [0,1,3,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
输出：[1.00000,3.00000,2.37500,2.50000,3.00000]
解释：用count表示的样本为[1,2,2,2,3,3,3,3]。
最小值和最大值分别为1和3。
均值是(1+2+2+2+3+3+3+3) / 8 = 19 / 8 = 2.375。
因为样本的大小是偶数，所以中位数是中间两个元素2和3的平均值，也就是2.5。
众数为3，因为它在样本中出现的次数最多。

示例 2：

输入：count = [0,4,3,2,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
输出：[1.00000,4.00000,2.18182,2.00000,1.00000]
解释：用count表示的样本为[1,1,1,1,2,2,3,3,3,4,4]。
最小值为1，最大值为4。
平均数是(1+1+1+1+2+2+2+3+3+4+4)/ 11 = 24 / 11 = 2.18181818…(为了显示，输出显示了整数2.18182)。
因为样本的大小是奇数，所以中值是中间元素2。
众数为1，因为它在样本中出现的次数最多。

 

提示：

• count.length == 256
• 0 <= count[i] <= 109
• 1 <= sum(count) <= 109
•  count 的众数是 唯一 的

解法

方法一：模拟

我们直接根据题目描述模拟即可，定义以下变量：

• 变量 $mi$ 表示最小值；
• 变量 $mx$ 表示最大值；
• 变量 $s$ 表示总和；
• 变量 $cnt$ 表示总个数；
• 变量 $mode$ 表示众数。

我们遍历数组 $count$，对于当前遍历到的数字 $count[k]$，如果 $count[k] \gt 0$，那么我们做以下更新操作：

• 更新 $mi = \min(mi, k)$；
• 更新 $mx = \max(mx, k)$；
• 更新 $s = s + k \times count[k]$；
• 更新 $cnt = cnt + count[k]$；
• 如果 $count[k] \gt count[mode]$，那么更新 $mode = k$。

遍历结束后，我们再根据 $cnt$ 的奇偶性来计算中位数 $median$，如果 $cnt$ 是奇数，那么中位数就是第 $\lfloor \frac{cnt}{2} \rfloor + 1$ 个数字，如果 $cnt$ 是偶数，那么中位数就是第 $\lfloor \frac{cnt}{2} \rfloor$ 和第 $\lfloor \frac{cnt}{2} \rfloor + 1$ 个数字的平均值。

这里我们通过一个简单的辅助函数 $find(i)$ 来找到第 $i$ 个数字，具体实现可以参考下面的代码。

最后，我们将 $mi, mx, \frac{s}{cnt}, median, mode$ 放入答案数组中返回即可。

时间复杂度 $O(n)$，其中 $n$ 是数组 $count$ 的长度。空间复杂度 $O(1)$。

class Solution:
  def sampleStats(self, count: List[int]) -> List[float]:
    def find(i: int) -> int:
      t = 0
      for k, x in enumerate(count):
        t += x
        if t >= i:
          return k

    mi, mx = inf, -1
    s = cnt = 0
    mode = 0
    for k, x in enumerate(count):
      if x:
        mi = min(mi, k)
        mx = max(mx, k)
        s += k * x
        cnt += x
        if x > count[mode]:
          mode = k

    median = (
      find(cnt // 2 + 1) if cnt & 1 else (find(cnt // 2) + find(cnt // 2 + 1)) / 2
    )
    return [mi, mx, s / cnt, median, mode]

class Solution {
  private int[] count;

  public double[] sampleStats(int[] count) {
    this.count = count;
    int mi = 1 << 30, mx = -1;
    long s = 0;
    int cnt = 0;
    int mode = 0;
    for (int k = 0; k < count.length; ++k) {
      if (count[k] > 0) {
        mi = Math.min(mi, k);
        mx = Math.max(mx, k);
        s += 1L * k * count[k];
        cnt += count[k];
        if (count[k] > count[mode]) {
          mode = k;
        }
      }
    }
    double median
      = cnt % 2 == 1 ? find(cnt / 2 + 1) : (find(cnt / 2) + find(cnt / 2 + 1)) / 2.0;
    return new double[] {mi, mx, s * 1.0 / cnt, median, mode};
  }

  private int find(int i) {
    for (int k = 0, t = 0;; ++k) {
      t += count[k];
      if (t >= i) {
        return k;
      }
    }
  }
}

class Solution {
public:
  vector<double> sampleStats(vector<int>& count) {
    auto find = [&](int i) -> int {
      for (int k = 0, t = 0;; ++k) {
        t += count[k];
        if (t >= i) {
          return k;
        }
      }
    };
    int mi = 1 << 30, mx = -1;
    long long s = 0;
    int cnt = 0, mode = 0;
    for (int k = 0; k < count.size(); ++k) {
      if (count[k]) {
        mi = min(mi, k);
        mx = max(mx, k);
        s += 1LL * k * count[k];
        cnt += count[k];
        if (count[k] > count[mode]) {
          mode = k;
        }
      }
    }
    double median = cnt % 2 == 1 ? find(cnt / 2 + 1) : (find(cnt / 2) + find(cnt / 2 + 1)) / 2.0;
    return vector<double>{(double) mi, (double) mx, s * 1.0 / cnt, median, (double) mode};
  }
};

func sampleStats(count []int) []float64 {
  find := func(i int) int {
    for k, t := 0, 0; ; k++ {
      t += count[k]
      if t >= i {
        return k
      }
    }
  }
  mi, mx := 1<<30, -1
  s, cnt, mode := 0, 0, 0
  for k, x := range count {
    if x > 0 {
      mi = min(mi, k)
      mx = max(mx, k)
      s += k * x
      cnt += x
      if x > count[mode] {
        mode = k
      }
    }
  }
  var median float64
  if cnt&1 == 1 {
    median = float64(find(cnt/2 + 1))
  } else {
    median = float64(find(cnt/2)+find(cnt/2+1)) / 2
  }
  return []float64{float64(mi), float64(mx), float64(s) / float64(cnt), median, float64(mode)}
}

function sampleStats(count: number[]): number[] {
  const find = (i: number): number => {
    for (let k = 0, t = 0; ; ++k) {
      t += count[k];
      if (t >= i) {
        return k;
      }
    }
  };
  let mi = 1 << 30;
  let mx = -1;
  let [s, cnt, mode] = [0, 0, 0];
  for (let k = 0; k < count.length; ++k) {
    if (count[k] > 0) {
      mi = Math.min(mi, k);
      mx = Math.max(mx, k);
      s += k * count[k];
      cnt += count[k];
      if (count[k] > count[mode]) {
        mode = k;
      }
    }
  }
  const median =
    cnt % 2 === 1 ? find((cnt >> 1) + 1) : (find(cnt >> 1) + find((cnt >> 1) + 1)) / 2;
  return [mi, mx, s / cnt, median, mode];
}