1093. 大样本统计

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 \(O(1)\)。

Python3JavaC++GoTypeScript

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];
}

1093. 大样本统计

题目描述

解法

方法一：模拟

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