# 1093. Statistics from a Large Sample

## Description

You are given a large sample of integers in the range [0, 255]. Since the sample is so large, it is represented by an array count where count[k] is the number of times that k appears in the sample.

Calculate the following statistics:

• minimum: The minimum element in the sample.
• maximum: The maximum element in the sample.
• mean: The average of the sample, calculated as the total sum of all elements divided by the total number of elements.
• median:
• If the sample has an odd number of elements, then the median is the middle element once the sample is sorted.
• If the sample has an even number of elements, then the median is the average of the two middle elements once the sample is sorted.
• mode: The number that appears the most in the sample. It is guaranteed to be unique.

Return the statistics of the sample as an array of floating-point numbers [minimum, maximum, mean, median, mode]. Answers within 10-5 of the actual answer will be accepted.

Example 1:

Input: 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]
Output: [1.00000,3.00000,2.37500,2.50000,3.00000]
Explanation: The sample represented by count is [1,2,2,2,3,3,3,3].
The minimum and maximum are 1 and 3 respectively.
The mean is (1+2+2+2+3+3+3+3) / 8 = 19 / 8 = 2.375.
Since the size of the sample is even, the median is the average of the two middle elements 2 and 3, which is 2.5.
The mode is 3 as it appears the most in the sample.


Example 2:

Input: 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]
Output: [1.00000,4.00000,2.18182,2.00000,1.00000]
Explanation: The sample represented by count is [1,1,1,1,2,2,2,3,3,4,4].
The minimum and maximum are 1 and 4 respectively.
The mean is (1+1+1+1+2+2+2+3+3+4+4) / 11 = 24 / 11 = 2.18181818... (for display purposes, the output shows the rounded number 2.18182).
Since the size of the sample is odd, the median is the middle element 2.
The mode is 1 as it appears the most in the sample.


Constraints:

• count.length == 256
• 0 <= count[i] <= 109
• 1 <= sum(count) <= 109
• The mode of the sample that count represents is unique.

## Solutions

• 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};
}
};

• 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]


• 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];
}