## Algorithm

Problem Name: 1093. Statistics from a Large Sample

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.

## Code Examples

### #1 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public double[] sampleStats(int[] num) {
Map map = new HashMap<>();
double sum = 0;
int currMin = Integer.MAX_VALUE;
int currMax = Integer.MIN_VALUE;
int currMod = -1;
int modCount = 0;
int count = 0;

for (int i = 0; i < num.length; i++) {
sum += num[i] * i;
if (num[i] > 0) {
currMin = Math.min(currMin, i);
currMax = Math.max(currMax, i);
map.put(i, num[i]);
if (map.get(i) > modCount) {
modCount = num[i];
currMod = i;
}
count += num[i];
}
}

return new double[]{
(double) currMin,
(double) currMax,
sum / count,
count % 2 == 0 ? getEvenMedian(num, count) : getOddMedian(num, count),
(double) currMod
};
}

private double getEvenMedian(int[] num, int count) {
int p1 = count/2;
int p2 = count/2 + 1;
double sum = 0;
int currCount = 0;
int i = 0;

while (i < num.length) {
if (num[i] != 0) {
currCount += num[i];
if (currCount >= p1 && currCount >= p2) {
return (double) i;
}

if (currCount >= p1 && currCount < p2) {
sum = i;
i++;
break;
}
}

i++;
}

while (i < num.length) {
if (num[i] != 0) {
currCount += num[i];
if (currCount >= p2) {
sum += i;
break;
}
}

i++;
}

return sum / 2;
}

private double getOddMedian(int[] num, int count) {
int p1 = count/2 + 1;
int currCount = 0;
for (int i = 0; i < num.length; i++) {
if (num[i] != 0) {
currCount += num[i];
if (currCount >= p1) {
return (double) i;
}
}
}

return (double) 1;
}
}
``````
Copy The Code &

Input

cmd
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

cmd
[1.00000,3.00000,2.37500,2.50000,3.00000]

### #2 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const sampleStats = function(count) {
const n = count.length
let numElems = 0
let sum = 0
let min = n - 1
let max = 0
let modVal = 0
let modIdx = 0
for (let i = 0; i < n; i++) {
if (count[i]) {
min = Math.min(i, min)
max = Math.max(i, max)
if (count[i] > modVal) {
modVal = count[i]
modIdx = i
}
sum += i * count[i]
numElems += count[i]
}
}
const half = Math.floor(numElems / 2)
let median
for (let i = 0, c = 0, last = 0; i < n; i++) {
if (count[i]) {
c += count[i]
if (c > half) {
if (numElems % 2 === 0 && c - count[i] === half) {
median = (i + last) / 2
} else {
median = i
}
break
}
last = i
}
}
return [min, max, sum / numElems, median, modIdx]
}
``````
Copy The Code &

Input

cmd
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

cmd
[1.00000,3.00000,2.37500,2.50000,3.00000]

### #3 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def sampleStats(self, count: List[int]) -> List[float]:
arr = [(i, c * 1.0) for i, c in enumerate(count) if c]
acc = list(itertools.accumulate(count, lambda x, y: x + y))
mean = sum(i * c for i, c in arr) / acc[-1]
mode = max(arr, key = lambda x: x[1])[0] * 1.0
m1 = bisect.bisect(acc, (acc[-1] - 1) // 2)
m2 = bisect.bisect(acc, acc[-1] // 2)
return [arr[0][0] * 1.0, arr[-1][0] * 1.0, mean, (m1 + m2) / 2.0, mode]

``````
Copy The Code &

Input

cmd
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

cmd
[1.00000,3.00000,2.37500,2.50000,3.00000]