1093. Statistics from a Large Sample LeetCode Solution
In this guide, you will get 1093. Statistics from a Large Sample LeetCode Solution with the best time and space complexity. The solution to Statistics from a Large Sample problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of 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.
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.
Complexity Analysis
Time Complexity:
Space Complexity:
1093. Statistics from a Large Sample LeetCode Solution in C++
class Solution {
public:
vector<double> sampleStats(vector<int>& count) {
const int n = accumulate(count.begin(), count.end(), 0);
const double mode = ranges::max_element(count) - count.begin();
return {
getMinimum(count),
getMaximum(count),
getMean(count, n),
(getLeftMedian(count, n) + getRightMedian(count, n)) / 2.0,
mode,
};
}
private:
double getMinimum(const vector<int>& count) {
for (int i = 0; i < count.size(); ++i)
if (count[i])
return i;
return -1;
}
double getMaximum(const vector<int>& count) {
for (int i = count.size() - 1; i >= 0; --i)
if (count[i])
return i;
return -1;
}
double getMean(const vector<int>& count, double n) {
double mean = 0;
for (long i = 0; i < count.size(); ++i)
mean += (i * count[i]) / n;
return mean;
}
double getLeftMedian(const vector<int>& count, double n) {
int numCount = 0;
for (int i = 0; i < count.size(); ++i) {
numCount += count[i];
if (numCount >= n / 2)
return i;
}
return -1;
}
double getRightMedian(const vector<int>& count, double n) {
int numCount = 0;
for (int i = count.size() - 1; i >= 0; --i) {
numCount += count[i];
if (numCount >= n / 2)
return i;
}
return -1;
}
}; /* code provided by PROGIEZ */
1093. Statistics from a Large Sample LeetCode Solution in Java
class Solution {
public double[] sampleStats(int[] count) {
final int n = Arrays.stream(count).sum();
return new double[] {
getMinimum(count), //
getMaximum(count), //
getMean(count, n), //
(getLeftMedian(count, n) + getRightMedian(count, n)) / 2.0, //
getMode(count),
};
}
private double getMinimum(int[] count) {
for (int i = 0; i < count.length; ++i)
if (count[i] > 0)
return i;
return -1;
}
private double getMaximum(int[] count) {
for (int i = count.length - 1; i >= 0; --i)
if (count[i] > 0)
return i;
return -1;
}
private double getMean(int[] count, double n) {
double mean = 0;
for (int i = 0; i < count.length; ++i)
mean += ((long) i * (long) count[i]) / n;
return mean;
}
private double getLeftMedian(int[] count, double n) {
int numCount = 0;
for (int i = 0; i < count.length; ++i) {
numCount += count[i];
if (numCount >= n / 2)
return i;
}
return -1;
}
private double getRightMedian(int[] count, double n) {
int numCount = 0;
for (int i = count.length - 1; i >= 0; --i) {
numCount += count[i];
if (numCount >= n / 2)
return i;
}
return -1;
}
private double getMode(int[] count) {
int mode = -1;
int maxCount = 0;
for (int i = 0; i < count.length; ++i)
if (count[i] > maxCount) {
maxCount = count[i];
mode = i;
}
return mode;
}
} // code provided by PROGIEZ
1093. Statistics from a Large Sample LeetCode Solution in Python
class Solution:
def sampleStats(self, count: list[int]) -> list[float]:
minimum = next((i for i, num in enumerate(count) if num), None)
maximum = next((i for i, num in reversed(
list(enumerate(count))) if num), None)
n = sum(count)
mean = sum(i * c / n for i, c in enumerate(count))
mode = count.index(max(count))
numCount = 0
leftMedian = 0
for i, c in enumerate(count):
numCount += c
if numCount >= n / 2:
leftMedian = i
break
numCount = 0
rightMedian = 0
for i, c in reversed(list(enumerate(count))):
numCount += c
if numCount >= n / 2:
rightMedian = i
break
return [minimum, maximum, mean, (leftMedian + rightMedian) / 2, mode] # code by PROGIEZ
