Statistics from a Large Sample

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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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the values in the `count` array, and what is the maximum size of the `count` array itself?
Is it possible for the input `count` array to be empty, or for any of its elements to be zero?
In the case of the median calculation, if the total number of elements is even, should I return the average of the two middle values, or is there a specific rounding rule I should follow?
If there are multiple modes (values with the same highest frequency), should I return any one of them, the smallest one, or is there another rule for selecting the mode?
Can I assume that the input `count` array is always sorted in ascending order by the value it represents?

Brute Force Solution

Approach

To find statistics like the minimum, maximum, mean, and median from a large dataset, the brute force approach directly examines every single number in the set. It's like looking at each item individually to figure out these statistics. This method is straightforward but can be inefficient for very large datasets.

Here's how the algorithm would work step-by-step:

To find the smallest value, start by assuming the first number is the smallest.
Then, compare this 'smallest' number with the next number in the dataset.
If the next number is smaller, update your 'smallest' number.
Keep repeating this comparison with every number in the dataset.
After checking all the numbers, the one you're holding as the 'smallest' is the true smallest.
Repeat a similar process to find the largest value, but instead of looking for smaller numbers, look for bigger ones.
To find the mean (average), add up every single number in the dataset to get the total sum.
Then, count how many numbers are in the dataset.
Finally, divide the total sum by the number of numbers to get the average.
To find the median (the middle value), first, make a copy of the data and sort it from smallest to largest.
If there's an odd number of values, the median is the middle value.
If there's an even number of values, the median is the average of the two middle values.

Code Implementation

def calculate_statistics(numbers):
    if not numbers:
        return None, None, None

    smallest_number = numbers[0]

    # Iterate through numbers to find smallest
    for current_number in numbers:

        if current_number < smallest_number:
            smallest_number = current_number

    largest_number = numbers[0]

    # Iterate through numbers to find largest
    for current_number in numbers:
        if current_number > largest_number:
            largest_number = current_number

    total = 0

    for current_number in numbers:
        total += current_number

    average = total / len(numbers)

    return smallest_number, largest_number, average

Big(O) Analysis

Time Complexity

O(n log n) – Finding the minimum and maximum each involve iterating through all n elements in the dataset once. Calculating the mean also requires iterating through all n elements to compute the sum. However, finding the median requires sorting the dataset, which typically takes O(n log n) time. Therefore, the dominant operation is the sorting step, making the overall time complexity O(n log n).

Space Complexity

O(N) – The space complexity is primarily determined by the median calculation, which involves creating a sorted copy of the input data. This copy requires additional memory proportional to the number of elements in the dataset, N. While other operations like finding the minimum, maximum, and mean use constant space, sorting the data leads to an auxiliary array of size N. Therefore, the overall auxiliary space complexity is O(N).

Optimal Solution

Approach

When we have a very large dataset summarized in a specific format, instead of manually calculating the average, minimum, and maximum, we can use the properties of a normal distribution to estimate these values. The core idea is to focus on the confidence interval to extrapolate properties of the entire dataset from the information provided.

Here's how the algorithm would work step-by-step:

Understand what information you're given about the dataset. It includes things like the average value, the range around it (confidence interval), and how sure we are about the average (confidence level).
Use the average as your starting point for the 'middle' of your data.
Based on the confidence interval and the confidence level, estimate how spread out the data is. This gives you a sense of how 'wide' your data distribution is.
Because we're assuming the data looks like a normal distribution, we can use this spread to estimate the likely minimum and maximum values of the actual dataset.
Adjust your estimated minimum and maximum values. It is more reasonable to keep the minimum and maximum values given by the data samples if they fall within the calculated bounds.
Return these calculated values (the average, minimum, and maximum) as estimates for the whole dataset.

Code Implementation

def sample_stats(counts):
    total_sample_size = sum(counts)
    minimum_value = next(i for i, count in enumerate(counts) if count > 0)
    maximum_value = next(i for i in range(len(counts) - 1, -1, -1) if counts[i] > 0)

    mean_sum = 0
    for value, count in enumerate(counts):
        mean_sum += value * count
    mean_value = mean_sum / total_sample_size

    # Determine median by finding middle value(s)
    halfway_point = total_sample_size / 2
    current_count = 0
    median_values = []

    for value, count in enumerate(counts):
        if current_count <= halfway_point and current_count + count >= halfway_point:
            median_values.append(value)

            #Check if sample size is even.
            if total_sample_size % 2 == 0 and current_count + count == halfway_point:
                #Must find the next value for the median calculation.
                next_value = next((i for i in range(value+1,len(counts)) if counts[i] > 0), None)
                if next_value is not None:
                    median_values.append(next_value)
                break
            elif total_sample_size % 2 != 0:
                break

        current_count += count

    #Calculate the median.
    if len(median_values) == 1:
        median_value = median_values[0]
    else:
        # If even, take average of the two middle values.
        median_value = (median_values[0] + median_values[1]) / 2

    return [minimum_value, maximum_value, mean_value, median_value]

Big(O) Analysis

Time Complexity

O(1) – The algorithm performs a fixed number of calculations based on the given sample data (average, confidence interval, and confidence level). It involves simple arithmetic operations and comparisons to determine the estimated minimum, maximum, and average. The number of operations does not depend on the size of the original dataset. Therefore, the time complexity is constant.

Space Complexity

O(1) – The algorithm, based on the provided plain English explanation, focuses on calculations and estimations based on the input data (average, confidence interval, confidence level). It does not describe the creation of any auxiliary data structures like lists, maps, or recursive call stacks. Therefore, it appears to use a fixed amount of extra memory for variables to store intermediate calculation results (e.g., estimated minimum, maximum). The space used is independent of the input size, so the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty input array (count)	Return null or an error, as statistics cannot be computed from an empty sample.
All counts are zero.	Return null or an error, indicating an invalid or empty sample.
Very large counts leading to integer overflow when calculating the sum of values or number of elements.	Use long data type or appropriate data type for the summation and element count to prevent overflow.
Large input array (count) causing memory issues or exceeding time limits.	Optimize the algorithm to avoid unnecessary memory allocations and ensure linear time complexity where possible.
Counts represent values that could result in floating-point precision issues when calculating the mean.	Use double data type for the mean calculation and consider rounding the result to an appropriate number of decimal places.
An input array where all counts are identical, simplifying median/mode calculation.	The algorithm should correctly handle this case and identify the appropriate median and mode.
When the total number of elements in the sample is even, correctly compute the average of the two middle values for the median.	Ensure that the algorithm correctly identifies and averages the two middle values for accurate median calculation.
Multiple modes exist (values with the same highest frequency).	Return any one of the modes, or define a clear convention (e.g., return the smallest) and implement accordingly.

Empty input array (count)

How to Handle:

Return null or an error, as statistics cannot be computed from an empty sample.

All counts are zero.

How to Handle:

Return null or an error, indicating an invalid or empty sample.

Very large counts leading to integer overflow when calculating the sum of values or number of elements.

How to Handle:

Use long data type or appropriate data type for the summation and element count to prevent overflow.

Large input array (count) causing memory issues or exceeding time limits.

How to Handle:

Optimize the algorithm to avoid unnecessary memory allocations and ensure linear time complexity where possible.

Counts represent values that could result in floating-point precision issues when calculating the mean.

How to Handle:

Use double data type for the mean calculation and consider rounding the result to an appropriate number of decimal places.

An input array where all counts are identical, simplifying median/mode calculation.

How to Handle:

The algorithm should correctly handle this case and identify the appropriate median and mode.

When the total number of elements in the sample is even, correctly compute the average of the two middle values for the median.

How to Handle:

Ensure that the algorithm correctly identifies and averages the two middle values for accurate median calculation.

Multiple modes exist (values with the same highest frequency).

How to Handle:

Return any one of the modes, or define a clear convention (e.g., return the smallest) and implement accordingly.