Introduction

In this blog post, we will explore how to solve the LeetCode problem "1093" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's the solution to the problem:

• Calculate Sum and Count: Iterate through the count array to calculate the sum of all elements in the sample and the total number of elements.
• Find Minimum, Maximum, and Mode: During the same iteration, identify the minimum, maximum, and mode. The mode is determined by keeping track of the element with the highest count.

• Calculate Median: Find the median using cumulative counts. This avoids sorting the entire sample.

• Runtime Complexity: O(256) which simplifies to O(1). Storage Complexity: O(1)

Code

    def sampleStats(count):
    n = len(count)
    total_sum = 0
    total_count = 0
    minimum = -1
    maximum = -1
    mode = -1
    max_count = 0

    for i in range(n):
        if count[i] > 0:
            if minimum == -1:
                minimum = i
            maximum = i
            total_sum += i * count[i]
            total_count += count[i]
            if count[i] > max_count:
                max_count = count[i]
                mode = i

    mean = total_sum / total_count

    median = 0.0
    mid = total_count // 2
    count_so_far = 0
    median_found = False
    first_median = -1

    for i in range(n):
        if count[i] > 0:
            count_so_far += count[i]
            if not median_found:
                if total_count % 2 == 1:
                    if count_so_far > mid:
                        median = float(i)
                        median_found = True
                        break
                else:
                    if count_so_far > mid - 1 and first_median == -1:
                        first_median = i
                        if count_so_far > mid:
                            median = float(i)
                            median_found = True
                            break
                    elif count_so_far > mid:
                        median = (first_median + i) / 2.0
                        median_found = True
                        break
    return [float(minimum), float(maximum), mean, median, float(mode)]