Statistics from a Large Sample - Practice Coding Problems

Statistics from a Large Sample - Problem

Imagine you're analyzing a massive dataset of pixel intensities from millions of images, where each pixel value ranges from 0 (black) to 255 (white). The dataset is so large that instead of storing individual values, you have a frequency count array where count[k] represents how many times the value k appears.

Your task is to calculate five key statistical measures from this compressed representation:

Minimum: The smallest value that appears at least once
Maximum: The largest value that appears at least once
Mean: The average of all values (sum of all elements ÷ total count)
Median: The middle value when all elements are sorted (or average of two middle values for even counts)
Mode: The most frequently occurring value (guaranteed to be unique)

Return these statistics as an array of floating-point numbers: [minimum, maximum, mean, median, mode]

Note: Answers within 10^-5 of the actual answer will be accepted.

Input & Output

example_1.py — Basic Case

$ 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]

💡 Note: Dataset: [1,2,2,2,3,3,3,3]. Min=1, Max=3, Mean=(1+6+12)/8=2.375, Median=(2+3)/2=2.5, Mode=3 (appears 4 times)

example_2.py — Single Element

$ 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]

💡 Note: Dataset: [1,1,1,1,2,2,2,3,3,4,4]. Mode=1 (appears 4 times), total 11 elements so median is 6th element which is 2

example_3.py — Edge Case

$ Input: count = [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1]

› Output: [0.00000, 255.00000, 127.50000, 127.50000, 0.00000]

💡 Note: Dataset: [0,255]. Min=0, Max=255, Mean=(0+255)/2=127.5, Median=(0+255)/2=127.5, Mode=0 (both appear once, but 0 comes first)

Constraints

count.length == 256
0 ≤ count[i] ≤ 10⁹
1 ≤ sum(count) ≤ 10⁹
It is guaranteed that mode is unique
Answers within 10^-5 of the actual answer will be accepted

Visualization

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Understanding the Visualization

Frequency Table Analysis

Extract min, max, mode, and calculate weighted sum in single pass

Mean Calculation

Use formula: Σ(value × frequency) ÷ total_count

Median via Cumulative Frequency

Find middle position(s) using running frequency totals

Key Takeaway

🎯 Key Insight: Frequency data contains all information needed for statistics - no reconstruction required! Use weighted sums for mean and cumulative frequencies for median.

Asked in

G Google 42 a Amazon 35 f Meta 28 ⊞ Microsoft 31

The optimal solution calculates all statistics directly from the frequency array without reconstructing the dataset. Key insights: use weighted sum for mean, cumulative frequency for median, and single pass for min/max/mode. This achieves O(1) time and space complexity regardless of dataset size.

Common Approaches

Approach	Time	Space	Notes
✓ Direct Frequency Analysis (Optimal)	O(1)	O(1)	Calculate all statistics directly from frequency counts without reconstruction
Brute Force (Reconstruct Full Dataset)	O(n log n)	O(n)	Reconstruct the entire dataset and calculate statistics normally

Direct Frequency Analysis (Optimal) — Algorithm Steps

Calculate total count and find min/max by scanning non-zero frequencies
Calculate mean using weighted sum: Σ(value × frequency) / total_count
Find mode as the index with maximum frequency
For median, use cumulative frequencies to locate middle element(s)
Handle even/odd total count cases for median calculation

Visualization

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Step-by-Step Walkthrough

Single Pass Scan

Extract min, max, mode, and weighted sum in one pass

Calculate Mean

Use weighted sum: Σ(value × frequency) / total

Find Median

Use cumulative frequency to locate middle element(s)

Code -

solution.c — C

double* sampleStats(int* count, int countSize, int* returnSize) {
    *returnSize = 5;
    double* result = (double*)malloc(5 * sizeof(double));
    
    // Calculate total count
    int totalCount = 0;
    for (int i = 0; i < 256; i++) {
        totalCount += count[i];
    }
    
    if (totalCount == 0) {
        result[0] = result[1] = result[2] = result[3] = result[4] = 0.0;
        return result;
    }
    
    // Find min, max, mode, and calculate weighted sum
    int minimum = -1, maximum = -1, mode = -1;
    long long weightedSum = 0;
    int maxFreq = 0;
    
    for (int i = 0; i < 256; i++) {
        if (count[i] > 0) {
            if (minimum == -1) minimum = i;
            maximum = i;
            weightedSum += (long long)i * count[i];
            
            if (count[i] > maxFreq) {
                maxFreq = count[i];
                mode = i;
            }
        }
    }
    
    // Calculate mean
    double mean = (double)weightedSum / totalCount;
    
    // Find median using cumulative frequency
    double median = 0.0;
    if (totalCount % 2 == 1) {
        // Odd count: find middle element
        int target = totalCount / 2;
        int cumulative = 0;
        for (int i = 0; i < 256; i++) {
            cumulative += count[i];
            if (cumulative > target) {
                median = i;
                break;
            }
        }
    } else {
        // Even count: average of two middle elements
        int target1 = totalCount / 2 - 1;
        int target2 = totalCount / 2;
        int cumulative = 0;
        int medianVals[2];
        int found = 0;
        
        for (int i = 0; i < 256 && found < 2; i++) {
            if (cumulative <= target1 && target1 < cumulative + count[i]) {
                medianVals[found++] = i;
            }
            if (cumulative <= target2 && target2 < cumulative + count[i] && found < 2) {
                medianVals[found++] = i;
            }
            cumulative += count[i];
        }
        
        median = (medianVals[0] + medianVals[1]) / 2.0;
    }
    
    result[0] = minimum;
    result[1] = maximum;
    result[2] = mean;
    result[3] = median;
    result[4] = mode;
    
    return result;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Always process exactly 256 elements regardless of dataset size

✓ Linear Growth

Space Complexity

O(1)

Only use a few variables, no additional data structures

✓ Linear Space

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Medium Frequency

~18 min Avg. Time

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Direct Frequency Analysis (Optimal) — Algorithm Steps

Visualization

Code -

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