Maximum Gap

class StorageBin {
public:
    bool isActive = false;
    int minValue = numeric_limits<int>::max();  // equivalent to INT_MAX
    int maxValue = numeric_limits<int>::min();  // equivalent to INT_MIN
};

class GapCalculator {
public:
    int findMaxGap(vector<int>& elements) {
        if (elements.empty() || elements.size() < 2) return 0;

        int smallest = *min_element(elements.begin(), elements.end()),
            largest = *max_element(elements.begin(), elements.end());

        int binSize =
            max(1, (largest - smallest) /
                       ((int)elements.size() - 1));  // bin size or capacity
        int numBins = (largest - smallest) / binSize + 1;  // number of bins
        vector<StorageBin> bins(numBins);

        for (auto&& element : elements) {
            int binIndex =
                (element - smallest) / binSize;  // finding correct bin
            bins[binIndex].isActive = true;
            bins[binIndex].minValue = min(element, bins[binIndex].minValue);
            bins[binIndex].maxValue = max(element, bins[binIndex].maxValue);
        }

        int lastBinMax = smallest, maximumGap = 0;
        for (auto&& bin : bins) {
            if (!bin.isActive) continue;

            maximumGap = max(maximumGap, bin.minValue - lastBinMax);
            lastBinMax = bin.maxValue;
        }

        return maximumGap;
    }
};

The provided C++ code implements a solution to the "Maximum Gap" problem. It utilizes a data structuring strategy known as bucket or bin sorting to efficiently compute the maximum difference between consecutive numbers in an unsorted array. Here's a concise explanation of how the solution works:

• StorageBin Class: Defines a structure for the bins. Each bin tracks whether it has received any numbers (isActive), and maintains the smallest (minValue) and largest (maxValue) values inserted into the bin.

• GapCalculator Class: Contains the method findMaxGap which performs the main operations:

  1. Handles edge cases—returns a gap of 0 if the array is empty or has only one element.
  2. Calculates the range (smallest to largest) of the numbers in the array.
  3. Determines the number of bins (numBins) needed and their size (binSize).
  4. Distributes each element into an appropriate bin by calculating the correct bin index based on the element's value.
  5. For each element placed in its respective bin, it updates the bin's minimum and maximum values.
  6. Iterates through the bins in sequence to calculate the maximum gap based on the difference between the minimum value of the current active bin and the maximum value of the last active bin encountered.

The technical challenges addressed include dealing with varying input sizes, optimizing the gap calculation by using dynamically sized bins based on input range, and ensuring efficient updates within each bin. This solution is efficient for large data sets where conventional sorting would be too slow, making it suitable for problems emphasizing large range or distribution differences within the array.

public class Solution {
    static class Interval {
        public boolean isFilled = false;
        public int least = Integer.MAX_VALUE;
        public int greatest = Integer.MIN_VALUE;
    }

    public int maxDifference(int[] elements) {
        if (elements == null || elements.length < 2) return 0;

        int smallest = Arrays.stream(elements).min().getAsInt(),
            largest = Arrays.stream(elements).max().getAsInt();

        int gapSize = Math.max(1, (largest - smallest) / (elements.length - 1));
        int intervalCount = (largest - smallest) / gapSize + 1;
        Interval[] intervals = new Interval[intervalCount];

        for (int item : elements) {
            int index = (item - smallest) / gapSize;
            if (intervals[index] == null) intervals[index] = new Interval();

            intervals[index].isFilled = true;
            intervals[index].least = Math.min(item, intervals[index].least);
            intervals[index].greatest = Math.max(item, intervals[index].greatest);
        }

        int previousMax = smallest, largestGap = 0;
        for (Interval interval : intervals) {
            if (interval == null || !interval.isFilled) continue;

            largestGap = Math.max(largestGap, interval.least - previousMax);
            previousMax = interval.greatest;
        }

        return largestGap;
    }
}

The provided Java code implements a solution to the Maximum Gap problem, where you need to find the largest difference between two consecutive elements in a sorted version of the given array. The process in the code is as follows:

1. First, check if the input array is null or contains fewer than two elements, in which case, return 0.
2. Identify the smallest and largest elements in the array using the stream operations for minimum and maximum.
3. Calculate an initial gap size, which is the minimum value between 1 and the division of the range (largest - smallest) over the number of elements minus one.
4. Determine the number of intervals or buckets to create by dividing the range by the gap size and adding one.
5. Create an array of Interval objects to hold details about the smallest and largest values that fall within each range bucket.
6. Loop through the original array, placing each element into the appropriate bucket according to its value.
7. For each element in a bucket, update the bucket’s smallest and greatest values accordingly to track the range within the bucket.
8. Initialize variables to track the previous maximum value and the largest found gap.
9. Iterate through the bucket array:
   • Skip empty buckets.
   • For each filled bucket, calculate the possible gap between the smallest value of the current bucket and the greatest value of the previous filled bucket.
   • Update the largestGap variable if the current calculated gap is larger.
10. Finally, return the value of largestGap, which represents the largest difference between consecutive numbers in the input array elements when sorted.

This approach ensures efficient processing by leveraging a bucket sort-like mechanism where each number is placed in a dynamically calculated interval, thus avoiding the overhead of a full sort operation, which leads to a potentially improved performance over simple sorting and iterating methods, especially with large datasets.

#define LOWER(i, j) ((i) < (j) ? (i) : (j))
#define HIGHER(i, j) ((i) > (j) ? (i) : (j))

typedef struct {
    int utilized;
    int smallest;
    int largest;
} StorageBin;

int findMaximumDifference(int* ptrNums, int size) {
    if (!ptrNums || size < 2) return 0;

    int smallest = ptrNums[0], largest = ptrNums[0];
    for (int index = 0; index < size; index++) {
        smallest = LOWER(smallest, ptrNums[index]);
        largest = HIGHER(largest, ptrNums[index]);
    }

    int binSize =
        HIGHER(1, (largest - smallest) / (size - 1));      // size of each storage bin
    int binCount = (largest - smallest) / binSize + 1;  // total bins needed
    StorageBin* bins = (StorageBin*)malloc(sizeof(StorageBin) * binCount);
    for (int i = 0; i < binCount; i++) {
        bins[i].utilized = 0;
        bins[i].smallest = INT_MAX;
        bins[i].largest = INT_MIN;
    }

    for (int index = 0; index < size; index++) {
        int binIndex =
            (ptrNums[index] - smallest) / binSize;  // calculate the appropriate bin
        bins[binIndex].utilized = 1;
        bins[binIndex].smallest = LOWER(ptrNums[index], bins[binIndex].smallest);
        bins[binIndex].largest = HIGHER(ptrNums[index], bins[binIndex].largest);
    }

    int lastBinMax = smallest, greatestDiff = 0;
    for (int i = 0; i < binCount; i++) {
        if (!bins[i].utilized) continue;

        greatestDiff = HIGHER(greatestDiff, bins[i].smallest - lastBinMax);
        lastBinMax = bins[i].largest;
    }

    free(bins);
    return greatestDiff;
}

The provided C code defines a function to calculate the maximum difference between successive elements when all integers are sorted from a given list. The method employed here is the bucket sort technique modified for optimum performance by utilizing storage bins. Each bin stores the largest and smallest values and a utilization flag. Here's how the approach is implemented:

• Determine the smallest and largest elements in the array to find the range of values.
• Calculate the required number of bins and the size of each bin to optimally distribute all numbers within the array.
• Allocate memory for the bins, initialized with maximal and minimal possible integer values alongside a flag indicating if a bin is used.
• Place each element in the proper bin while updating the bin's smallest and largest values.
• Traverse the bins to find the maximum gap between the largest value of a previously utilized bin and the smallest value of the current bin.
• Free allocated memory resources used for bins to avoid memory leaks.

The solution effectively handles large input sizes by reducing the problem to O(n) complexity using the bucket sort foundational principles tailored for the scenario. Hence, you ensure efficiency and scalability in obtaining the maximum gap between elements when sorted.

class BucketContainer {
    constructor() {
        this.isFilled = false;
        this.smallest = Number.MAX_SAFE_INTEGER;
        this.largest = Number.MIN_SAFE_INTEGER;
    }
}

var findMaximumGap = function (numbers) {
    if (!numbers || numbers.length < 2) return 0;

    let minVal = Math.min(...numbers),
        maxVal = Math.max(...numbers);

    let sizeOfBucket = Math.max(1, Math.floor((maxVal - minVal) / (numbers.length - 1)));
    let totalBuckets = Math.floor((maxVal - minVal) / sizeOfBucket) + 1;
    let allBuckets = Array.from({ length: totalBuckets }, () => new BucketContainer());

    for (let number of numbers) {
        let index = Math.floor((number - minVal) / sizeOfBucket);
        allBuckets[index].isFilled = true;
        allBuckets[index].smallest = Math.min(number, allBuckets[index].smallest);
        allBuckets[index].largest = Math.max(number, allBuckets[index].largest);
    }

    let previousMax = minVal,
        greatestGap = 0;
    for (let bucket of allBuckets) {
        if (!bucket.isFilled) continue;

        greatestGap = Math.max(greatestGap, bucket.smallest - previousMax);
        previousMax = bucket.largest;
    }

    return greatestGap;
};

The given JavaScript program solves the problem of finding the maximum gap between successive elements in a sorted version of an array. It employs a bucket-sort-like strategy for optimal performance, particularly useful when there is a large difference between the minimum and maximum values in the array. Here's how it functions:

• Initializes BucketContainer to store the smallest and largest values of numbers that fall into each bucket.
• Validates input to handle edge cases where the array has less than two numbers.
• Determines the range covered by each bucket based on the minimum and maximum values in the array.
• Allocates buckets to cover the entire range of numbers.
• Places each number into its corresponding bucket and updates the smallest and largest values within those buckets.
• Finally, iterates through the buckets to find the largest gap between the largest value of the previous bucket and the smallest value of the current bucket.

The result is the maximum gap found, ensuring the solution is efficient by only looping through the numbers and buckets a constant number of times relative to the size of the input.

class DataBucket:
    def __init__(self):
        self.is_filled = False
        self.minimum = float("inf")
        self.maximum = float("-inf")

class MaxGapCalculator:
    def calculateMaxGap(self, elements):
        if len(elements) < 2:
            return 0

        low, high = min(elements), max(elements)

        sizeOfBucket = max(1, (high - low) // (len(elements) - 1))
        numberBuckets = (high - low) // sizeOfBucket + 1
        buckets = [DataBucket() for _ in range(numberBuckets)]

        for element in elements:
            idx = (element - low) // sizeOfBucket
            buckets[idx].is_filled = True
            buckets[idx].minimum = min(element, buckets[idx].minimum)
            buckets[idx].maximum = max(element, buckets[idx].maximum)

        lastBucketMax = low
        greatestGap = 0
        for bucket in buckets:
            if not bucket.is_filled:
                continue

            greatestGap = max(greatestGap, bucket.minimum - lastBucketMax)
            lastBucketMax = bucket.maximum

        return greatestGap

The given Python3 code defines a solution for finding the maximum gap (the largest difference) between consecutive elements when sorted in a sequence. It employs the concept of bucket sort to effectively find this gap.

• The code consists of two classes:

  • DataBucket: Maintains minimum and maximum values within a bucket and tracks whether the bucket is filled.
  • MaxGapCalculator: Contains the method calculateMaxGap to compute the maximum gap.

• The calculateMaxGap function includes the following steps:

  1. Handle the edge case where the number of elements is less than two, returning a gap of zero.
  2. Determine the minimum (low) and maximum (high) values in the input list.
  3. Compute the appropriate size for each bucket such that elements are evenly distributed based on their value range.
  4. Initialize the bucket array using the predefined size.
  5. Assign each element to its respective bucket and update the bucket's minimum and maximum values.
  6. Iterate through each bucket to find the maximum consecutive gap. This is done by comparing the minimum value of a filled bucket to the maximum of the last checked filled bucket.

• This algorithm improves efficiency over a simple sort and scan approach by reducing the number of comparisons needed to find the maximum gap. It handles edge cases effectively and leverages a structured bucket mechanism to categorize and compare relevant data points efficiently.