Missing Ranges

class Solution {
public:
    vector<vector<int>> getMissingIntervals(vector<int>& nums, int lower, int upper) {
        int len = nums.size();
        vector<vector<int>> missingIntervals;
        if (len == 0) {
            missingIntervals.push_back(vector<int>{lower, upper});
            return missingIntervals;
        }
    
        if (lower < nums[0]) {
            missingIntervals.push_back(vector<int>{lower, nums[0] - 1});
        }
    
        for (int i = 0; i < len - 1; i++) {
            if (nums[i + 1] - nums[i] <= 1) {
                continue;
            }
            missingIntervals.push_back(vector<int>{nums[i] + 1, nums[i + 1] - 1});
        }
    
        if (upper > nums[len - 1]) {
            missingIntervals.push_back(vector<int>{nums[len - 1] + 1, upper});
        }
    
        return missingIntervals;
    }
};

The provided C++ solution aims to identify missing ranges within a given array of integers, where the missing ranges are those intervals that do not include any elements from the array but fall within a specified range from lower to upper.

• At the beginning, the solution defines a method getMissingIntervals that takes an array of integers nums and two integers lower and upper.
• The method first determines the length of the nums array. If nums is empty, it immediately returns the interval from lower to upper as the missing range.
• If the array isn't empty and the lower bound is less than the first element of the array, it includes the range from lower to one less than the first element as a missing interval.
• The solution iterates through the array and checks consecutive elements. If the gap between two consecutive numbers is greater than 1, the interval between these numbers minus the boundaries is identified as missing and added to the result list.
• After iterating through the array, if upper is greater than the last element, it adds the range from one more than the last element in nums to upper as a missing range.
• Finally, the method returns a list of vectors, each representing a missing range.

This function provides an efficient way to identify and return all missing numerical ranges between given boundaries, considering gaps within a sorted array of numbers.

class Solution {
    public List<List<Integer>> getMissingRanges(int[] values, int start, int end) {
        List<List<Integer>> ranges = new ArrayList<>();
        int size = values.length;
    
        if (size == 0) {
            ranges.add(Arrays.asList(start, end));
            return ranges;
        }
    
        if (start < values[0]) {
            ranges.add(Arrays.asList(start, values[0] - 1));
        }
    
        for (int i = 0; i < size - 1; i++) {
            if (values[i + 1] - values[i] > 1) {
                ranges.add(Arrays.asList(values[i] + 1, values[i + 1] - 1));
            }
        }
    
        if (end > values[size - 1]) {
            ranges.add(Arrays.asList(values[size - 1] + 1, end));
        }
    
        return ranges;
    }
}

The Java solution presented focuses on identifying and listing all missing ranges in a sorted array of integers. Given two integers, start and end, which define the inclusive boundaries of a range, the function determines all the sub-ranges that are not covered by the integers in the values array. Here's how the solution operates:

• A new ArrayList of type List<List<Integer>> named ranges is created to store the missing ranges.
• The function first checks if the values array is empty. If true, it adds the complete range from start to end to ranges since no numbers are present to create any sub-ranges.
• If the values array is not empty, the function assesses the space before the first element of values:
  • If the start value is less than the first value in values, then the range from start to values[0]-1 is added to ranges.
• A loop then iterates over the values array (up to the second last element) checking for gaps between consecutive elements:
  • If the difference between the current element and the next is greater than 1, it identifies a missing range which is then added to ranges.
• Post loop, the function examines the difference between the last element of values and the end:
  • If the last value in values is less than end, the range from values[size - 1] + 1 to end is added to the ranges.
• Finally, the ranges list containing all identified missing sub-ranges is returned.

This solution is optimal and concise, providing a clear method to identify all gaps within any given range, accommodating for arrays of any size including empty arrays. It utilizes basic array manipulation and conditional checks to determine the presence of gaps efficiently.

struct Interval {
    int low, high;
};
    
struct Interval** detectMissingIntervals(int* elements, int elementCount, int lowBound, int highBound,
                                         int* sizeOfReturn, int** sizeColumns) {
    int totalElements = elementCount;
    
    struct Interval** missingIntervals = malloc((totalElements + 2) * sizeof(struct Interval*));
    *sizeColumns = malloc((totalElements + 2) * sizeof(int));
    *sizeOfReturn = 0;
    
    if (totalElements == 0) {
        missingIntervals[*sizeOfReturn] = malloc(2 * sizeof(int));
        missingIntervals[*sizeOfReturn]->low = lowBound;
        missingIntervals[*sizeOfReturn]->high = highBound;
        (*sizeColumns)[*sizeOfReturn] = 2;
        (*sizeOfReturn)++;
        return missingIntervals;
    }
    
    if (lowBound < elements[0]) {
        missingIntervals[*sizeOfReturn] = malloc(2 * sizeof(int));
        missingIntervals[*sizeOfReturn]->low = lowBound;
        missingIntervals[*sizeOfReturn]->high = elements[0] - 1;
        (*sizeColumns)[*sizeOfReturn] = 2;
        (*sizeOfReturn)++;
    }
    
    for (int idx = 0; idx < totalElements - 1; idx++) {
        if (elements[idx + 1] - elements[idx] <= 1) {
            continue;
        }
        missingIntervals[*sizeOfReturn] = malloc(2 * sizeof(int));
        missingIntervals[*sizeOfReturn]->low = elements[idx] + 1;
        missingIntervals[*sizeOfReturn]->high = elements[idx + 1] - 1;
        (*sizeColumns)[*sizeOfReturn] = 2;
        (*sizeOfReturn)++;
    }
    
    if (highBound > elements[totalElements - 1]) {
        missingIntervals[*sizeOfReturn] = malloc(2 * sizeof(int));
        missingIntervals[*sizeOfReturn]->low = elements[totalElements - 1] + 1;
        missingIntervals[*sizeOfReturn]->high = highBound;
        (*sizeColumns)[*sizeOfReturn] = 2;
        (*sizeOfReturn)++;
    }
    
    return missingIntervals;
}

The provided C code defines a function designed to identify missing ranges in a sequence of numbers. Here's how it works:

• Function Definition: detectMissingIntervals takes several parameters: a pointer to an elements array of integers, an elementCount indicating the number of elements, boundary values lowBound and highBound, pointers sizeOfReturn and sizeColumns to hold the size of the resulting array and its columns, respectively.

• Memory Allocation: Initializes an array of Interval structures to store the missing intervals and the sizes of columns in the result.

• Handling Empty Element List: If no elements are provided, the function sets the missing interval to the entire range from lowBound to highBound and updates the return size.

• Finding Missing Intervals:

  • Checks if there are any gaps before the first element and after the last element in the sequence.
  • Iterates through the elements array to determine gaps between consecutive elements that are larger than one, thereby identifying missing intervals.

• Memory Allocation For Each Missing Interval: For each identified interval, it allocates memory and assigns the interval's lower and upper bounds before updating the return size.

• Return the Result: Returns the dynamically allocated array of intervals, allowing the caller to access the missing ranges.

Keep in mind that the memory allocated for storing these intervals and the return size information needs to be managed (i.e., freed) externally to avoid memory leaks.ing or adding debug logs at key points in the code can help trace the value changes and behavior, crucial for validating modifications or enhancements.

var calculateMissingRanges = function (values, start, end) {
    let length = values.length;
    let missing = [];
    if (length === 0) {
        missing.push([start, end]);
        return missing;
    }
    
    // Filling missed ranges from the low boundary to the first element
    if (start < values[0]) {
        missing.push([start, values[0] - 1]);
    }
    
    // Missing ranges between consecutive elements
    for (let index = 0; index < length - 1; index++) {
        if (values[index + 1] - values[index] <= 1) {
            continue;
        }
        missing.push([values[index] + 1, values[index + 1] - 1]);
    }
    
    // Missing numbers after the last element till the upper boundary
    if (end > values[length - 1]) {
        missing.push([values[length - 1] + 1, end]);
    }
    
    return missing;
};

The provided JavaScript function, calculateMissingRanges, efficiently finds the missing ranges of numbers between a specified start and end, against a list of sorted values. Here's a concise understanding of how it operates:

• First, it initializes a variable, missing, to store the resultant ranges of missing numbers.
• It checks if the values array is empty. If so, the entire range from start to end is missing, and it returns this range.
• It handles the scenario where the range starts before the first element in the values array by adding the range from start to one less than the first element in values.
• Next, it loops through the values array, examining gaps between consecutive numbers that are greater than one, signaling a missing range. It adds these ranges to the missing array.
• It ends by checking if there are missing numbers after the last element up to the end. If so, it adds this range from one greater than the last number in values to end.
• The function returns the list of missing ranges.

This implementation offers a straightforward approach to determine the absent sequences in a given numerical interval, making it suitable for various applications that require such numerical gap analyses.

class Solution:
    def findGaps(self, numbers: List[int], low: int, high: int) -> List[List[int]]:
        length = len(numbers)
        gaps = []
        if length == 0:
            gaps.append([low, high])
            return gaps
    
        # Handle the gap before the first number
        if low < numbers[0]:
            gaps.append([low, numbers[0] - 1])
    
        # Handle gaps between numbers in the array
        for index in range(length - 1):
            if numbers[index + 1] - numbers[index] > 1:
                gaps.append([numbers[index] + 1, numbers[index + 1] - 1])
    
        # Handle the gap after the last number
        if high > numbers[-1]:
            gaps.append([numbers[-1] + 1, high])
    
        return gaps

The solution provided involves finding the missing ranges in a given sorted list of integers, denoted by the input parameters numbers, low, and high. The function findGaps implemented in Python ensures you identify those missing sequences efficiently.

• Start by initializing a list gaps to store the missing ranges as sublists.
• Check if the numbers list is empty; if so, the entire range from low to high is missing, and this range is appended to gaps before returning it.
• Address the potential gap that might exist before the first number in the list. If the lower bound, low, is less than the first element of numbers, append the range from low to numbers[0] - 1 to gaps.
• Iterate through the given numbers to find and append gaps between consecutive numbers. Specifically, for each pair of consecutive elements where the difference is greater than one, append the range starting from numbers[index] + 1 to numbers[index + 1] - 1 to gaps.
• Identify and append the missing range after the last number if necessary. If the upper bound, high, is greater than the last element of numbers, append the range from numbers[-1] + 1 to high to gaps.
• Return the completed gaps list that now contains all missing ranges.

By following these steps, the function effectively fills the gaps with each missing range, thus providing a full view of all absent sequences between the specified bounds.