Merge Intervals

class Solution {
public:
    vector<vector<int>> combine(vector<vector<int>>& segments) {
        sort(segments.begin(), segments.end());

        vector<vector<int>> combinedSegments;
        for (auto seg : segments) {
            if (combinedSegments.empty() || combinedSegments.back()[1] < seg[0]) {
                combinedSegments.push_back(seg);
            }
            else {
                combinedSegments.back()[1] = max(combinedSegments.back()[1], seg[1]);
            }
        }
        return combinedSegments;
    }
};

This C++ solution addresses the problem of merging overlapping intervals. The code defines a class Solution with a member function combine that accepts a vector of vector integers representing segments and returns a vector of combined intervals.

• Start by sorting the segments vector based on the starting times of intervals to organize them in ascending order.
• Initialize a new vector called combinedSegments to store the merged intervals.
• Iterate through the sorted segments:
  • If combinedSegments is empty or the last added interval does not overlap with the current interval (seg), add seg directly to combinedSegments.
  • If the last interval in combinedSegments overlaps with seg (the end of the last interval is greater than or equal to the start of seg), merge them by extending the end of the last interval in combinedSegments to the maximum end of both intervals.
• Finally, return combinedSegments containing the merged intervals.

By following this approach, the program efficiently merges all overlapping intervals into minimal disjoint intervals.

class Solution {
    public int[][] mergeIntervals(int[][] intervals) {
        Arrays.sort(intervals, (x, y) -> Integer.compare(x[0], y[0]));
        LinkedList<int[]> mergedList = new LinkedList<>();
        for (int[] range : intervals) {
            if (mergedList.isEmpty() || mergedList.getLast()[1] < range[0]) {
                mergedList.add(range);
            } else {
                mergedList.getLast()[1] = Math.max(mergedList.getLast()[1], range[1]);
            }
        }
        return mergedList.toArray(new int[mergedList.size()][]);
    }
}

The provided Java solution for merging overlapping intervals involves sorting the interval list and using a LinkedList to hold the merged intervals efficiently. Below is a comprehensive step-by-step approach to understand the solution:

1. Sort the array of intervals based on the starting times. This ensures that any intervals which might need to be merged are adjacent in the array, making it easier to merge them in a single pass.

2. Initialize a LinkedList named mergedList to store the merged intervals. LinkedList is preferred here for dynamic addition and efficient retrieval of the last element.

3. Iterate through each interval in the sorted array:

   • If mergedList is empty or if the last interval in mergedList does not overlap with the current interval (i.e., the end of the last interval in mergedList is less than the start of the current interval), simply add the current interval to mergedList.
   • If there is an overlap (the end of the last interval in mergedList is greater than or equal to the start of the current interval), merge the current interval with the last interval in mergedList. This is done by updating the end of the last interval in mergedList to be the maximum of its own end and the end of the current interval.

4. Convert mergedList back to a 2D array to match the expected return type of the function.

This method ensures an optimal approach to the merge intervals problem by reducing unnecessary comparisons and making efficient use of data structures to handle the intervals dynamically.

int minimum(int x, int y) { return x < y ? x : y; }
int maximum(int x, int y) { return x > y ? x : y; }
int customCompare(const void* x, const void* y) {
    int* ix = *(int**)x;
    int* iy = *(int**)y;
    return ix[0] == iy[0] ? iy[1] - ix[1] : ix[0] - iy[0];
}
int** combineIntervals(int** intervals, int size, const int* colSize,
                       int* outSize, int** outColSizes) {
    int count = size;
    if (count == 0) {
        *outSize = 0;
        return NULL;
    }
    qsort(intervals, count, sizeof(int*), customCompare);
    int** combined = malloc(sizeof(int*) * count);
    for (int i = 0; i < count; ++i) combined[i] = malloc(sizeof(int) * 2);
    memcpy(combined[0], intervals[0], 2 * sizeof(int));
    int idx = 0;  // track the last index in the combined array
    for (int i = 1; i < count; ++i) {
        int start = intervals[i][0], end = intervals[i][1];
        if (start <= combined[idx][1]) {
            combined[idx][1] = maximum(combined[idx][1], end);
        } else {
            memcpy(combined[++idx], intervals[i], 2 * sizeof(int));
        }
    }
    ++idx;
    *outSize = idx;
    *outColSizes = malloc(sizeof(int) * idx);
    for (int i = 0; i < idx; ++i) {
        (*outColSizes)[i] = 2;
    }
    return combined;
}

This summary explains the solution to the problem of merging overlapping intervals using C programming language.

The provided code defines several functions for handling, comparing, and merging intervals:

• minimum and maximum functions are utility functions used for returning the lesser and greater of two integers, respectively, ensuring correct interval boundaries during merging.

• customCompare function is defined to help in sorting the intervals. It makes use of the qsort standard function's capability by providing it a way to compare intervals primarily by their starting points. If starting points are the same, it sorts by ending points.

• combineIntervals is the principal function that handles the logic of merging intervals. It starts by handling the base case—returning NULL if no intervals are present. Intervals are then sorted using qsort with customCompare.

  After sorting, memory is allocated dynamically for storing merged intervals. The merging happens in a loop; if the current interval's start time is less than or equal to the previously stored interval's end time, the intervals overlap and are merged by adjusting the end time to the maximum of the two end times. If they do not overlap, the interval is added to the result as a new non-overlapping interval.

  The end of this process updates the output count and adjusts output column sizes to reflect pairs of integers (start and end times for each interval).

In summary, the solution involves sorting the intervals based on their start (and conditionally end) times, then merging overlapping intervals iteratively, storing the result in dynamically allocated memory, and finally outputting the merged intervals and their count. This method ensures efficient handling of potentially large sets of intervals with overlapping times.

var combineIntervals = function (ranges) {
    ranges.sort((first, second) => first[0] - second[0]);
    let combined = [];
    for (let range of ranges) {
        if (combined.length === 0 || combined[combined.length - 1][1] < range[0]) {
            combined.push(range);
        } else {
            combined[combined.length - 1][1] = Math.max(
                combined[combined.length - 1][1],
                range[1],
            );
        }
    }
    return combined;
};

The solution provided addresses the problem of merging overlapping intervals. The JavaScript function combineIntervals takes an array of intervals as an argument, where each interval is represented as a two-element array. The function merges all overlapping intervals and returns the merged intervals as an array. Here's how the solution works:

• The intervals are first sorted based on their starting points to simplify the merging process.
• Initialize an empty array combined to store the merged intervals.
• Iterate through each interval in the sorted array:
  • If the combined array is empty or if the last interval in combined does not overlap with the current interval, simply add the current interval to combined.
  • If there is an overlap, merge the current interval with the last interval in combined by updating the end point of the last interval to be the maximum of the end points of the last interval in combined and the current interval.
• Finally, return the combined array containing all the merged intervals.

This approach ensures that all overlapping intervals are merged into a minimal set of intervals.

class Solution:
    def merge_intervals(self, ranges: List[List[int]]) -> List[List[int]]:

        # First, sort the ranges based on the starting values of each interval.
        ranges.sort(key=lambda interval: interval[0])

        # Initialize an empty list to store the merged intervals
        combined_intervals = []
        for current_interval in ranges:
            # Check if the combined_intervals list is empty or if the last interval does not overlap with current one
            if not combined_intervals or combined_intervals[-1][1] < current_interval[0]:
                combined_intervals.append(current_interval)
            else:
                # There is an overlap, so we merge the current interval with the last one in the list
                combined_intervals[-1][1] = max(combined_intervals[-1][1], current_interval[1])

        return combined_intervals

The solution implements a function to merge overlapping intervals given a list of interval ranges. Follow these steps to understand the process executed by the Python code:

1. Sorts the intervals by their start times to ensure they are in order and easier to compare.

2. Initializes an empty list combined_intervals to store the merged intervals.

3. Iterates through the sorted ranges. For each current_interval:

   • Checks if combined_intervals is empty or if the last interval in combined_intervals does not overlap with current_interval. If true, appends current_interval to combined_intervals.
   • If there is an overlap (meaning the end of the last interval in combined_intervals is greater than or equal to the start of current_interval), merges the intervals by updating the end of the last interval in combined_intervals to be the maximum end value of both overlapping intervals.

4. Finally, the function returns the combined_intervals which contains all the merged intervals.

This approach ensures that all overlapping intervals are combined into single intervals, and non-overlapping intervals are appended as distinct entries. The resultant list provides merged ranges in a minimal set.