🚀Day 4. Overlapping Intervals 🧠

The problem can be found at the following link: Problem Link

💡 Problem Description:

Given an array of intervals arr[][], where each interval arr[i] = [starti, endi] represents a range of integers, merge all overlapping intervals.

The output should return an array of the merged intervals sorted by their starting points.

🔍 Example Walkthrough:

Input: arr = [[1,3],[2,4],[6,8],[9,10]] Output: [[1,4],[6,8],[9,10]]

Explanation:

• [1,3] and [2,4] overlap, so they merge into [1,4].

Input: arr = [[6,8],[1,9],[2,4],[4,7]] Output: [[1,9]]

Explanation:

• All intervals overlap with [1,9], so they merge into [1,9].

Constraints:

• $1 ≤ arr.size() ≤ 10^5$

• $0 ≤ starti ≤ endi ≤ 10^5$

🎯 My Approach:

1. Sorting Intervals:

   • The first step is to sort the intervals by their starting points.

   • This ensures that we can linearly process intervals and merge overlapping ones efficiently.

2. Merging Logic:

   • Maintain a result list and iterate through the sorted intervals.

   • For each interval:

     • If it overlaps with the last interval in the result, merge them by updating the end point.

     • Otherwise, add the interval to the result as it doesn't overlap with any previous intervals.

3. Final Output:

   • The resulting intervals are guaranteed to be sorted and non-overlapping.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(n log n), where n is the size of the array. The sorting step dominates the time complexity.

• Expected Auxiliary Space Complexity: O(n), as we use additional space for storing the merged intervals.

📝 Solution Code

Code (C)

int compare(const void* a, const void* b) {
    return ((Interval*)a)->start - ((Interval*)b)->start;
}

int mergeOverlap(Interval arr[], int n, Interval result[]) {
    if (n == 0) return 0;

    qsort(arr, n, sizeof(Interval), compare);

    result[0] = arr[0];
    int index = 0;

    for (int i = 1; i < n; i++) {
        if (arr[i].start <= result[index].end) {
            result[index].end = result[index].end > arr[i].end ? result[index].end : arr[i].end;
        } else {
            index++;
            result[index] = arr[i];
        }
    }
    return index + 1;
}

Code (Cpp)

class Solution {
public:
    vector<vector<int>> mergeOverlap(vector<vector<int>>& intervals) {
        if (intervals.empty())
            return {};

        sort(intervals.begin(), intervals.end());

        vector<vector<int>> merged;
        merged.push_back(intervals[0]);

        for (int i = 1; i < intervals.size(); i++) {
            auto& last = merged.back();

            if (intervals[i][0] <= last[1]) {
                last[1] = max(last[1], intervals[i][1]);
            } else {
                merged.push_back(intervals[i]);
            }
        }

        return merged;
    }
};

Code (Java)

class Solution {
    public List<int[]> mergeOverlap(int[][] intervals) {
        List<int[]> merged = new ArrayList<>();
        if (intervals.length == 0) return merged;

        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);

        int[] current = intervals[0];
        merged.add(current);

        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] <= current[1]) {
                current[1] = Math.max(current[1], intervals[i][1]);
            } else {
                current = intervals[i];
                merged.add(current);
            }
        }

        return merged;
    }
}

Code (Python)

class Solution:
    def mergeOverlap(self, intervals):
        if not intervals:
            return []

        intervals.sort(key=lambda x: x[0])
        merged = [intervals[0]]

        for interval in intervals[1:]:
            last = merged[-1]
            if interval[0] <= last[1]:
                last[1] = max(last[1], interval[1])
            else:
                merged.append(interval)

        return merged

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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