LeetCode 352: Data Stream as Disjoint Intervals Solution

Master LeetCode problem 352 (Data Stream as Disjoint Intervals), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

352. Data Stream as Disjoint Intervals

Hard

Binary Search Design Ordered Set

Problem Explanation

Explanation

To solve this problem, we can maintain a list of disjoint intervals as they are added to the stream. Whenever a new number is added, we check if it can be merged with any existing intervals. We then merge intervals accordingly and maintain the intervals in sorted order based on their start values.

Algorithm:

Initialize an empty list to store the intervals.
When a new number is added, iterate through the list to check if it can be merged with any existing intervals.
If the number can be merged, update the interval accordingly.
If the number cannot be merged with any existing interval, add a new interval to the list.
Sort the intervals based on their start values.

Time Complexity:

Adding a number takes O(n) time, where n is the number of intervals.
Merging intervals and sorting the intervals take O(nlogn) time.
Overall, the time complexity is O(nlogn).

Space Complexity:

The space complexity is O(n) to store the intervals.

Solution Code

import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;

class SummaryRanges {
    List<int[]> intervals;

    public SummaryRanges() {
        intervals = new ArrayList<>();
    }

    public void addNum(int val) {
        int[] newInterval = {val, val};
        List<int[]> newIntervals = new ArrayList<>();
        boolean inserted = false;

        for (int[] interval : intervals) {
            if (interval[1] + 1 < newInterval[0]) {
                newIntervals.add(interval);
            } else if (interval[0] - 1 > newInterval[1]) {
                if (!inserted) {
                    newIntervals.add(newInterval);
                    inserted = true;
                }
                newIntervals.add(interval);
                newInterval = interval;
            } else {
                newInterval[0] = Math.min(newInterval[0], interval[0]);
                newInterval[1] = Math.max(newInterval[1], interval[1]);
            }
        }

        if (!inserted) {
            newIntervals.add(newInterval);
        }

        intervals = newIntervals;
    }

    public int[][] getIntervals() {
        return intervals.toArray(new int[intervals.size()][]);
    }
}

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