0352. Data Stream as Disjoint Intervals

https://leetcode.com/problems/data-stream-as-disjoint-intervals

Description

Given a data stream input of non-negative integers a1, a2, ..., an, summarize the numbers seen so far as a list of disjoint intervals.

Implement the SummaryRanges class:

• SummaryRanges() Initializes the object with an empty stream.

• void addNum(int val) Adds the integer val to the stream.

• int[][] getIntervals() Returns a summary of the integers in the stream currently as a list of disjoint intervals [starti, endi].

Example 1:

**Input**
["SummaryRanges", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals"]
[[], [1], [], [3], [], [7], [], [2], [], [6], []]
**Output**
[null, null, [[1, 1]], null, [[1, 1], [3, 3]], null, [[1, 1], [3, 3], [7, 7]], null, [[1, 3], [7, 7]], null, [[1, 3], [6, 7]]]
**Explanation**
SummaryRanges summaryRanges = new SummaryRanges();
summaryRanges.addNum(1);      // arr = [1]
summaryRanges.getIntervals(); // return [[1, 1]]
summaryRanges.addNum(3);      // arr = [1, 3]
summaryRanges.getIntervals(); // return [[1, 1], [3, 3]]
summaryRanges.addNum(7);      // arr = [1, 3, 7]
summaryRanges.getIntervals(); // return [[1, 1], [3, 3], [7, 7]]
summaryRanges.addNum(2);      // arr = [1, 2, 3, 7]
summaryRanges.getIntervals(); // return [[1, 3], [7, 7]]
summaryRanges.addNum(6);      // arr = [1, 2, 3, 6, 7]
summaryRanges.getIntervals(); // return [[1, 3], [6, 7]]

Constraints:

• 0 <= val <= 104

• At most 3 * 104 calls will be made to addNum and getIntervals.

Follow up: What if there are lots of merges and the number of disjoint intervals is small compared to the size of the data stream?

ac

class SummaryRanges {
    TreeMap<Integer, Integer> tmap;
    /** Initialize your data structure here. */
    public SummaryRanges() {
        tmap = new TreeMap<>();
    }
    
    public void addNum(int val) {
        Integer f = tmap.lowerKey(val); // floor
        Integer c = tmap.ceilingKey(val); // ceil
        // 4 situations
        if (f != null && tmap.get(f) + 1 == val && c != null && val + 1 == c) { // concatenate both
            tmap.put(f, tmap.get(c));
            tmap.remove(c);
        } else if (f != null && tmap.get(f) + 1 == val && (c == null || val + 1 < c)) { // concatenate floor
            tmap.put(f, val);
        } else if ((f == null || tmap.get(f) + 1 < val) && c != null && val + 1 == c) { // concatenate ceil
            tmap.put(val, tmap.get(c));
            tmap.remove(c);
        } else if ((f == null || tmap.get(f) + 1 < val) && (c == null || val + 1 < c)) { // independent
            tmap.put(val, val);
        }
    }
    
    public List<Interval> getIntervals() {
        List<Interval> res = new ArrayList<>();
        for (int k : tmap.keySet()) {
            res.add(new Interval(k, tmap.get(k)));
        }
        return res;
    }
}

/**
 * Your SummaryRanges object will be instantiated and called as such:
 * SummaryRanges obj = new SummaryRanges();
 * obj.addNum(val);
 * List<Interval> param_2 = obj.getIntervals();
 */

/*
1) TreeMap, key:start, value:end; 2) when adding, check floor(val) and ceil(val);  3) 5 situations: if floor.end + 1 = val and val+1 = ceil.start, concatenate and delete ceil. concatenate both, concatenate floor, concatenate ceil, independent, already in range can't add
*/