Java 2ms solution that beats 98%, binary search


  • 0
    Y

    There are three cases:

    1. when the newInterval is a subspan of one interval, no insertion is needed;
    2. when the newInterval is disjoint from all intervals, directly insert it without merge;
    3. insertion with merge; using binary search to locate the intervals that newInterval.start and newInterval.end can be merged into.
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        if (intervals == null || newInterval == null) {
            return intervals;
        }
        List<Interval> result = new ArrayList<Interval>();
        // binary search for the intervals
        int left = 0;
        int right = intervals.size() - 1;
        int mergeStart = -1; // the interval that can be merged with newInterval.start
        int mergeEnd = intervals.size(); // the interval that can be merged with newInterval.end
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (newInterval.start > intervals.get(mid).end) {
                left = mid + 1;
            } else if (newInterval.end < intervals.get(mid).start) {
                right = mid - 1;
            } else if (newInterval.start >= intervals.get(mid).start 
                    && newInterval.end <= intervals.get(mid).end) {
                // no insertion is needed
                return intervals;
            } else {
                mergeStart = mid;
                mergeEnd = mid;
                // find the right boundary
                while (mergeEnd + 1 < intervals.size() 
                    && newInterval.end >= intervals.get(mergeEnd + 1).start ) {
                    mergeEnd++;
                }
                // find the left boundary
                while (mergeStart - 1 >= 0 
                    && newInterval.start <= intervals.get(mergeStart - 1).end ) {
                    mergeStart--;
                }
                break;
            }
        }
        if (right == -1) { 
            // newInterval is smaller than the first interval
            mergeEnd = -1;
        } else if (left == intervals.size()) { 
            // newInterval is larger than the last interval
            mergeStart = intervals.size();
        } else if (left > right) { 
            // can be inserted between two intervals, no merge is needed
            intervals.add(left, newInterval);
            return intervals;
        }
    
        // three steps to populate result list
        // 1. add intervals before newInterval
        // 2. add newInterval, with possible merge
        // 3. add intervals after newInterval
        for (int i = 0; i < mergeStart; i++) {
            result.add(intervals.get(i));
        }
        if (mergeStart >= 0 && mergeStart < intervals.size()) {
            newInterval.start = Math.min(newInterval.start, intervals.get(mergeStart).start);
        }
        if (mergeEnd >= 0 && mergeEnd < intervals.size()) {
            newInterval.end = Math.max(newInterval.end, intervals.get(mergeEnd).end);
        }
        result.add(newInterval);
        for (int i = mergeEnd+1; i < intervals.size(); i++) {
            result.add(intervals.get(i));
        }
        return result;
    }

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