Sunday, October 30, 2016

LeetCode 436 - Find Right Interval


http://bookshadow.com/weblog/2016/10/30/leetcode-find-right-interval/
Given a set of intervals, for each of the interval i, check if there exists an interval j whose start point is bigger than or equal to the end point of the interval i, which can be called that j is on the "right" of i.
For any interval i, you need to store the minimum interval j's index, which means that the interval j has the minimum start point to build the "right" relationship for interval i. If the interval j doesn't exist, store -1 for the interval i. Finally, you need output the stored value of each interval as an array.
Note:
  1. You may assume the interval's end point is always bigger than its start point.
  2. You may assume none of these intervals have the same start point.
Example 1:
Input: [ [1,2] ]

Output: [-1]

Explanation: There is only one interval in the collection, so it outputs -1.
Example 2:
Input: [ [3,4], [2,3], [1,2] ]

Output: [-1, 0, 1]

Explanation: There is no satisfied "right" interval for [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point;
For [1,2], the interval [2,3] has minimum-"right" start point.
Example 3:
Input: [ [1,4], [2,3], [3,4] ]

Output: [-1, 2, -1]

Explanation: There is no satisfied "right" interval for [1,4] and [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point.
X. Use TreeMap to store intervals
Use TreeMap<Integer, PriorityQueue<Integer|Interval>> to handle duplicate start
https://discuss.leetcode.com/topic/65817/java-clear-o-n-logn-solution-based-on-treemap
    public int[] findRightInterval(Interval[] intervals) {
        int[] result = new int[intervals.length];
        java.util.NavigableMap<Integer, Integer> intervalMap = new TreeMap<>();
        
        for (int i = 0; i < intervals.length; ++i) {
            intervalMap.put(intervals[i].start, i);    
        }
        
        for (int i = 0; i < intervals.length; ++i) {
            Map.Entry<Integer, Integer> entry = intervalMap.ceilingEntry(intervals[i].end);
            result[i] = (entry != null) ? entry.getValue() : -1;
        }
        
        return result;
    }
- When there is duplicate start
Given that, you are using a treemap, which derived from Map. if you put the same key with a different value it will replace the original one with the newer one. In such case, if you have case like
[[4,6],[1,2],[4,8]]
the correct answer should be
[-1,0,-1]
However, the answer you provide will give
[-1,2,1]
Because in the treeMap the [4,8] will store as 4->2 ,which replace the
previously inserted [4,6] stored as 4->0 for the Map property.
Anyway, you guys show me how to use treemap, still be a brilliant solution.
Updated:
Here attached my edited code which can fix this problem while passing all the test case.
  public int[] findRightInterval(Interval[] intervals) {
         TreeMap<Integer, PriorityQueue<Integer>> map = new TreeMap<>();
         int[] res = new int[intervals.length];
         for(int i=0;i<intervals.length;i++){ 
          if(map.get(intervals[i].start)==null){
           PriorityQueue<Integer> pq=new PriorityQueue<>();
           pq.add(i);
           map.put(intervals[i].start, pq);
          }
          else{
           map.get(intervals[i].start).add(i);
          }
         }
         for(int i=0;i<intervals.length;i++) {
             Integer key = map.ceilingKey(intervals[i].end);
             res[i] = key!=null ?map.get(key).peek() : -1;
         }
         return res;
     }

X. 排序(Sort)+ 二分查找(Binary Search) 按照区间起点排序,然后二分查找即可。
https://discuss.leetcode.com/topic/67399/java-concise-binary-search
http://www.cnblogs.com/javanerd/p/6061042.html
把这些interval按照start从小到大排序,然后对每一个interval用其end去在排好序的队列里面做二分查找,找到符合要求的一个interval
public int[] findRightInterval(Interval[] intervals){
        Interval[] sortedIntervals = Arrays.copyOf(intervals,intervals.length);
        Arrays.sort(sortedIntervals,(o1, o2) -> o1.start - o2.start);
        int[] result = new int[intervals.length];
        for (int i = 0; i < intervals.length; i++) {
            Interval current = intervals[i];
            int insertIndex = Arrays.binarySearch(sortedIntervals, current, (o1, o2) -> o1.start - o2.end);
            if (insertIndex < 0){
                insertIndex = -insertIndex - 1;
            }
            if (insertIndex == intervals.length){
                result[i] = -1;
            }else {
                Interval match = sortedIntervals[insertIndex];
                for (int j = 0; j < intervals.length; j++){// bad....
                    if (i != j && match.start == intervals[j].start && match.end == intervals[j].end){
                       // System.out.println(",old index:"+j);
                        result[i] = j;
                    }
                }
            }

        }
        return result;
    }

If we are not allowed to use TreeMap:
  1. Sort starts
  2. For each end, find leftmost start using binary search
  3. To get the original index, we need a map
public int[] findRightInterval(Interval[] intervals) {
    Map<Integer, Integer> map = new HashMap<>();
    List<Integer> starts = new ArrayList<>();
    for (int i = 0; i < intervals.length; i++) {
        map.put(intervals[i].start, i);
        starts.add(intervals[i].start);
    }
    
    Collections.sort(starts);
    int[] res = new int[intervals.length];
    for (int i = 0; i < intervals.length; i++) {
        int end = intervals[i].end;
        int start = binarySearch(starts, end);
        if (start < end) {
            res[i] = -1;
        } else {
            res[i] = map.get(start);
        }
    }
    return res;
}

public int binarySearch(List<Integer> list, int x) {
    int left = 0, right = list.size() - 1;
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (list.get(mid) < x) { 
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    return list.get(left);
}
https://discuss.leetcode.com/topic/65641/commented-java-o-n-logn-solution-sort-binary-search
Time compexity: n*log(n)
n*log(n) for sorting
log(n) for binary search X n times is n*log(n)
Space complexity: n
n for auxilliary array
Algorithm:
  1. Clone intervals and update end with index.
  2. Sort clone-intervals by start
  3. Iterate over each interval and find the right by binary searching the clone-intervals.
  4. If found, shove the end i.e., the original index of the right interval from clone-intervals into the output array.
public int[] findRightInterval(Interval[] intervals) {
        
        int n;
        // boundary case
        if (intervals == null || (n = intervals.length) == 0) return new int[]{};
        
        // output
        int[] res = new int[intervals.length];
        // auxilliary array to store sorted intervals
        Interval[] sintervals = new Interval[n];
        
        // sintervals don't have any use of 'end', so let's use it for tracking original index
        for (int i = 0; i < n; ++i) {
            sintervals[i] = new Interval(intervals[i].start, i);
        }
        
        // sort
        Arrays.sort(sintervals, (a, b)->a.start-b.start);
        
        int i = 0;
        for (; i < n; ++i) {
            int key = intervals[i].end;
            // binary search in sintervals for key
            int l = 0, r = n - 1;
            int right = -1;
            while (l <= r) {
                int m = l + (r - l) / 2;
                if (sintervals[m].start == key) {
                    right = sintervals[m].end; // original index is stored in end
                    break;
                } else if (sintervals[m].start < key) {
                    l = m + 1;
                } else {
                    r = m - 1;
                }
            }
            
            // if we haven't found the key, try looking for 'start' that's just greater
            if ((right == -1) && (l < n) && (sintervals[l].start > key)) {
                right = sintervals[l].end; // original index is stored in end
            }
            
            res[i] = right;
        }
        
        return res;
    }
X.
https://discuss.leetcode.com/topic/65585/java-sweep-line-solution-o-nlogn
  1. wrapper class: Point
    1. value
    2. flag: 1 indicates start, 2 indicates end
    3. index: original pos in intervals array
    4. Comparable: sort by value ascending, end in front of start if they have same value.
  1. Iterate intervals array and fill a points list, then sort it
  2. Iterate points list, since the sequence will be "order by position, and end will come before start".
    1. whenever meet a end point, keep a list(prevIdxs) before next start, save original index of curr interval to the list.
    2. whenever meet a start point, this start point is the right interval to the intervals in the list (prevIdxs). Take out each index in it and update to result.
class Point implements Comparable<Point>{
    int val;
    int flag; //1 start, 0 end
    int index;
    public Point(int val, int flag, int index) {
        this.val = val;
        this.flag = flag;
        this.index = index;
    }
    public int compareTo(Point o) {
        if (this.val == o.val) return this.flag - o.flag; //end in front of start
        return this.val - o.val;
    }
}
public int[] findRightInterval(Interval[] intervals) {
    if (intervals == null || intervals.length == 0) return new int[]{};
    
    int[] res = new int[intervals.length];
    Arrays.fill(res, -1);
    
    List<Point> points = new ArrayList<>();
    for (int i = 0; i < intervals.length; i++) {
        points.add(new Point(intervals[i].start, 1, i));
        points.add(new Point(intervals[i].end, 0, i));
    }
    
    Collections.sort(points);
    
    int prevEnd = 0;
    List<Integer> prevIdxs = new ArrayList<>();
    
    for (Point point: points) {
        if (point.flag == 1) {
                for (Integer prevIdx: prevIdxs) {
                   res[prevIdx] = point.index; 
                }
                prevIdxs = new ArrayList<>();
        } else {
            prevEnd = point.val;
            prevIdxs.add(point.index);
        }
    }
    
    return res;
}
http://www.cnblogs.com/grandyang/p/6018581.html

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