[Leetcode] Binary search--436. Find Right Interval

Posted 2020-09-21 安新

 

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.

 

Solution:

 

1.  1st thing I come up with is to use binary search, but it has LTE problem

 

  (1) maintain the index of each interval by create a tuple

  (2) sort the list by the start point

  (3) iterate each interval, find the end point endElement

  (4) binary search the right most insertion position of each endElement with each rest interval start point, and then find the index

 

        def binarySearchStartPoint(targetLst, ele):
            if len(targetLst) == 1:            #only one element
                if targetLst[0][0].start >= ele:
                    return targetLst[0][1]
                else: 
                    return -1
            l = 0
            h = len(targetLst)-1
            while (l <= h):
                mid = (l + h)/2
                if ele == targetLst[mid][0].start:
                    return targetLst[mid][1]                    #index
                elif ele < targetLst[mid][0].start:
                    h = mid - 1
                else:
                    l = mid + 1
            #print ( ‘ddd : ‘, len(targetLst), l)
            if l >= len(targetLst):
                return -1
            return targetLst[l][1]
                
        intersIndexLst = [(intl, i) for i, intl in enumerate(intervals)]

        intersIndexSortedLst = sorted(intersIndexLst, key = lambda k: k[0].start)
        i = 0
        ansLst = [0] * len(intersIndexSortedLst)
        
        while (i < len(intersIndexSortedLst)-1):
            endElement = intersIndexSortedLst[i][0].end
            targetLst = intersIndexSortedLst[i+1:]
            currInd = intersIndexSortedLst[i][1]
            indRight = binarySearchStartPoint(targetLst, endElement)
            #print (‘targetLst: ‘, endElement, indRight)
            ansLst[currInd] = indRight
            i += 1
        ansLst[intersIndexSortedLst[-1][1]] = -1          #the last emelement in intersIndexSortedLst
        return ansLst

 

 

2. 

I refer to other‘s solution, which makes the question so simple.
(1) only need to maintain the start point with a tuple
(2) bisect can be used in that way,
(3) after sorted, directly compare the end with the original interval list to find the insertion position (index)

 

       ansLst = []
        intersIndexLst = [(intl.start, i) for i, intl in enumerate(intervals)]
        intersIndexSortedLst = sorted(intersIndexLst)
        for intl in intervals:
            ind = bisect_left(intersIndexSortedLst, (intl.end, ))
            ansLst.append(intersIndexSortedLst[ind][1] if ind < len(intervals) else - 1)
        return ansLst

 

--reference:

https://discuss.leetcode.com/topic/65596/python-o-nlogn-short-solution-with-explanation