O(nlogn) Python solution, binary search, easy to understand


  • 1
    K
    def findLongestChain(self, pairs):
        """
        :type pairs: List[List[int]]
        :rtype: int
        """
        # sort by x for pairs (x1, y1), (x2, y2), (x3, y3)...
        pairs.sort()
            
        # min_end_y[i] is the ending tuple minimum y of length=i chain
        min_end_y = [float('inf')] * len(pairs)
        for x, y in pairs:
            # since (a, b) can chain (c, d) iff b < c, use bisect_left
            i = bisect.bisect_left(min_end_y, x)
            # greedy method, for the same length chain, the smaller y the better
            min_end_y[i] = min(min_end_y[i], y)  
        
        return bisect.bisect_left(min_end_y, float('inf'))
    

    See also https://discuss.leetcode.com/topic/28814/another-o-n-log-n-python


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