LC436 Find Right Interval

class Solution:
    def findRightInterval(self, intervals: List[List[int]]) -> List[int]:
        """
            step1,  check edge case if  length is less than 2, if yes, return -1 based on length 
            step2, initialize res array filled with -1 and length is same as intervals
            step3,  use tuple (interval) as key, index as val to save original index for later sort 
            step4, sort by starting time 
            step5, loop through sorted interval, find the earliest interval that meet inter[i][1] <= inter[j][0],
            if found, break current loop, find their index correspoindingly and update res array.
        """
        # time O(N^2)for 2 loop  space O(N) dictionary 
        if len(intervals) < 2:
            return [-1]*len(intervals)
        d = {}
        res = [-1] * len(intervals)
        for idx,inte in enumerate(intervals):
            d[tuple(inte)] = idx # here using tuple because python doesn't accept list as key since list is unhashable
        intervals.sort(key= lambda k: k[0])
        for i in range(len(intervals)-1):
            for j in range(i+1,len(intervals)):
                if intervals[i][1]<=intervals[j][0]:
                    res[d[tuple(intervals[i])]] = d[tuple(intervals[j])]
                    break
        return res