LC57 Insert Interval

class Solution:
    def insert(self, A: List[List[int]], B: List[int]) -> List[List[int]]:
        """
        main idea: for non-overlap intevals, if interval is smaller than B, updating pointer to be inserted; if interval is bigger than B, just push it into res;    ===  for overlaped intevals, keep updating longest inteval overlap with B, update B
        after iteration A is complete, inset target index for findal B 
        tc O(2N) sc O(1)
        """
        res = [] 
        n = len(A)
        if not A:
            return [B]
        cur = 0
        res = []
        for i,pair in enumerate(A):
            s,e = pair
            # non-overlap (1) end < new_start (2) start > new_end 
            if e < B[0] :
                res.append(pair)
                cur += 1 
            elif s > B[1]:
                res.append(pair)
            else:
                B = [min(s,B[0]),max(e,B[1])]
        res.insert(cur,B)
        return res

class Solution:
    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
        """ time O(N) space: O(1)  
        main idea: use 3 loops to seperate different non-overlap inteverse
        edge case:  intvs is empty 
        case 1: intv end < new_start 
        case_2: intv_start <= new_end, note intv_start == new_end will meet overlap condition, don missout 
        case_3: intv_start > new_start 
        """
        if not A:
            return [B]
        n,i = len(A),0
        res = []
        while i < n and A[i][1] < B[0]:
            res.append(A[i])
            i += 1
            
        while i < n and A[i][0] <= B[1]:# and B[0]<=A[i][1]: # cross check if not necessary since we are extending B with overlapping inteval insead of finding intersections 
            B = [min(B[0],A[i][0]),max(A[i][1],B[1])]
            i += 1
        # note, keep updating B until there is no overlapping, then put final B into res 
        res.append(B)
        
        while i < n and A[i][0] > B[1]:
            res.append(A[i])
            i += 1 
        return res