134 Gas Station

problem definition

gas = [1,2,3,4,5] cost = [3,4,5,1,2] -> 3

gas[i] is the number of gas cost[i] is the cost to off the position

if it is possible to pass all the ways in clockwise without out of gas, then return the start index but -1.

1st solution

for every single input, it checks the previous inputs

M + (M - 1) + (M - 2) … = M(M + 1)/2 = O(M^2)

def reconstructQueue(self, people: [[int]]) -> [[int]]:
    people.sort(key=lambda p: p[1], reverse=False)
    seq = []

    for hi, ki in people:
        i = 0
        temp_ki = ki

        while seq and i < len(seq):
            if seq[i][0] < hi:
                i += 1
            elif temp_ki:
                temp_ki -= 1
                i += 1
            else:
                break

        seq.insert(i, [hi, ki])

    return seq

 

optimized one

fill the value from the most highest height and the lowest number of people in front

def reconstructQueue(self, people: List[List[int]]) -> List[List[int]]:
    q = []
    for hi, ki in people:
        heapq.heappush(q, (-hi, ki))

    seq = []
    while q:
        hi, ki = heapq.heappop(q)
        seq.insert(ki, (-hi, ki))

    return seq

 

TC

print([[5,0],[7,0],[5,2],[6,1],[4,4],[7,1]] == Solution().reconstructQueue([[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]))