codility 5-2. PassingCars

|

문제출처

문제

A non-empty zero-indexed array A consisting of N integers is given. The consecutive elements of array A represent consecutive cars on a road. Array A contains only 0s and/or 1s: 0 represents a car traveling east, 1 represents a car traveling west. The goal is to count passing cars. We say that a pair of cars (P, Q), where 0 ≤ P < Q < N, is passing when P is traveling to the east and Q is traveling to the west. For example, consider array A such that: A[0] = 0 A[1] = 1 A[2] = 0 A[3] = 1 A[4] = 1 We have five pairs of passing cars: (0, 1), (0, 3), (0, 4), (2, 3), (2, 4). Write a function: def solution(A) that, given a non-empty zero-indexed array A of N integers, returns the number of pairs of passing cars. The function should return −1 if the number of pairs of passing cars exceeds 1,000,000,000. For example, given: A[0] = 0 A[1] = 1 A[2] = 0 A[3] = 1 A[4] = 1 the function should return 5, as explained above. Assume that: N is an integer within the range [1..100,000]; each element of array A is an integer that can have one of the following values: 0, 1. Complexity: expected worst-case time complexity is O(N); expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments). 

풀이코드 - O(n^2)

  • Detected time complexity: O(N)
  • 조합 가능한 (0,1) pair의 갯수를 찾는다
  • 조건
    • 0의 index < 1의 index
    • pair의 갯수가 100만이 넘으면 -1 리턴
    • 시간 복잡도 O(N)
def solution(A): index_zero = [i for i, x in enumerate(A) if x == 0] result = 0 for i in index_zero: result += A[i+1:].count(1) if result > 1000000000: return -1 return result 

다른사람 풀이

  • 리스트를 순회하면서, 0의 갯수를 누적해서 더하고, 1을 만나면 누적 값을 result에 합산한다
def solution(A): result = 0 count_zero = 0 for i in A: if i == 1 and count_zero == 0: continue elif i == 0: count_zero += 1 elif i == 1: result += count_zero if result > 1000000000: return -1 return result