Lesson 5 - Prefix Sum

A non-empty 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:

class Solution { public int solution(int[] A); }

that, given a non-empty 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.

Write an efficient algorithm for the following assumptions:

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.

---

내 답안

역시나 요구사항에 충실하게 구현 O(N²)으로 퍼포먼스에서 모두 실패하였다.

class Solution {
    public int solution(int[] A) {
        int count = 0;

        for(int i=0; i<A.length; i++) {
            if (A[i] == 0) {
                for(int j=(i+1); j<A.length; j++) {
                    if (A[j] == 1) {
                        count++;
                    }
                }
            }
        }

        if (count > 1000000000) {
            count = -1;
        }

        return count;
    }
}

---

베스트 답안

0숫자를 따로 카운트해놓고 마주치는 1만큼 누적해서 합산하면 되는것이였다.

index로 보여주자면

index 0 => 0 ( 0숫자 카운트 1, 1숫자 카운트 0)
index 1 => 1 ( 0 숫자 카운트 1, 1숫자 카운트 1)
index 2 => 0 ( 0 숫자 카운트 2, 1숫자 카운트 1)
index 3 => 1 ( 0 숫자 카운트 2, 1숫자 카운트 3)
index 4 => 1 ( 0 숫자 카운트 2, 1숫자 카운트 5)

기본적으로 숫자가 각각 어떻게 증가하는지 책같은데에 적어봤다면 규칙을 찾을 수 있었던거 같다. 노력해야 겠다.

class Solution {
	public int solution(int[] A) {
        int pairs = 0;
        int count_zero = 0;

        for (int n : A) {
            if (n == 0) {
                count_zero++;
            } else if (n == 1) {
                pairs += count_zero;
            }

            if (pairs > 1000000000) {
                return -1;
            }
        }

        return pairs;
    }
}