문제
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.
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나의 풀이
풀이 1 --> 50% (성능 이슈)
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) {
return -1;
}
}
}
}
}
return count;
}
}풀이 2 --> 100%
class Solution {
public int solution(int[] A) {
int result = 0;
int sum = 0;
for (int i = 0; i < A.length; i++) {
if (A[i] == 0) {
sum++;
} else {
result += sum;
if (result > 1000000000) {
return -1;
}
}
}
return result;
}
}