[Lesson 5] PassingCars ( score : 90 )
by 브이담곰
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:
int solution(vector<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.
풀이
첫번째 제출 score : 90
// you can use includes, for example:
// #include <algorithm>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int solution(vector<int> &A) {
// Implement your solution here
int size = A.size();
int sum[100003];
int count = 0;
//prefix sum
sum[0] = A[0];
for(int i = 1; i < size; i++)
{
if(A[i] == 1)
{
sum[i] += (sum[i-1] + 1);
}
else
{
sum[i] += sum[i-1];
}
}
//counting
for(int i = 0; i < size; i++)
{
if(count > 1000000000) return -1;
if(A[i] == 0)
{
count += (sum[size-1] - sum[i]);
}
}
return count;
}
'Coding Test > Codility' 카테고리의 다른 글
[Lesson 6] Distinct (0) | 2024.02.21 |
---|---|
[Lesson 4] FrogRiverOne (0) | 2024.02.21 |
[Lesson 3] TapeEquilibrium (0) | 2024.02.21 |
[Lesson 3] PermMissingElem (0) | 2024.02.20 |
[Lesson 3] FrogJmp (0) | 2024.02.20 |
블로그의 정보
농담곰담곰이의곰담농
브이담곰