An array A consisting of N integers is given. A triplet (P, Q, R) is triangular if 0 ≤ P < Q < R < N and:
A[P] + A[Q] > A[R],
A[Q] + A[R] > A[P],
A[R] + A[P] > A[Q].
For example, consider array A such that:
A[0] = 10 A[1] = 2 A[2] = 5
A[3] = 1 A[4] = 8 A[5] = 20
Triplet (0, 2, 4) is triangular.
Write a function:
int solution(vector<int> &A);
that, given an array A consisting of N integers, returns 1 if there exists a triangular triplet for this array and returns 0 otherwise.
For example, given array A such that:
A[0] = 10 A[1] = 2 A[2] = 5
A[3] = 1 A[4] = 8 A[5] = 20
the function should return 1, as explained above. Given array A such that:
A[0] = 10 A[1] = 50 A[2] = 5
A[3] = 1
the function should return 0.
Write an efficient algorithm for the following assumptions:
N is an integer within the range [0..100,000];
each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].
A[P] + A[Q] > A[R],
A[Q] + A[R] > A[P],
A[R] + A[P] > A[Q].
For example, consider array A such that:
A[0] = 10 A[1] = 2 A[2] = 5
A[3] = 1 A[4] = 8 A[5] = 20
Triplet (0, 2, 4) is triangular.
Write a function:
int solution(vector<int> &A);
that, given an array A consisting of N integers, returns 1 if there exists a triangular triplet for this array and returns 0 otherwise.
For example, given array A such that:
A[0] = 10 A[1] = 2 A[2] = 5
A[3] = 1 A[4] = 8 A[5] = 20
the function should return 1, as explained above. Given array A such that:
A[0] = 10 A[1] = 50 A[2] = 5
A[3] = 1
the function should return 0.
Write an efficient algorithm for the following assumptions:
N is an integer within the range [0..100,000];
each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].
#include <algorithm> bool is_triangle(int64_t a, int64_t b, int64_t c) { return (a + b > c) && (b + c > a) && (c + a > b); } int solution(vector<int> &A) { // write your code in C++11 (g++ 4.8.2) if (A.size() < 3) return 0; sort(A.begin(), A.end()); int res = 0; for (int i = 0; i < int(A.size() - 2); i++) { if (is_triangle(A[i], A[i + 1], A[i + 2])) return 1; } return 0; }
Không có nhận xét nào:
Đăng nhận xét