Tech It Yourself
tháng 1 20, 2020
0
An array A consisting of N integers is given. A triplet (P, Q, R) is triangular if it is possible to build a triangle with sides of lengths A[P], A[Q] and A[R]. In other words, 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] = 12
There are four triangular triplets that can be constructed from elements of this array, namely (0, 2, 4), (0, 2, 5), (0, 4, 5), and (2, 4, 5).
Write a function:
int solution(vector<int> &A);
that, given an array A consisting of N integers, returns the number of triangular triplets in this array.
For example, given array A such that:
A[0] = 10 A[1] = 2 A[2] = 5
A[3] = 1 A[4] = 8 A[5] = 12
the function should return 4, as explained above.
Write an efficient algorithm for the following assumptions:
N is an integer within the range [0..1,000];
each element of array A is an integer within the range [1..1,000,000,000].
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] = 12
There are four triangular triplets that can be constructed from elements of this array, namely (0, 2, 4), (0, 2, 5), (0, 4, 5), and (2, 4, 5).
Write a function:
int solution(vector<int> &A);
that, given an array A consisting of N integers, returns the number of triangular triplets in this array.
For example, given array A such that:
A[0] = 10 A[1] = 2 A[2] = 5
A[3] = 1 A[4] = 8 A[5] = 12
the function should return 4, as explained above.
Write an efficient algorithm for the following assumptions:
N is an integer within the range [0..1,000];
each element of array A is an integer within the range [1..1,000,000,000].
#include <algorithm> int solution(vector<int> &A) { // write your code in C++11 (g++ 4.8.2) int s = int(A.size()); if (s < 3) return 0; sort(A.begin(), A.end()); if (A[0] + A[1] > A[s - 1]) { return s * (s - 1) * (s -2) / 6; } int num = 0; int start = 0; int end = 0; for (int i = 0; i < s -2; i++) { for (int j = i + 1; j < s - 1; j++) { for (int k = j + 1; k < s; k++) { if (A[i] + A[j] > A[k]) { num++; } else { break; } } } } return num; }
