Tech It Yourself
tháng 1 20, 2020
0
We draw N discs on a plane. The discs are numbered from 0 to N − 1. An array A of N non-negative integers, specifying the radiuses of the discs, is given. The J-th disc is drawn with its center at (J, 0) and radius A[J].
We say that the J-th disc and K-th disc intersect if J ≠ K and the J-th and K-th discs have at least one common point (assuming that the discs contain their borders).
The figure below shows discs drawn for N = 6 and A as follows:
A[0] = 1
A[1] = 5
A[2] = 2
A[3] = 1
A[4] = 4
A[5] = 0
There are eleven (unordered) pairs of discs that intersect, namely:
discs 1 and 4 intersect, and both intersect with all the other discs;
disc 2 also intersects with discs 0 and 3.
Write a function:
int solution(vector<int> &A);
that, given an array A describing N discs as explained above, returns the number of (unordered) pairs of intersecting discs. The function should return −1 if the number of intersecting pairs exceeds 10,000,000.
Given array A shown above, the function should return 11, as explained above.
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 [0..2,147,483,647].
We say that the J-th disc and K-th disc intersect if J ≠ K and the J-th and K-th discs have at least one common point (assuming that the discs contain their borders).
The figure below shows discs drawn for N = 6 and A as follows:
A[0] = 1
A[1] = 5
A[2] = 2
A[3] = 1
A[4] = 4
A[5] = 0
There are eleven (unordered) pairs of discs that intersect, namely:
discs 1 and 4 intersect, and both intersect with all the other discs;
disc 2 also intersects with discs 0 and 3.
Write a function:
int solution(vector<int> &A);
that, given an array A describing N discs as explained above, returns the number of (unordered) pairs of intersecting discs. The function should return −1 if the number of intersecting pairs exceeds 10,000,000.
Given array A shown above, the function should return 11, as explained above.
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 [0..2,147,483,647].
#include <algorithm> int solution(vector<int>& A) { int len = A.size(); if (0 == len)return 0; vector<std::pair<long long, long long> > intervals; for (int i = 0; i < len; ++i) intervals.push_back(make_pair(i - (long long)A[i], i + (long long)A[i])); sort(intervals.begin(), intervals.end()); int l, h, m, count=0; for (int i = 0; i < len - 1; ++i) { l = i + 1; h = len - 1; while (l <= h) { m = l + (h - l) / 2; if (intervals[i].second >= intervals[m].first) l = m + 1; else h = m - 1; } count += l - 1 - i; if (count > 10000000) return -1; } return count; }
