On a 6 sided dice, what is the probability of rolling a number less than 3 and 6. P(n<3 and 6) 3/6
1/18
1/2
1/12​

Step-by-step explanation:

Iterate the tags string and populate the containers to hold the tags and distance from (0, 0) (sqrt(X[i] * X[i] + Y[i] * Y[i]))Find the max distance of repeated tags while calculating distance of all tag points and keep the max distance so far.Now max distance among from the repeated tags which would be condition whether the points can be included in the circle or not. Iterate the X and Y axis array/vector , validate the distance against max distance. if the distance is less than max distance then point can be included into the circle.Time complexity approximately  O(2N). Correct me if it wrong.

/ C++ Solution /

#include <iostream>

#include <string>

#include <cmath>

#include <algorithm>

#include <vector>

#include <climits>

using namespace std;

int solution(string &S, vector<int> & X, vector<int> & Y)

{  

   vector<char> tags;

   vector<int> tagsDis;

   

   int sameTagDis = 0;

   int noOfPoints = 0;

   int maxDis = INT_MIN;

   

   size_t N = X.size();

   vector<char> ::iterator itr;

   

   for(size_t i = 0; i < S.length() ; i++) {

       int tmp = 0;

       if(tags.size() == 0) {

           tags.push_back(S.at(i));

           tmp = sqrt((X[i] * X[i]) + (Y[i] * Y[i]));

           tagsDis.push_back(tmp);

           noOfPoints++;

       } else {

           

           itr = std::find(tags.begin(), tags.end(), S.at(i));          

           if (itr != tags.end()) {

               int idx = std::distance(tags.begin(), itr);

               tmp = sqrt((X[i] * X[i]) + (Y[i] * Y[i]));

               tmp = std::max(tmp, tagsDis[idx]);

               maxDis = std::max(maxDis, tmp);

               tagsDis.push_back(tmp);

           } else {

               int tmpDis = sqrt((X[i] * X[i]) + (Y[i] * Y[i]));

               tags.push_back(S.at(i));

               noOfPoints++;

           }

           

       }

   }

   

   noOfPoints = 0;

   for(size_t i = 0; i < N; i++) {

       int tmpDis = sqrt((X[i] * X[i]) + (Y[i] * Y[i]));

       if(maxDis > tmpDis) {

           noOfPoints++;

       }

   }

   return noOfPoints;

}

int main()

{

 /*string S("ABDCA");

 vector <int> X {2, -1,-4, -3, 3};

 vector <int> Y {2, -2, 4, 1, -3};*/

 

 /*string S("ABB");

 vector <int> X {1, -2, 2};

 vector <int> Y {1, -2, 2};*/

 

 string S("ABDCAD");

 vector <int> X {2, -1,-4, -3, 3, 2};

 vector <int> Y {2, -2, 4, 1, -3, 2};

 

 int N = solution(S, X, Y);

 

 cout << "N : " <<  N << "\n";

}

//

Better solution will be appriciated.