People Whose List of Favorite Companies Is Not a Subset of Another List in C++

C++Server Side ProgrammingProgramming

Suppose we have an array called favorite companies where favoriteCompanies[i] is the list of favorites companies of the ith person. We have to find the indices of people whose list of favorite companies is not a subset of any other list of favorites companies.

So, if the input is like favoriteCompanies = [["TCS", "google", "facebook"], ["google","microsoft"], ["google", "facebook"], ["google"], ["amazon"]], then the output will be [0,1,4], this is because person with index=2 has ["google", "facebook"] which is a subset of favoriteCompanies[0]= ["TCS", "google", "facebook"] corresponding to the person with index 0.

Now person with index=3 has ["google"] which is a subset of favoriteCompanies[0]= ["TCS", "google", "facebook"] and favoriteCompanies[1]= ["google", "microsoft"]. Other lists of favorite companies are not a subset of another list, so, the answer is [0,1,4].

To solve this, we will follow these steps −

  • Define a function ok(), this will take an array a, an array b,

  • cnt := 0, i := 0, j := 0

  • while (i < size of a and j < size of b), do −

    • if a[i] is same as b[j], then −

      • (increase i, j and cnt by 1)

    • otherwise when a[i] < b[j], then −

      • (increase i by 1)

    • Otherwise

      • (increase j by 1)

  • return true when cnt < size of a

  • From the main method do the following −

  • Define one set s

  • n := size of f

  • for initialize i := 0, when i < n, update (increase i by 1), do −

    • sort the array f[i]

  • for initialize i := 0, when i < n, update (increase i by 1), do −

    • c := true

    • for initialize j := 0, when j < n, update (increase j by 1), do −

      • if i is same as j, then −

        • Ignore the following part, skip to the next iteration

      • c := c AND ok(f[i], f[j])

    • if c is non-zero, then −

      • insert i into s

  • return elements of s as an array

Example 

Let us see the following implementation to get a better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<int> v){
   cout << "[";
   for(int i = 0; i<v.size(); i++){
      cout << v[i] << ", ";
   }
   cout << "]"<<endl;
}
class Solution {
public:
   bool ok(vector<string>& a, vector<string>& b){
      int cnt = 0;
      int i = 0;
      int j = 0;
      while (i < a.size() && j < b.size()) {
         if (a[i] == b[j]) {
            i++;
            j++;
            cnt++;
         }
         else if (a[i] < b[j]) {
            i++;
         }
         else {
            j++;
         }
      }
      return cnt < a.size();
   }
   vector<int> peopleIndexes(vector<vector<string> >& f){
      set<int> s;
      int n = f.size();
      for (int i = 0; i < n; i++) {
         sort(f[i].begin(), f[i].end());
      }  
      for (int i = 0; i < n; i++) {
         bool c = true;
         for (int j = 0; j < n; j++) {
            if (i == j)
               continue;
            c &= ok(f[i], f[j]);
         }
         if (c)
            s.insert(i);
      }
      return vector<int>(s.begin(), s.end());
   }
};
main(){
   Solution ob;
   vector<vector<string>> v = {{"TCS","google","facebook"},{"google","microsoft"},{"google","facebo
ok"},{"google"},{"amazon"}};
print_vector(ob.peopleIndexes(v));
}

Input

{{"TCS","google","facebook"},{"google","microsoft"},{"google","facebook"},{"google"},{"amazon"}}

Output

[0, 1, 4, ]
raja
Updated on 17-Nov-2020 12:33:21

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