Minimum Index Sum of Two Lists in C++

Suppose there are two fiends Amal and Bimal want to choose a restaurant for dinner, now they both have a list of favorite restaurants represented by strings. We have to help them to find out their common interest with the least list index sum. If there is a choice tie between different answers, then return all of them with no order requirement.

So, if the input is like ["ABC","PQR","MNO","XYZ"], and ["TUV","GHI","KLM","ABC"], then the output will be ["ABC"]

To solve this, we will follow these steps −

• Define one map mp

• least := inf

• for initialize i := 0, when i < size of l1, update (increase i by 1), do −

  • for initialize j := 0, when j < size of l2, update (increase j by 1), do −

    • if l1[i] is same as l2[j], then −

      • insert l1[i] at the end of mp[i + j]

• Define an array res

• it = first element of mp

• res := value of it

• return res

Example 

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto> v){
   cout << "[";
   for(int i = 0; i<v.size(); i++){
      cout << v[i] << ", ";
   }
   cout << "]"<<endl;
}
class Solution {
public:
   vector<string> findRestaurant(vector<string>& l1, vector<string>& l2) {
      map<int, vector<string> > mp;
      int least = INT_MAX;
      for (int i = 0; i < l1.size(); i++)
         for (int j = 0; j < l2.size(); j++)
            if (l1[i] == l2[j])
               mp[i + j].push_back(l1[i]);
      vector<string> res;
      auto it = mp.begin();
      res = it->second;
      return res;
   }
};
main(){
   Solution ob;
   vector<string> v = {"ABC","PQR","MNO","XYZ"}, v1 = {"TUV","GHI","KLM","ABC"};
   print_vector(ob.findRestaurant(v, v1));
}

Input

{"ABC","PQR","MNO","XYZ"}, {"TUV","GHI","KLM","ABC"}

Output

[ABC, ]