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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 −
#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, ]
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