# TV Shows in C++

C++Server Side ProgrammingProgramming

Suppose we have a list of TV shows, and another list of duration, and an integer k, here shows[i] and duration[i] shows the name and duration watched by the ith person, we have to find the total duration watched of the k most watched shows.

So, if the input is like shows: ["Castle Play", "Fairy Tale Series", "Castle Play", "Jerry Mouse", "Rich Boy"], duration: [6, 4, 6, 14, 5] and k = 2, then the output will be 26.

To solve this, we will follow these steps −

• Define one map m

• n := size of v

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

• m[v[i]] := m[v[i]] + d[i]

• define an array arr

• for each key-value pair it of m

• insert value of it at the end of arr

• sort the array arr in reverse order

• ret := 0

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

• ret := ret + arr[i]

• return ret

Let us see the following implementation to get better understanding −

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int solve(vector<string>& v, vector<int>& d, int k) {
map <string, int> m;
int n = v.size();
for(int i = 0; i < n; i++){
m[v[i]] += d[i];
}
vector < int > arr;
for(auto it : m){
arr.push_back(it.second);
}
sort(arr.rbegin(), arr.rend());
int ret = 0;
for(int i = 0; i < k; i++){
ret += arr[i];
}
return ret;
}
};
int main(){
vector<string> v = {"Castle Play", "Fairy Tale Series", "Castle
Play", "Jerry Mouse", "Rich Boy"};
vector<int> v1 = {6, 4, 6, 14, 5};
Solution ob;
cout << (ob.solve(v, v1, 2));
}

## Input

{"Castle Play", "Fairy Tale Series", "Castle Play", "Jerry Mouse",
"Rich Boy"}, {6, 4, 6, 14, 5}, 2

## Output

26