Find K items with the lowest values

Given a list of items and their values. The task is to find k items with the lowest value. It is possible that two items have the same value, in that case item whose name comes first (lexicographically) will be given higher priority.

Examples:

Input : items[] = {Bat, Gloves, Wickets, Ball}, 
        values[] = {100, 50, 200, 100}
        k = 2
Output : Gloves Ball
Explanation : 
Gloves has the lowest value.
Ball and Bat has the same value but Ball comes first lexicographically.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
This question can be solved by picking the items greedily according to the values. We will sort use the items list in the ascending order of the values and in case of the same values items will be sorted lexicographical order increasing order.
We will store the data in the form of pairs in a vector and will use an inbuilt sort function with boolean compactor function which will be used to compare two items.

Below is the implementation of the above approach:

// C++ implementation of above approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Boolean Comparator Function 
// to compare two pairs of item-value 
bool comp(pair<string, int> A, pair<string, int> B) 
{ 
    // Compare the name if the values are equal 
    if (A.second == B.second) 
        return A.first < B.first; 
  
    // Else compare values 
    return A.second < B.second; 
} 
  
// Driver code 
int main() 
{ 
    int k = 2; 
    int n = 3; 
  
    // Store data in a vector of Item-Value Pair 
    vector<pair<string, int> > items; 
  
    // inserting items-value pairs in the vector 
    items.push_back(make_pair("Bat", 100)); 
    items.push_back(make_pair("Gloves", 50)); 
    items.push_back(make_pair("Wickets", 200)); 
    items.push_back(make_pair("Ball", 100)); 
  
    // Sort items using Inbuilt function 
    sort(items.begin(), items.end(), comp); 
  
    // Print top k values 
    // or if n is less than k 
    // Print all n items 
    for (int i = 0; i < min(n, k); ++i) { 
        cout << items[i].first << '\n'; 
    } 
  
    return 0; 
} 