Students with maximum average score of three subjects

Given a file containing data of student name and marks scored by him/her in 3 subjects. The task is to find the list of students having the maximum average score. 

Note: If more than one student has the maximum average score, print any one of them.

Examples:

Input: N = 2, file = {{“Shrikanth”, “20” ,”30″, “10”}, {“Ram”, “100”, “50”, “10”}}
Output: Ram 53
Explanation:
Shrikanth has an average of 20, whereas
Ram has a average of 53. So, Ram has the
maximum average.

Input: N = 3, file = {{“Adam”, “50”, “10”, “40”}, {“Rocky”, “100”, “90”, “10”}, {“Suresh”, “10”, “90” ,”100″}}
Output: Rocky 66
Explanation:
Rocky and Suresh both have an average of 66, which is the
highest in the class.

Approach :

1. Traverse the file data and store average scores for each student.
2. Now, find the maximum average score and search for all the students with this maximum average score.
3. Print the maximum average score and names as per the order in the file.

Below is the implementation of the above approach: 

C++

// C++ program to find the
// list of students having maximum average score
#include <bits/stdc++.h>
using namespace std;

string studentRecord(vector<vector<string> >& S, int N)
{
    // code here

    int maxi = INT_MIN;
    string result = "";
    for (int i = 0; i < N; i++) {
        int avg = (stoi(S[i][1]) + stoi(S[i][2])
                   + stoi(S[i][3]))
                  / 3;
        if (avg > maxi) {
            maxi = avg;
            result = S[i][0];
        }
        else if (avg == maxi) {
            result = result + " " + S[i][0];
        }
    }
    return result + " " + to_string(maxi);
}

// Driver code
int main()
{
    int N = 2;
    vector<vector<string>> file
        = { {"Shrikanth", "20", "30" ,"10"}, {"Ram" ,"100", "50", "10"}};

    cout << studentRecord(file, N);
}

Java

/*package whatever //do not write package name here */
import java.util.*;
class GFG {

  static String studentRecord(String[][] S, int N)
  {

    // code here
    int maxi = Integer.MIN_VALUE;
    String result = "";
    for (int i = 0; i < N; i++) {
      int avg = (Integer.parseInt(S[i][1]) + Integer.parseInt(S[i][2]) + Integer.parseInt(S[i][3]))/ 3;
      if (avg > maxi) {
        maxi = avg;
        result = S[i][0];
      }
      else if (avg == maxi) {
        result = result + " " + S[i][0];
      }
    }
    return result + " " + maxi;
  }
  public static void main (String[] args) {
    int N = 2;
    String[][] file= { {"Shrikanth", "20", "30" ,"10"},
                      {"Ram" ,"100", "50", "10"}};

    System.out.println(studentRecord(file, N));
  }
}

// This code is contributed by aadityaburujwale.

Python3

def studentRecord(S, N):
  maxi = float("-inf")
  result = ""
  for i in range(N):
      avg = (int(S[i][1]) + int(S[i][2]) + int(S[i][3])) / 3
  if avg > maxi:
      maxi = avg
      result = S[i][0]
  elif avg == maxi:
      result = result + " " + S[i][0]
  print(result +" " + str(int(maxi)))

N = 2
file = [["Shrikanth", "20", "30", "10"], ["Ram", "100", "50", "10"]]

studentRecord(file, N)

# This code is contributed by akashish__

C#

// C# program to find the
// list of students having maximum average score
using System;
class GFG {

  static string studentRecord(String[, ] S, int N)
  {

    // code here
    int maxi = Int32.MinValue;
    String result = "";
    for (int i = 0; i < N; i++) {
      int avg = (Int32.Parse(S[i, 1])
                 + Int32.Parse(S[i, 2])
                 + Int32.Parse(S[i, 3]))
        / 3;
      if (avg > maxi) {
        maxi = avg;
        result = S[i, 0];
      }
      else if (avg == maxi) {
        result = result + " " + S[i, 0];
      }
    }
    return result + " " + maxi;
  }
  public static void Main(string[] args)
  {
    int N = 2;
    String[, ] file
      = { { "Shrikanth", "20", "30", "10" },
         { "Ram", "100", "50", "10" } };

    Console.WriteLine(studentRecord(file, N));
  }
}

// This Code is contributed by karandeep1234

Javascript

function studentRecord( S,  N)
{
    // code here
    let maxi =Number.MIN_VALUE;
    let result = "";
    for (let i = 0; i < N; i++) {
        let avg = (parseInt(S[i][1]) + parseInt(S[i][2])
                   + parseInt(S[i][3]))
                  / 3;
        if (avg > maxi) {
            maxi = avg;
            result = S[i][0];
        }
        else if (avg == maxi) {
            result = result + " " + S[i][0];
        }
    }
    console.log(result + " " + parseInt(maxi));
}

// Driver code
    let N = 2;
   let file
        = [["Shrikanth", "20", "30" ,"10"], ["Ram" ,"100", "50", "10"]];

     studentRecord(file, N);

// This code is contributed by garg28harsh.

Output

Ram 53

Time complexity: O(N) where N is the number of students in the given string array.
Auxiliary space: O(N)