Given three arrays names[], marks[] and updates[] where:
- names[] contains the names of students.
- marks[] contains the marks of the same students.
- updates[] contains the integers by which the marks of these students are to be updated.
The task is find the name of the student with maximum marks after updation and the jump in the student’s rank i.e. previous rank – current rank.
Note: The details of the students are in descending order of their marks and if more than two students scored equal marks (also the highest) then choose the one who had a lower rank previously.
Examples:
Input: names[] = {“sam”, “ram”, “geek”, “sonu”},
marks[] = {99, 75, 70, 60},
updates[] = {-10, 5, 9, 39}
Output: Name: sonu, Jump: 3
Updated marks are {89, 80, 79, 99}, its clear that sonu has the highest marks with jump of 3Input: names[] = {“sam”, “ram”, “geek”},
marks[] = {80, 79, 75},
updates[] = {0, 5, -9}
Output: Name: ram, Jump: 1
Approach: Create a structure struct Student to store the information of each student, each of which will have 3 attributes name of the student, marks of the student, prev_rank of the student.
Now, updated the marks according to the content of the updates[] then with a single traversal of the array, find the student who has the highest marks after updation.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Structure to store the information of // students struct Student {
string name;
int marks = 0;
int prev_rank = 0;
}; // Function to print the name of student who // stood first after updation in rank void nameRank(string names[], int marks[],
int updates[], int n)
{ // Array of students
Student x[n];
for ( int i = 0; i < n; i++) {
// Store the name of the student
x[i].name = names[i];
// Update the marks of the student
x[i].marks = marks[i] + updates[i];
// Store the current rank of the student
x[i].prev_rank = i + 1;
}
Student highest = x[0];
for ( int j = 1; j < n; j++) {
if (x[j].marks >= highest.marks)
highest = x[j];
}
// Print the name and jump in rank
cout << "Name: " << highest.name
<< ", Jump: "
<< abs (highest.prev_rank - 1)
<< endl;
} // Driver code int main()
{ // Names of the students
string names[] = { "sam" , "ram" , "geek" };
// Marks of the students
int marks[] = { 80, 79, 75 };
// Updates that are to be done
int updates[] = { 0, 5, -9 };
// Number of students
int n = sizeof (marks) / sizeof (marks[0]);
nameRank(names, marks, updates, n);
} |
// Java implementation of the approach import java.util.*;
class GFG{
static class Student
{ String name;
int marks, prev_rank;
Student()
{
this .marks = 0 ;
this .prev_rank = 0 ;
}
} // Function to print the name of student who // stood first after updation in rank static void nameRank(String []names, int []marks,
int []updates, int n)
{ // Array of students
Student []x = new Student[n];
for ( int i = 0 ; i < n; i++)
{
x[i] = new Student();
// Store the name of the student
x[i].name = names[i];
// Update the marks of the student
x[i].marks = marks[i] + updates[i];
// Store the current rank of the student
x[i].prev_rank = i + 1 ;
}
Student highest = x[ 0 ];
for ( int j = 1 ; j < n; j++)
{
if (x[j].marks >= highest.marks)
highest = x[j];
}
// Print the name and jump in rank
System.out.print( "Name: " + highest.name +
", Jump: " +
Math.abs(highest.prev_rank - 1 ));
} // Driver code public static void main(String[] args)
{ // Names of the students
String []names = { "sam" , "ram" , "geek" };
// Marks of the students
int []marks = { 80 , 79 , 75 };
// Updates that are to be done
int []updates = { 0 , 5 , - 9 };
// Number of students
int n = marks.length;
nameRank(names, marks, updates, n);
} } // This code is contributed by pratham76 |
# Python3 implementation of the approach # Function to print the name of student who # stood first after updation in rank def nameRank(names, marks, updates, n):
# Array of students
x = [[ 0 for j in range ( 3 )] for i in range (n)]
for i in range (n):
# Store the name of the student
x[i][ 0 ] = names[i]
# Update the marks of the student
x[i][ 1 ] = marks[i] + updates[i]
# Store the current rank of the student
x[i][ 2 ] = i + 1
highest = x[ 0 ]
for j in range ( 1 , n):
if (x[j][ 1 ] > = highest[ 1 ]):
highest = x[j]
# Print the name and jump in rank
print ( "Name: " , highest[ 0 ], ", Jump: " ,
abs (highest[ 2 ] - 1 ), sep = "")
# Driver code # Names of the students names = [ "sam" , "ram" , "geek" ]
# Marks of the students marks = [ 80 , 79 , 75 ]
# Updates that are to be done updates = [ 0 , 5 , - 9 ]
# Number of students n = len (marks)
nameRank(names, marks, updates, n) # This code is contributed by SHUBHAMSINGH10 |
// C# implementation of the approach using System;
class GFG{
public class Student
{ public string name;
public int marks, prev_rank;
public Student()
{
this .marks = 0;
this .prev_rank = 0;
}
} // Function to print the name of student who // stood first after updation in rank static void nameRank( string []names, int []marks,
int []updates, int n)
{ // Array of students
Student []x = new Student[n];
for ( int i = 0; i < n; i++)
{
x[i] = new Student();
// Store the name of the student
x[i].name = names[i];
// Update the marks of the student
x[i].marks = marks[i] + updates[i];
// Store the current rank of the student
x[i].prev_rank = i + 1;
}
Student highest = x[0];
for ( int j = 1; j < n; j++)
{
if (x[j].marks >= highest.marks)
highest = x[j];
}
// Print the name and jump in rank
Console.Write( "Name: " + highest.name +
", Jump: " +
Math.Abs(highest.prev_rank - 1));
} // Driver code public static void Main( string [] args)
{ // Names of the students
string []names = { "sam" , "ram" , "geek" };
// Marks of the students
int []marks = { 80, 79, 75 };
// Updates that are to be done
int []updates = { 0, 5, -9 };
// Number of students
int n = marks.Length;
nameRank(names, marks, updates, n);
} } // This code is contributed by rutvik_56 |
<script> // Javascript implementation of the approach // Function to print the name of student who // stood first after updation in rank function nameRank(names, marks, updates, n)
{ // Array of students
let x = new Array(n);
for (let i = 0; i < n; i++)
{
x[i] = new Array(3);
for (let j = 0; j < 3; j++)
{
x[i][j] = 0;
}
}
for (let i = 0; i < n; i++)
{
// Store the name of the student
x[i][0] = names[i]
// Update the marks of the student
x[i][1] = marks[i] + updates[i]
// Store the current rank of the student
x[i][2] = i + 1
}
let highest = x[0];
for (let j = 1; j < n; j++)
{
if (x[j][1] >= highest[1])
highest = x[j]
}
// Print the name and jump in rank
document.write( "Name: " , highest[0], ", Jump: " ,
Math.abs(highest[2] - 1), sep = "" )
} // Driver code // Names of the students let names = [ "sam" , "ram" , "geek" ];
// Marks of the students let marks = [ 80, 79, 75 ]; // Updates that are to be done let updates = [ 0, 5, -9 ]; // Number of students let n = marks.length; nameRank(names, marks, updates, n) // This code is contributed by unknown2108 </script> |
Name: ram, Jump: 1
Complexity Analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(N)