Dadas tres arrays nombres[] , marcas[] y actualizaciones[] donde:
- names[] contiene los nombres de los estudiantes.
- notes[] contiene las notas de los mismos estudiantes.
- actualizaciones[] contiene los números enteros por los cuales se actualizarán las calificaciones de estos estudiantes.
La tarea es encontrar el nombre del estudiante con las máximas calificaciones después de la actualización y el salto en el rango del estudiante, es decir, rango anterior – rango actual .
Nota: Los detalles de los estudiantes están en orden descendente de sus calificaciones y si más de dos estudiantes obtuvieron calificaciones iguales (también la más alta), entonces elija el que tuvo una calificación más baja anteriormente.
Ejemplos:
Entrada: nombres[] = {“sam”, “ram”, “geek”, “sonu”},
marcas[] = {99, 75, 70, 60},
actualizaciones[] = {-10, 5, 9, 39}
Salida: Nombre: sonu, Salto: 3
Las marcas actualizadas son {89, 80, 79, 99}, está claro que sonu tiene las marcas más altas con un salto de 3Entrada: nombres[] = {“sam”, “ram”, “geek”},
marcas[] = {80, 79, 75},
actualizaciones[] = {0, 5, -9}
Salida: Nombre: ram, Saltar: 1
Enfoque: Cree una estructura struct Student para almacenar la información de cada estudiante, cada uno de los cuales tendrá 3 atributos nombre del estudiante, calificaciones del estudiante, prev_rank del estudiante.
Ahora, actualice las calificaciones de acuerdo con el contenido de las actualizaciones [] luego, con un solo recorrido de la array, encuentre al estudiante que tiene las calificaciones más altas después de la actualización.
A continuación se muestra la implementación del enfoque anterior:
C++
// 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
// 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
# 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#
// 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
Javascript
<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
Complejidad temporal: O(N)
Espacio auxiliar: O(N)
Publicación traducida automáticamente
Artículo escrito por mohit kumar 29 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA