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 3

Input: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++

`// 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); ` `} ` |

*chevron_right*

*filter_none*

## Python3

`# Python3 implementation of the approach ` ` ` `# Function to prthe 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 ` |

*chevron_right*

*filter_none*

**Output:**

Name: ram, Jump: 1

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.

## Recommended Posts:

- Find the total marks obtained according to given marking scheme
- Jump Search
- Jump Pointer Algorithm
- Count number of ways to jump to reach end
- Number of cyclic elements in an array where we can jump according to value
- C program to store Student records as Structures and Sort them by Name
- Rank of an element in a stream
- Rank of all elements in an array
- Replace each element of Array with it's corresponding rank
- Find the minimum of maximum length of a jump required to reach the last island in exactly k jumps
- Rank of all elements in a Stream in descending order when they arrive
- Minimum cost to reach end of array array when a maximum jump of K index is allowed
- Nth Term of a Fibonacci Series of Primes formed by concatenating pairs of Primes in a given range
- Absolute distinct count in a Linked List

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.