# Number of players whose rank is equal to or less than a given cutoff rank

Given an array** arr[]** consisting of **N** integers and an integer **R**, denoting the cutoff rank, the task is to count the number of array elements with rank **at most R** such that the equal array element are ranked the same and distinct array elements are ranked based on their positions in the array **arr[]**.

**Examples:**

Input:arr[] = {100, 50, 50, 25}, R = 3Output:3Explanation:

The players are ranked as: {1, 2, 2, 4}. The players having ranked at most R(= 3) is {1, 2, 2}. Therefore, the total count is 3.

Input:arr[] = {2, 2, 3, 4, 5}, R = 4Output:5

**Approach:** The given problem can be solved by using the concept of Sorting. Follow the below step to solve this problem:

- Sort the given array
**arr[]**in decreasing order. - Initialize two variables, say
**rank**as**1**to store the rank of the array elements and say**count**as**0**to store the required result. - Traverse the given array
**arr[]**, using the variable**i**, and perform the following steps:- If the
**arr[i]**is equal to the previous element then assign the same rank as the previous rank to the current element. - Otherwise, assign the value of
**(count + 1)**rank to the current element.^{th} - If the
**rank**is greater than**R**then break. Otherwise, increment the**count**by**1**.

- If the
- After completing the above steps, print the value of
**count**as the answer.

Below is the implementation of the above approach:

## C++

`// C++ program for above approach` `#include <algorithm>` `#include <iostream>` `using` `namespace` `std;` `// Function to find the count of array` `// elements having rank at most R` `int` `countElements(` `int` `R, ` `int` `N, ` `int` `arr[])` `{` ` ` `// Sort the array arr[] in the` ` ` `// decreasing order` ` ` `sort(arr, arr + N, greater<` `int` `>());` ` ` `// Stores the rank and required` ` ` `// count of array elements` ` ` `int` `rank = 1, count = 0;` ` ` `// store the previou element` ` ` `int` `prevScore = arr[0], score;` ` ` `// Traverse the array` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` `score = arr[i];` ` ` `// If score is less than the` ` ` `// prevScore` ` ` `if` `(score < prevScore) {` ` ` `rank = count + 1;` ` ` `}` ` ` `// If the rank is greater than R` ` ` `if` `(rank > R) {` ` ` `break` `;` ` ` `}` ` ` `// Increment count by 1` ` ` `count++;` ` ` `// update prevscore` ` ` `prevScore = score;` ` ` `}` ` ` `// return count` ` ` `return` `count;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = { 100, 50, 50, 25 };` ` ` `int` `R = 2;` ` ` `int` `N = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `cout << countElements(R, N, arr);` ` ` `return` `0;` `}` `// This code is contributed by Parth Manchanda` |

## Java

`// Java program for the above approach` `import` `java.util.*;` `class` `GFG` `{` ` ` `static` `void` `reverse(` `int` `a[])` ` ` `{` ` ` `int` `n = a.length;` ` ` `int` `[] b = ` `new` `int` `[n];` ` ` `int` `j = n;` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) {` ` ` `b[j - ` `1` `] = a[i];` ` ` `j = j - ` `1` `;` ` ` `}` ` ` `}` ` ` `// Function to find the count of array` ` ` `// elements having rank at most R` ` ` `static` `int` `countElements(` `int` `R, ` `int` `N, ` `int` `[] arr)` ` ` `{` ` ` `// Sort the array arr[] in the` ` ` `// decreasing order` ` ` `Arrays.sort(arr);` ` ` `reverse(arr);` ` ` `// Stores the rank and required` ` ` `// count of array elements` ` ` `int` `rank = ` `1` `;` ` ` `int` `count = -` `1` `;` ` ` `// Stores the previous element` ` ` `int` `prevScore = arr[` `0` `];` ` ` `// Traverse the array` ` ` `for` `(` `int` `score : arr)` ` ` `{` ` ` `// If score is less than the` ` ` `// prevScore` ` ` `if` `(score < prevScore)` ` ` `rank = count + ` `1` `;` ` ` `// If the rank is greater than R` ` ` `if` `(rank > R)` ` ` `break` `;` ` ` `// Increment count by 1` ` ` `count = count + ` `1` `;` ` ` `// Update prevScore` ` ` `prevScore = score;` ` ` `}` ` ` `// Return the result` ` ` `return` `count;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `[] arr = { ` `100` `, ` `50` `, ` `50` `, ` `25` `};` ` ` `int` `R = ` `2` `;` ` ` `int` `N = arr.length;` ` ` `// Function Call` ` ` `System.out.println(countElements(R, N, arr));` ` ` `}` `}` `// This code is contributed by sanjoy_62.` |

## Python3

`# Python program for the above approach` `# Function to find the count of array` `# elements having rank at most R` `def` `countElements(R, N, arr):` ` ` `# Sort the array arr[] in the` ` ` `# decreasing order` ` ` `arr.sort(reverse ` `=` `True` `)` ` ` `# Stores the rank and required` ` ` `# count of array elements` ` ` `rank ` `=` `1` ` ` `count ` `=` `0` ` ` `# Stores the previous element` ` ` `prevScore ` `=` `arr[` `0` `]` ` ` `# Traverse the array` ` ` `for` `score ` `in` `arr:` ` ` `# If score is less than the` ` ` `# prevScore` ` ` `if` `score < prevScore:` ` ` `rank ` `=` `count ` `+` `1` ` ` `# If the rank is greater than R` ` ` `if` `rank > R:` ` ` `break` ` ` ` ` `# Increment count by 1` ` ` `count ` `+` `=` `1` ` ` `# Update prevScore` ` ` `prevScore ` `=` `score` ` ` `# Return the result` ` ` `return` `count` `# Driver Code` `arr ` `=` `[` `100` `, ` `50` `, ` `50` `, ` `25` `]` `R ` `=` `2` `N ` `=` `len` `(arr)` `# Function Call` `print` `(countElements(R, N, arr))` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG{` ` ` `// Function to find the count of array` `// elements having rank at most R` `static` `int` `countElements(` `int` `R, ` `int` `N, ` `int` `[] arr)` `{` ` ` ` ` `// Sort the array arr[] in the` ` ` `// decreasing order` ` ` `Array.Sort(arr);` ` ` `Array.Reverse(arr);` ` ` `// Stores the rank and required` ` ` `// count of array elements` ` ` `int` `rank = 1;` ` ` `int` `count = 0;` ` ` `// Stores the previous element` ` ` `int` `prevScore = arr[0];` ` ` `// Traverse the array` ` ` `foreach` `(` `int` `score ` `in` `arr)` ` ` `{` ` ` ` ` `// If score is less than the` ` ` `// prevScore` ` ` `if` `(score < prevScore)` ` ` `rank = count + 1;` ` ` `// If the rank is greater than R` ` ` `if` `(rank > R)` ` ` `break` `;` ` ` `// Increment count by 1` ` ` `count = count + 1;` ` ` `// Update prevScore` ` ` `prevScore = score;` ` ` `}` ` ` ` ` `// Return the result` ` ` `return` `count;` `}` `// Driver code` `static` `public` `void` `Main()` `{` ` ` `int` `[] arr = { 100, 50, 50, 25 };` ` ` `int` `R = 2;` ` ` `int` `N = arr.Length;` ` ` `// Function Call` ` ` `Console.WriteLine(countElements(R, N, arr));` `}` `}` `// This code is contributed by target_2.` |

## Javascript

`<script>` `// JavaScript program for the above approach` `// Function to find the count of array` `// elements having rank at most R` `function` `countElements(R, N, arr)` `{` ` ` ` ` `// Sort the array arr[] in the` ` ` `// decreasing order` ` ` `arr.sort(` `function` `(a, b){ ` `return` `b - a; });` ` ` `// Stores the rank and required` ` ` `// count of array elements` ` ` `let rank = 1;` ` ` `let count = 0;` ` ` `// Stores the previous element` ` ` `let prevScore = arr[0];` ` ` `// Traverse the array` ` ` `for` `(let score of arr)` ` ` `{` ` ` ` ` `// If score is less than the` ` ` `// prevScore` ` ` `if` `(score < prevScore)` ` ` `rank = count + 1;` ` ` `// If the rank is greater than R` ` ` `if` `(rank > R)` ` ` `break` `;` ` ` `// Increment count by 1` ` ` `count = count + 1;` ` ` `// Update prevScore` ` ` `prevScore = score;` ` ` `}` ` ` ` ` `// Return the result` ` ` `return` `count;` `}` `// Driver Code` `let arr = [ 100, 50, 50, 25 ];` `let R = 2;` `let N = arr.length;` `// Function Call` `document.write(countElements(R, N, arr));` `// This code is contributed by lokeshpotta20` `</script>` |

**Output:**

3

**Time Complexity:** O(N*log N)**Auxiliary Space:** O(1)

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