# Minimize count of array elements to be removed to maximize difference between any pair up to K

Given an array arr[] and an integer K, the task is to count the number of elements to be removed from the array such that the difference of the maximum and the minimum number left does not exceed K.
Examples:

Input: K = 1, arr[] = {1, 2, 3, 4, 5}
Output:
Explanation:
Removal of {5, 4, 3} modifies array to {1, 2} where the maximum difference is 1(= K).

Input: K = 3, arr[] = {1, 2, 3, 4, 5}
Output:
Explanation:
Removal of {5} modifies array to {1, 2, 3, 4} where the maximum difference is 3( = K).

Approach:
The task is to find the difference between the maximum and minimum array element which should not exceed K.

• Sort the array in ascending order and initialize a variable to a minimum value.
• Iterate over the array to generate all possible pairs and check if the difference between any pair is less than or equal to K.
• Update the minimum number of removals for each pair.
• Finally, print the minimum removals obtained.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement ` `// the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to count the number of ` `// elements to be removed from the ` `// array based on the given condition ` `int` `min_remove(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``// Sort the array ` `    ``sort(arr, arr + n); ` ` `  `    ``/// Initialize the variable ` `    ``int` `ans = INT_MAX; ` ` `  `    ``// Iterate for all possible pairs ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``for` `(``int` `j = i; j < n; j++) { ` ` `  `            ``// Check the difference ` `            ``// between the numbers ` `            ``if` `(arr[j] - arr[i] <= k) { ` ` `  `                ``// Update the minimum removals ` `                ``ans = min(ans, n - j + i - 1); ` `            ``} ` `        ``} ` `    ``} ` `    ``// Return the answer ` `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `k = 3; ` `    ``int` `arr[] = { 1, 2, 3, 4, 5 }; ` ` `  `    ``int` `n = ``sizeof` `arr / ``sizeof` `arr; ` ` `  `    ``cout << min_remove(arr, n, k); ` `    ``return` `0; ` `} `

## Java

 `// Java Program to implement ` `// the above approach ` `import` `java.util.*; ` `class` `GFG{ ` ` `  `// Function to count the number of ` `// elements to be removed from the ` `// array based on the given condition ` `static` `int` `min_remove(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``// Sort the array ` `    ``Arrays.sort(arr); ` ` `  `    ``/// Initialize the variable ` `    ``int` `ans = Integer.MAX_VALUE; ` ` `  `    ``// Iterate for all possible pairs ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` `        ``for` `(``int` `j = i; j < n; j++) ` `        ``{ ` ` `  `            ``// Check the difference ` `            ``// between the numbers ` `            ``if` `(arr[j] - arr[i] <= k)  ` `            ``{ ` ` `  `                ``// Update the minimum removals ` `                ``ans = Math.min(ans, n - j + i - ``1``); ` `            ``} ` `        ``} ` `    ``} ` `    ``// Return the answer ` `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `k = ``3``; ` `    ``int` `arr[] = { ``1``, ``2``, ``3``, ``4``, ``5` `}; ` ` `  `    ``int` `n = arr.length; ` ` `  `    ``System.out.print(min_remove(arr, n, k)); ` `} ` `} ` ` `  `// This code is contributed by sapnasingh4991`

## Python3

 `# Python3 program to implement ` `# the above approach ` `import` `sys ` ` `  `# Function to count the number of ` `# elements to be removed from the ` `# array based on the given condition ` `def` `min_remove(arr, n, k): ` ` `  `    ``# Sort the array ` `    ``arr.sort() ` ` `  `    ``# Initialize the variable ` `    ``ans ``=` `sys.maxsize ` ` `  `    ``# Iterate for all possible pairs ` `    ``for` `i ``in` `range``(n): ` `        ``for` `j ``in` `range``(i, n): ` ` `  `            ``# Check the difference ` `            ``# between the numbers ` `            ``if` `(arr[j] ``-` `arr[i] <``=` `k): ` ` `  `                ``# Update the minimum removals ` `                ``ans ``=` `min``(ans, n ``-` `j ``+` `i ``-` `1``) ` `     `  `    ``# Return the answer ` `    ``return` `ans ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``k ``=` `3` `    ``arr ``=` `[ ``1``, ``2``, ``3``, ``4``, ``5` `] ` ` `  `    ``n ``=` `len``(arr) ` ` `  `    ``print` `(min_remove(arr, n, k)) ` ` `  `# This code is contributed by chitranayal `

## C#

 `// C# Program to implement ` `// the above approach ` `using` `System; ` `class` `GFG{ ` ` `  `// Function to count the number of ` `// elements to be removed from the ` `// array based on the given condition ` `static` `int` `min_remove(``int` `[]arr, ``int` `n, ``int` `k) ` `{ ` `    ``// Sort the array ` `    ``Array.Sort(arr); ` ` `  `    ``/// Initialize the variable ` `    ``int` `ans = ``int``.MaxValue; ` ` `  `    ``// Iterate for all possible pairs ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        ``for` `(``int` `j = i; j < n; j++) ` `        ``{ ` ` `  `            ``// Check the difference ` `            ``// between the numbers ` `            ``if` `(arr[j] - arr[i] <= k)  ` `            ``{ ` ` `  `                ``// Update the minimum removals ` `                ``ans = Math.Min(ans, n - j + i - 1); ` `            ``} ` `        ``} ` `    ``} ` `    ``// Return the answer ` `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `k = 3; ` `    ``int` `[]arr = { 1, 2, 3, 4, 5 }; ` ` `  `    ``int` `n = arr.Length; ` ` `  `    ``Console.Write(min_remove(arr, n, k)); ` `} ` `} ` ` `  `// This code is contributed by sapnasingh4991`

Output:

```1
```

Time Complexity: O(N2
Auxiliary Space: O(1)

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