# Minimize count of Subsets with difference between maximum and minimum element not exceeding K

• Last Updated : 19 May, 2021

Given an array arr[ ] and an integer K, the task is to split the given array into minimum number of subsets having the difference between the maximum and the minimum element ≤ K.

Examples:

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Input: arr[ ] = {1, 3, 7, 9, 10}, K = 3
Output: 2
Explanation:
One of the possible subsets of arr[] are {1, 3} and {7, 9, 10} where the difference between maximum and minimum element does not greater than K i.e, 3.

Input: arr[ ] = {1, 10, 8, 3, 9}, K =3
Output: 2.

Approach: Follow the steps below to solve the problem:

1. Sort the array in ascending order.
2. Iterate over the array, setting currMin as the first element of the array and keep updating currMax with the elements traversed.
3. If at any index, the difference between currMax and currMin exceeds K, increment answer by 1 and set currMax and currMin to arr[i].

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement``// above approach``#include ``using` `namespace` `std;` `// Function to find the minimum count``// of subsets of required type``int` `findCount(``int` `arr[], ``int` `N, ``int` `K)``{``    ``sort(arr, arr + N);` `    ``// Stores the result``    ``int` `result = 1;` `    ``// Store the maximum and minimum``      ``// element of the current subset``    ``int` `cur_max = arr;``    ``int` `cur_min = arr;``  ` `    ``for` `(``int` `i = 1; i < N; i++) {``      ` `        ``// Update current maximum``        ``cur_max = arr[i];``      ` `        ``// If difference exceeds K``        ``if` `(cur_max - cur_min > K) {``          ` `            ``// Update count``            ``result++;` `            ``// Update maximum and minimum``            ``// to the current subset``            ``cur_max = arr[i];``            ``cur_min = arr[i];``        ``}``    ``}``  ` `    ``return` `result;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 1,10, 8, 3, 9 };``    ``int` `K = 3;``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << findCount(arr, N, K);` `    ``return` `0;``}`

## Java

 `// Java program to implement``// above approach``import` `java.util.*;` `class` `GFG{` `// Function to find the minimum count``// of subsets of required type``static` `int` `findCount(``int` `arr[], ``int` `N, ``int` `K)``{``    ``Arrays.sort(arr);` `    ``// Stores the result``    ``int` `result = ``1``;` `    ``// Store the maximum and minimum``    ``// element of the current subset``    ``int` `cur_max = arr[``0``];``    ``int` `cur_min = arr[``0``];` `    ``for``(``int` `i = ``1``; i < N; i++)``    ``{``        ` `        ``// Update current maximum``        ``cur_max = arr[i];``    ` `        ``// If difference exceeds K``        ``if` `(cur_max - cur_min > K)``        ``{``        ` `            ``// Update count``            ``result++;` `            ``// Update maximum and minimum``            ``// to the current subset``            ``cur_max = arr[i];``            ``cur_min = arr[i];``        ``}``    ``}``    ``return` `result;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``1``, ``10``, ``8``, ``3``, ``9` `};``    ``int` `K = ``3``;``    ``int` `N = arr.length;``    ` `    ``System.out.print(findCount(arr, N, K));``}``}` `// This code is contributed by amal kumar choubey`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to find the minimum count``# of subsets of required type``def` `findCount(arr, N, K):` `    ``arr.sort()` `    ``# Stores the result``    ``result ``=` `1` `    ``# Store the maximum and minimum``    ``# element of the current subset``    ``cur_max ``=` `arr[``0``]``    ``cur_min ``=` `arr[``0``]` `    ``for` `i ``in` `range``(``1``, N):` `        ``# Update current maximum``        ``cur_max ``=` `arr[i]` `        ``# If difference exceeds K``        ``if``(cur_max ``-` `cur_min > K):` `            ``# Update count``            ``result ``+``=` `1` `            ``# Update maximum and minimum``            ``# to the current subset``            ``cur_max ``=` `arr[i]``            ``cur_min ``=` `arr[i]` `    ``return` `result` `# Driver Code``arr ``=` `[ ``1``, ``10``, ``8``, ``3``, ``9` `]``K ``=` `3``N ``=` `len``(arr)` `# Function call``print``(findCount(arr, N, K))` `# This code is contributed by Shivam Singh`

## C#

 `// C# program to implement``// above approach``using` `System;``class` `GFG{` `// Function to find the minimum count``// of subsets of required type``static` `int` `findCount(``int` `[]arr,``                     ``int` `N, ``int` `K)``{``    ``Array.Sort(arr);` `    ``// Stores the result``    ``int` `result = 1;` `    ``// Store the maximum and minimum``    ``// element of the current subset``    ``int` `cur_max = arr;``    ``int` `cur_min = arr;` `    ``for``(``int` `i = 1; i < N; i++)``    ``{``        ` `        ``// Update current maximum``        ``cur_max = arr[i];``    ` `        ``// If difference exceeds K``        ``if` `(cur_max - cur_min > K)``        ``{``        ` `            ``// Update count``            ``result++;` `            ``// Update maximum and minimum``            ``// to the current subset``            ``cur_max = arr[i];``            ``cur_min = arr[i];``        ``}``    ``}``    ``return` `result;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]arr = { 1, 10, 8, 3, 9 };``    ``int` `K = 3;``    ``int` `N = arr.Length;``    ` `    ``Console.Write(findCount(arr, N, K));``}``}` `// This code is contributed by gauravrajput1`

## Javascript

 ``
Output:
`2`

Time Complexity: O(NLog(N))
Auxiliary Space: O(1)

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