# Split array into minimum number of subsets having difference between maximum and minimum element at most K

Last Updated : 22 Sep, 2021

Given an array arr[] consisting of N integers and an integer K, the task is to find the minimum number of sets, the array elements can be divided into such that the difference between the maximum and minimum element of each set is at most K.

Examples:

Input: arr[] = {1, 2, 3, 4, 5}, K = 2
Output: 2
Explanation:
The given array can be divided into two sets {1, 2, 3} having the difference between maximum and minimum as 3 – 1= 2 and {4, 5} having the difference between maximum and minimum as 5 – 4 = 1.

Input: arr[] = {5, 2, 9, 7, 3, 2, 4, 6, 14, 10}, K = 3
Output: 4

Approach: The given problem can be solved by sorting the given array and finding the minimum number of subarrays the array elements can be broken such that the difference between the maximum and minimum element at most K. Follow the steps below to solve the given problem:

• Sort the given array arr[] in non-decreasing order.
• Initialize two iterators begin and end as 0 representing the beginning and end of each set.
• Initialize a variable, say setCount as 1 that stores the resultant minimum number of breaking of array elements into subarrays.
• Iterate a loop until the value of end is less than N and perform the following steps:
1. If the value of arr[end] – arr[begin] <= K, then increment the value of end.
2. Otherwise, increment the value setCount by 1 and update the value of begin to end representing the new set.
• After completing the above steps, print the value of setCount as the result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to find the minimum number` `// of sets the array can be divided such` `// that for each set max-min <= K` `int` `minSetCount(``int` `arr[], ``int` `N, ``int` `K)` `{` `    ``// Sort the input array` `    ``sort(arr, arr + N);`   `    ``// Stores the count of set required` `    ``int` `setCount = 1;`   `    ``// Stores the beginning and ending` `    ``// of the current set` `    ``int` `begin = 0, end = 0;`   `    ``// Loop to iterate over the array` `    ``while` `(end < N) {`   `        ``// If arr[end] can be included` `        ``// in the current set else` `        ``// begin a new set` `        ``if` `(arr[end] - arr[begin] <= K) {` `            ``end++;` `        ``}` `        ``else` `{` `            ``// Increment the set count` `            ``setCount++;` `            ``begin = end;` `        ``}` `    ``}`   `    ``// Return answer` `    ``return` `setCount;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `arr[] = { 5, 2, 9, 7, 3, 2, 4, 6, 14, 10 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(``int``);` `    ``int` `K = 3;` `    ``cout << minSetCount(arr, N, K);`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach`   `import` `java.util.*;`   `class` `GFG {`   `    ``// Function to find the minimum number` `    ``// of sets the array can be divided such` `    ``// that for each set max-min <= K` `    ``static` `int` `minSetCount(``int``[] arr, ``int` `N, ``int` `K)` `    ``{` `        ``// Sort the input array` `        ``Arrays.sort(arr);`   `        ``// Stores the count of set required` `        ``int` `setCount = ``1``;`   `        ``// Stores the beginning and ending` `        ``// of the current set` `        ``int` `begin = ``0``, end = ``0``;`   `        ``// Loop to iterate over the array` `        ``while` `(end < N) {`   `            ``// If arr[end] can be included` `            ``// in the current set else` `            ``// begin a new set` `            ``if` `(arr[end] - arr[begin] <= K) {` `                ``end++;` `            ``}` `            ``else` `{` `                ``// Increment the set count` `                ``setCount++;` `                ``begin = end;` `            ``}` `        ``}`   `        ``// Return answer` `        ``return` `setCount;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int``[] arr = { ``5``, ``2``, ``9``, ``7``, ``3``, ``2``, ``4``, ``6``, ``14``, ``10` `};` `        ``int` `N = arr.length;` `        ``int` `K = ``3``;` `        ``System.out.print(minSetCount(arr, N, K));` `    ``}` `}`   `// This code is contributed by subham348.`

## Python3

 `# Python 3 program for the above approach`   `# Function to find the minimum number` `# of sets the array can be divided such` `# that for each set max-min <= K` `def` `minSetCount(arr, N, K):` `    ``# Sort the input array` `    ``arr.sort()`   `    ``# Stores the count of set required` `    ``setCount ``=` `1`   `    ``# Stores the beginning and ending` `    ``# of the current set` `    ``begin ``=` `0` `    ``end ``=` `0`   `    ``# Loop to iterate over the array` `    ``while` `(end < N):` `      `  `        ``# If arr[end] can be included` `        ``# in the current set else` `        ``# begin a new set` `        ``if` `(arr[end] ``-` `arr[begin] <``=` `K):` `            ``end ``+``=` `1` `        ``else``:` `            ``# Increment the set count` `            ``setCount ``+``=` `1` `            ``begin ``=` `end`   `    ``# Return answer` `    ``return` `setCount`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[``5``, ``2``, ``9``, ``7``, ``3``, ``2``, ``4``, ``6``, ``14``, ``10``]` `    ``N ``=` `len``(arr)` `    ``K ``=` `3` `    ``print``(minSetCount(arr, N, K))`   `    ``# This code is contributed by SURENDRA_GANGWAR.`

## C#

 `// C# program for the above approach` `using` `System;`   `public` `class` `GFG` `{`   `    ``// Function to find the minimum number` `    ``// of sets the array can be divided such` `    ``// that for each set max-min <= K` `    ``static` `int` `minSetCount(``int``[] arr, ``int` `N, ``int` `K)` `    ``{` `      `  `        ``// Sort the input array` `        ``Array.Sort(arr);`   `        ``// Stores the count of set required` `        ``int` `setCount = 1;`   `        ``// Stores the beginning and ending` `        ``// of the current set` `        ``int` `begin = 0, end = 0;`   `        ``// Loop to iterate over the array` `        ``while` `(end < N) {`   `            ``// If arr[end] can be included` `            ``// in the current set else` `            ``// begin a new set` `            ``if` `(arr[end] - arr[begin] <= K) {` `                ``end++;` `            ``}` `            ``else` `{` `                ``// Increment the set count` `                ``setCount++;` `                ``begin = end;` `            ``}` `        ``}`   `        ``// Return answer` `        ``return` `setCount;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `Main(``string``[] args)` `    ``{` `        ``int``[] arr = { 5, 2, 9, 7, 3, 2, 4, 6, 14, 10 };` `        ``int` `N = arr.Length;` `        ``int` `K = 3;` `        ``Console.WriteLine(minSetCount(arr, N, K));` `    ``}` `}`   `// This code is contributed by AnkThon`

## Javascript

 ``

Output:

`4`

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

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