# Partition a set into two subsets such that difference between max of one and min of other is minimized

Given an array **arr[]** of **N** integers, the task is to split the array into two subsets such that the absolute difference between the maximum of first subset and minimum of second subset is minimum.**Examples:**

Input:arr[] = {3, 1, 2, 6, 4}Output:1Explanation:

Splitting the given array in two subsets, A = [1, 2, 4], B = [3, 6]. Difference of maximum of first set is 4 and minimum of second set is 3 and their difference is 1.Input:arr[] = {2, 1, 3, 2, 4, 3}Output:0Explanation:

Splitting the given array in two subsets, A = [1, 2, 2, 3], B = [3, 4]. Difference of maximum of first set is 3 and minimum of second set is 3 and their difference is 0.

**Approach:** To solve the above problem we have to find the two integers such that **m** and **n** such that max of first set is **m** and the min of second set is **n**. The idea is to sort the given array ascending order and after sorting the array, the minimum difference between the consecutive element is the required minimum difference after partitioning the array elements into subsets.

Below is the implementation of above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to split the array` `int` `splitArray(` `int` `arr[], ` `int` `N)` `{` ` ` `// Sort the array in increasing order` ` ` `sort(arr, arr + N);` ` ` `int` `result = INT_MAX;` ` ` `// Calculating the max difference` ` ` `// between consecutive elements` ` ` `for` `(` `int` `i = 1; i < N; i++) {` ` ` `result = min(result,` ` ` `arr[i] - arr[i - 1]);` ` ` `}` ` ` `// Return the final minimum difference` ` ` `return` `result;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given array arr[]` ` ` `int` `arr[] = { 3, 1, 2, 6, 4 };` ` ` `// Size of array` ` ` `int` `N = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `// Function Call` ` ` `cout << splitArray(arr, N);` ` ` `return` `0;` `}` |

## Java

`// java program for the above approach` `import` `java.util.*;` `class` `GFG{` `// Function to split the array` `static` `int` `splitArray(` `int` `arr[], ` `int` `N)` `{` ` ` `// Sort the array in increasing order` ` ` `Arrays.sort(arr);` ` ` `int` `result = Integer.MAX_VALUE;` ` ` `// Calculating the max difference` ` ` `// between consecutive elements` ` ` `for` `(` `int` `i = ` `1` `; i < N; i++)` ` ` `{` ` ` `result = Math.min(result,` ` ` `arr[i] - arr[i - ` `1` `]);` ` ` `}` ` ` `// Return the final minimum difference` ` ` `return` `result;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `// Given array arr[]` ` ` `int` `arr[] = { ` `3` `, ` `1` `, ` `2` `, ` `6` `, ` `4` `};` ` ` `// Size of array` ` ` `int` `N = arr.length;` ` ` `// Function Call` ` ` `System.out.print(splitArray(arr, N));` `}` `}` `// This code is contributed by shivanisinghss2110` |

## Python3

`# Python3 program for the above approach` `# Function to split the array` `def` `splitArray(arr, N):` ` ` ` ` `# Sort the array in increasing` ` ` `# order` ` ` `arr ` `=` `sorted` `(arr)` ` ` `result ` `=` `10` `*` `*` `9` ` ` `# Calculating the max difference` ` ` `# between consecutive elements` ` ` `for` `i ` `in` `range` `(` `1` `, N):` ` ` `result ` `=` `min` `(result, arr[i] ` `-` `arr[i ` `-` `1` `])` ` ` `# Return the final minimum difference` ` ` `return` `result` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `# Given array arr[]` ` ` `arr ` `=` `[ ` `3` `, ` `1` `, ` `2` `, ` `6` `, ` `4` `]` ` ` `# Size of array` ` ` `N ` `=` `len` `(arr)` ` ` `# Function Call` ` ` `print` `(splitArray(arr, N))` `# This code is contributed by mohit kumar 29` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG{` `// Function to split the array` `static` `int` `splitArray(` `int` `[]arr, ` `int` `N)` `{` ` ` `// Sort the array in increasing order` ` ` `Array.Sort(arr);` ` ` `int` `result = Int32.MaxValue;` ` ` `// Calculating the max difference` ` ` `// between consecutive elements` ` ` `for` `(` `int` `i = 1; i < N; i++)` ` ` `{` ` ` `result = Math.Min(result,` ` ` `arr[i] - arr[i - 1]);` ` ` `}` ` ` `// Return the final minimum difference` ` ` `return` `result;` `}` `// Driver Code` `public` `static` `void` `Main()` `{` ` ` `// Given array arr[]` ` ` `int` `[]arr = { 3, 1, 2, 6, 4 };` ` ` `// Size of array` ` ` `int` `N = arr.Length;` ` ` `// Function Call` ` ` `Console.Write(splitArray(arr, N));` `}` `}` `// This code is contributed by Code_Mech` |

## Javascript

`<script>` ` ` `// Javascript program for the above approach` ` ` ` ` `// Function to split the array` ` ` `function` `splitArray(arr, N)` ` ` `{` ` ` `// Sort the array in increasing order` ` ` `arr.sort();` ` ` `let result = Number.MAX_VALUE;` ` ` `// Calculating the max difference` ` ` `// between consecutive elements` ` ` `for` `(let i = 1; i < N; i++) {` ` ` `result = Math.min(result,` ` ` `arr[i] - arr[i - 1]);` ` ` `}` ` ` `// Return the final minimum difference` ` ` `return` `result;` ` ` `}` ` ` ` ` `// Given array arr[]` ` ` `let arr = [ 3, 1, 2, 6, 4 ];` ` ` ` ` `// Size of array` ` ` `let N = arr.length;` ` ` ` ` `// Function Call` ` ` `document.write(splitArray(arr, N));` ` ` `</script>` |

**Output:**

1

**Time Complexity:** *O(N*log N)*

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. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend **live classes **with experts, please refer **DSA Live Classes for Working Professionals **and **Competitive Programming Live for Students**.