Given an array **arr[]** consisting of **N** positive integers, the task is to minimize the maximum difference between any pair of array elements by multiplying any odd array element by **2** and dividing any even array element by **2**.

**Examples:**

Input:arr[] = {4, 1, 5, 20, 3}Output:3Explanation:

Operation 1: Multiplying arr[1] by 2 modifies arr[] to {4, 2, 5, 20, 3}.

Operation 2: Dividing arr[3] by 2 modifies arr[] to {4, 2, 5, 10, 3}.

Operation 3: Dividing arr[3] by 2 modifies arr[] to {4, 2, 5, 5, 3}.

Therefore, the minimum of the maximum difference of any pair in the array = 5 – 2 = 3.

Input:arr[] = {1, 2, 5, 9}Output:7Explanation:

Operation 1: Multiplying arr[0] by 2 modifies arr[] to { 2, 2, 5, 9 }

Operation 2: Multiplying arr[2] by 2 modifies arr[] to {2, 2, 10, 9 }

Therefore, the minimum of the maximum difference of any pair in the array = 9 – 2 = 7.

**Approach:** Follow the steps below to solve the given problem:

- First, insert all array elements in a Set
**S**. If the array element is even, then insert it as it is. Otherwise, convert it into even by**multiplying it by 2**. - Store the difference between the last and first element in the Set
**S**in a variable, say**res**. - Traverse the set
**S**in the reverse order and do the following:- Update the value of res to the maximum of
**res**and the difference between the first and the current element of the set. - Remove the current element from the Set.
- Insert
**(current element) / 2**into the set. - If the value of the current element is odd, then no array more elements can make the maximum difference smaller. Hence break out of the loop.

- Update the value of res to the maximum of
- After completing the above steps, print the value of
**res**as the resultant difference.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to minimize the maximum` `// difference between any pair of elements` `// of the array by the given operations` `int` `minimumMaxDiff(vector<` `int` `>& nums)` `{` ` ` `set<` `int` `> s;` ` ` `// Traverse the array` ` ` `for` `(` `int` `i = 0; i < nums.size(); i++) {` ` ` `// If current element is even` ` ` `if` `(nums[i] % 2 == 0)` ` ` `// Insert it into even` ` ` `s.insert(nums[i]);` ` ` `// Otherwise` ` ` `else` ` ` `// Make it even by multiplying` ` ` `// by 2 and insert it into set` ` ` `s.insert(nums[i] * 2);` ` ` `}` ` ` `// Calculate difference between first` ` ` `// and the last element of the set` ` ` `int` `res = *s.rbegin() - *s.begin();` ` ` `// Iterate until difference is minimized` ` ` `while` `(*s.rbegin() % 2 == 0) {` ` ` `int` `x = *s.rbegin();` ` ` `// Erase the current element` ` ` `s.erase(x);` ` ` `// Reduce current element by half` ` ` `// and insert it into the Set` ` ` `s.insert(x / 2);` ` ` `// Update difference` ` ` `res = min(res, *s.rbegin()` ` ` `- *s.begin());` ` ` `}` ` ` `// Return the resultant difference` ` ` `return` `res;` `}` `// Driver Code` `int` `main()` `{` ` ` `vector<` `int` `> arr = { 1, 2, 5, 9 };` ` ` `cout << minimumMaxDiff(arr);` `}` |

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## Java

`// Java program for the above approach` `import` `java.io.*;` `import` `java.util.*;` `class` `GFG` `{` ` ` `// Function to minimize the maximum` ` ` `// difference between any pair of elements` ` ` `// of the array by the given operations` ` ` `static` `int` `minimumMaxDiff(` `int` `[] nums)` ` ` `{` ` ` `TreeSet<Integer> s = ` `new` `TreeSet<Integer>();` ` ` `// Traverse the array` ` ` `for` `(` `int` `i = ` `0` `; i < nums.length; i++) ` ` ` `{` ` ` `// If current element is even` ` ` `if` `(nums[i] % ` `2` `== ` `0` `)` ` ` `// Insert it into even` ` ` `s.add(nums[i]);` ` ` `// Otherwise` ` ` `else` ` ` `// Make it even by multiplying` ` ` `// by 2 and insert it into set` ` ` `s.add(nums[i] * ` `2` `);` ` ` `}` ` ` `// Calculate difference between first` ` ` `// and the last element of the set` ` ` `int` `res = s.last() - s.first();` ` ` `// Iterate until difference is minimized` ` ` `while` `(s.last() % ` `2` `== ` `0` `)` ` ` `{` ` ` `int` `x = s.last();` ` ` `// Erase the current element` ` ` `s.remove(x);` ` ` `// Reduce current element by half` ` ` `// and insert it into the Set` ` ` `s.add(x / ` `2` `);` ` ` `// Update difference` ` ` `res = Math.min(res, s.last() - s.first());` ` ` `}` ` ` `// Return the resultant difference` ` ` `return` `res;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `[] arr = ` `new` `int` `[] { ` `1` `, ` `2` `, ` `5` `, ` `9` `};` ` ` `System.out.print(minimumMaxDiff(arr));` ` ` `}` `}` `// This code is contributed by jithin` |

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## Python3

`# Python3 program for the above approach` `# Function to minimize the maximum` `# difference between any pair of elements` `# of the array by the given operations` `def` `minimumMaxDiff(nums):` ` ` ` ` `s ` `=` `{}` ` ` `# Traverse the array` ` ` `for` `i ` `in` `range` `(` `len` `(nums)):` ` ` `# If current element is even` ` ` `if` `(nums[i] ` `%` `2` `=` `=` `0` `):` ` ` `# Insert it into even` ` ` `s[nums[i]] ` `=` `1` ` ` `# Otherwise` ` ` `else` `:` ` ` `# Make it even by multiplying` ` ` `# by 2 and insert it into set` ` ` `s[nums[i] ` `*` `2` `] ` `=` `1` ` ` `# Calculate difference between first` ` ` `# and the last element of the set` ` ` `sr ` `=` `list` `(s.keys())` ` ` `res ` `=` `sr[` `-` `1` `] ` `-` `sr[` `0` `]` ` ` `# Iterate until difference is minimized` ` ` `while` `(` `list` `(s.keys())[` `-` `1` `] ` `%` `2` `=` `=` `0` `):` ` ` `r ` `=` `list` `(s.keys())` ` ` `x ` `=` `r[` `-` `1` `]` ` ` `# Erase the current element` ` ` `del` `s[x]` ` ` `# Reduce current element by half` ` ` `# and insert it into the Set` ` ` `s[x ` `/` `/` `2` `] ` `=` `1` ` ` `rr ` `=` `list` `(s.keys())` ` ` `# Update difference` ` ` `res ` `=` `min` `(res, rr[` `-` `1` `] ` `-` `r[` `0` `])` ` ` `# Return the resultant difference` ` ` `return` `res` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `arr ` `=` `[ ` `1` `, ` `2` `, ` `5` `, ` `9` `]` ` ` ` ` `print` `(minimumMaxDiff(arr))` ` ` `# This code is contributed by mohit kumar 29` |

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## C#

`// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;` `using` `System.Linq; ` `class` `GFG ` `{` ` ` `// Function to minimize the maximum` ` ` `// difference between any pair of elements` ` ` `// of the array by the given operations` ` ` `static` `int` `minimumMaxDiff(` `int` `[] nums)` ` ` `{` ` ` `HashSet<` `int` `> s = ` `new` `HashSet<` `int` `>();` ` ` `// Traverse the array` ` ` `for` `(` `int` `i = 0; i < nums.Length; i++) {` ` ` `// If current element is even` ` ` `if` `(nums[i] % 2 == 0)` ` ` `// Insert it into even` ` ` `s.Add(nums[i]);` ` ` `// Otherwise` ` ` `else` ` ` `// Make it even by multiplying` ` ` `// by 2 and insert it into set` ` ` `s.Add(nums[i] * 2);` ` ` `}` ` ` `// Calculate difference between first` ` ` `// and the last element of the set` ` ` `int` `res = s.Last() - s.First();` ` ` `// Iterate until difference is minimized` ` ` `while` `(s.Last() % 2 == 0) {` ` ` `int` `x = s.Last();` ` ` `// Erase the current element` ` ` `s.Remove(x);` ` ` `// Reduce current element by half` ` ` `// and insert it into the Set` ` ` `s.Add(x / 2);` ` ` `// Update difference` ` ` `res = Math.Min(res, s.Last() - s.First());` ` ` `}` ` ` `// Return the resultant difference` ` ` `return` `res;` ` ` `}` ` ` `// Driver code` ` ` `static` `public` `void` `Main()` ` ` `{` ` ` `int` `[] arr = ` `new` `int` `[] { 1, 2, 5, 9 }; ` ` ` `Console.WriteLine(minimumMaxDiff(arr));` ` ` `}` `}` `// This code is contributed by Dharanendra L V` |

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**Output:**

7

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

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