# Minimize difference after changing all odd elements to even

Given an array arr[] of N positive integers. We have to perform one operation on every odd element in the given array i.e., multiply every odd element by 2 in the given array, the task is to find the minimum difference between any two elements in the array after performing the given operation.

Examples:

Input: arr[] = {2, 8, 15, 29, 40}
Output: 1
Explanation:
Multiply the third element 15 by 2 so it will become 30.
Now you have 30 and 29, so the minimum difference will become 1.

Input: arr[] = { 3, 8, 13, 30, 50 }
Output : 2
Explanation:
Multiply 3 by 2 so it will become 6.
Now you have 6 and 8, so the minimum difference will become 2.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

1. Convert every odd number in the given array to even by multiplying it by 2.
2. Sort the given array in increasing order.
3. Find the minimum difference between any two consecutive element in the above sorted array.
4. The difference calculated above is the minimum difference between any two elements in the array after performing the given operation.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` `#define ll long long ` ` `  `// Function to minimize the difference ` `// between two elements of array ` `void` `minDiff(vector a, ``int` `n) ` `{ ` ` `  `    ``// Find all the element which are ` `    ``// possible by multiplying ` `    ``// 2 to odd numbers ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(a[i] % 2 == 1) ` `            ``a.push_back(a[i] * 2); ` `    ``} ` ` `  `    ``// Sort the array ` `    ``sort(a.begin(), a.end()); ` ` `  `    ``ll mindifference = a - a; ` ` `  `    ``// Find the minimum difference ` `    ``// Iterate and find which adjacent ` `    ``// elements have the minimum difference ` `    ``for` `(``int` `i = 1; i < a.size(); i++) { ` ` `  `        ``mindifference = min(mindifference, ` `                            ``a[i] - a[i - 1]); ` `    ``} ` ` `  `    ``// Print the minimum difference ` `    ``cout << mindifference << endl; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Given array ` `    ``vector arr = { 3, 8, 13, 30, 50 }; ` ` `  `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``// Function Call ` `    ``minDiff(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program for the above approach  ` `import` `java.util.*; ` `class` `GFG{ ` `     `  `// Function to minimize the difference  ` `// between two elements of array  ` `public` `static` `void` `minDiff(``long``[] a, ``int` `n)  ` `{  ` `     `  `    ``// Find all the element which are  ` `    ``// possible by multiplying  ` `    ``// 2 to odd numbers  ` `    ``for``(``int` `i = ``0``; i < n; i++) ` `    ``{  ` `       ``if` `(a[i] % ``2` `== ``1``)  ` `           ``a[i] *= ``2``;  ` `    ``}  ` ` `  `    ``// Sort the array  ` `    ``Arrays.sort(a);  ` ` `  `    ``long` `mindifference = a[``1``] - a[``0``];  ` ` `  `    ``// Find the minimum difference  ` `    ``// Iterate and find which adjacent  ` `    ``// elements have the minimum difference  ` `    ``for``(``int` `i = ``1``; i < a.length; i++) ` `    ``{  ` `       ``mindifference = Math.min(mindifference,  ` `                                ``a[i] - a[i - ``1``]);  ` `    ``}  ` `     `  `    ``// Print the minimum difference  ` `    ``System.out.println(mindifference);  ` `}  ` ` `  `// Driver Code  ` `public` `static` `void` `main(String []args)  ` `{  ` `     `  `    ``// Given array  ` `    ``long` `[] arr = { ``3``, ``8``, ``13``, ``30``, ``50` `};  ` ` `  `    ``int` `n = arr.length;  ` ` `  `    ``// Function call  ` `    ``minDiff(arr, n);  ` `}  ` `} ` ` `  `// This code is contributed by jrishabh99 `

## Python3

 `# Python3 program for the above approach ` ` `  `# Function to minimize the difference ` `# between two elements of array ` `def` `minDiff(a,n): ` ` `  `    ``# Find all the element which are ` `    ``# possible by multiplying ` `    ``# 2 to odd numbers ` `    ``for` `i ``in` `range``(n): ` `        ``if` `(a[i] ``%` `2` `=``=` `1``): ` `            ``a.append(a[i] ``*` `2``) ` ` `  `    ``# Sort the array ` `    ``a ``=` `sorted``(a) ` ` `  `    ``mindifference ``=` `a[``1``] ``-` `a[``0``] ` ` `  `    ``# Find the minimum difference ` `    ``# Iterate and find which adjacent ` `    ``# elements have the minimum difference ` `    ``for` `i ``in` `range``(``1``, ``len``(a)): ` ` `  `        ``mindifference ``=` `min``(mindifference, ` `                            ``a[i] ``-` `a[i ``-` `1``]) ` ` `  `    ``# Print the minimum difference ` `    ``print``(mindifference) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=` `[``3``, ``8``, ``13``, ``30``, ``50``] ` ` `  `    ``n ``=` `len``(arr) ` ` `  `    ``# Function Call ` `    ``minDiff(arr, n) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# program for the above approach  ` `using` `System; ` `class` `GFG{ ` `     `  `// Function to minimize the difference  ` `// between two elements of array  ` `public` `static` `void` `minDiff(``long``[] a, ``int` `n)  ` `{  ` `     `  `    ``// Find all the element which are  ` `    ``// possible by multiplying  ` `    ``// 2 to odd numbers  ` `    ``for``(``int` `i = 0; i < n; i++) ` `    ``{  ` `        ``if` `(a[i] % 2 == 1)  ` `            ``a[i] *= 2;  ` `    ``}  ` ` `  `    ``// Sort the array  ` `    ``Array.Sort(a);  ` ` `  `    ``long` `mindifference = a - a;  ` ` `  `    ``// Find the minimum difference  ` `    ``// Iterate and find which adjacent  ` `    ``// elements have the minimum difference  ` `    ``for``(``int` `i = 1; i < a.Length; i++) ` `    ``{  ` `        ``mindifference = Math.Min(mindifference,  ` `                                 ``a[i] - a[i - 1]);  ` `    ``}  ` `     `  `    ``// Print the minimum difference  ` `    ``Console.Write(mindifference);  ` `}  ` ` `  `// Driver Code  ` `public` `static` `void` `Main()  ` `{  ` `     `  `    ``// Given array  ` `    ``long` `[]arr = { 3, 8, 13, 30, 50 };  ` ` `  `    ``int` `n = arr.Length;  ` ` `  `    ``// Function call  ` `    ``minDiff(arr, n);  ` `}  ` `} ` ` `  `// This code is contributed by Code_Mech `

Output:

```2
```

Time Complexity: O(N*log2N)

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