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. 

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 elements 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 <bits/stdc++.h>
using namespace std;
#define ll long long

// Function to minimize the difference
// between two elements of array
void minDiff(vector<ll> 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[1] - a[0];

    // 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<ll> arr = { 3, 8, 13, 30, 50 };

    int n = sizeof(arr) / sizeof(arr[0]);

    // 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[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
    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

Javascript

<script>

// Javascript implementation of the above approach

// Function to minimize the difference
// between two elements of array
function minDiff(a, n)
{

    // Find all the element which are
    // possible by multiplying
    // 2 to odd numbers
    for(let i = 0; i < n; i++)
    {
       if (a[i] % 2 == 1)
           a[i] *= 2;
    }

    // Sort the array
    a.sort((a, b) => a - b);

    let mindifference = a[1] - a[0];

    // Find the minimum difference
    // Iterate and find which adjacent
    // elements have the minimum difference
    for(let i = 1; i < a.length; i++)
    {
       mindifference = Math.min(mindifference,
                                a[i] - a[i - 1]);
    }

    // Print the minimum difference
     document.write(mindifference);
}


  // Driver Code

    // Given array
    let arr = [ 3, 8, 13, 30, 50 ];

    let n = arr.length;

    // Function call
    minDiff(arr, n);

</script>

Output:

2

Time complexity: O(nlogn)

Auxiliary Space: O(1)