Maximize distance between smallest and largest Array elements by a single swap

Given an arr[] consisting of N elements in the range [1, N], the task is to maximize the distance between smallest and largest array element by a single swap.
Examples:

Input: arr[] = {1, 4, 3, 2} 
Output: 3 
Explanation:
Swapping of arr[1] and arr[3] maximizes the distance. 
Input: arr[] = {1, 6, 5, 3, 4, 7, 2} 
Output: 6 
Explanation:
Swapping of arr[5] and arr[6] maximizes the distance. 

Approach

1. Find the indices of 1 and N in the array.
2. Let minIdx and maxIdx be the minimum and maximum of the two indices, respectively.
3. Now, maxIdx – minIdx is the current distance between the two elements. It can be maximized by the maximum possible from the following two swaps: 
   • Swapping a[minIdx] with a[0] increasing the distance by minIdx.
   • Swapping a[maxIdx] with a[N – 1] increasing the distance by N – 1 – maxIdx.

Below is the implementation of the above approach:

C++

// C++ program maximize the
// distance between smallest
// and largest array element
// by a single swap
#include <bits/stdc++.h>
using namespace std;

// Function to maximize the distance
// between the smallest and largest
// array element by a single swap
int find_max_dist(int arr[], int N)
{

    int minIdx = -1, maxIdx = -1;

    for (int i = 0; i < N; i++) {
        if (arr[i] == 1 || arr[i] == N) {
            if (minIdx == -1)
                minIdx = i;
            else {
                maxIdx = i;
                break;
            }
        }
    }

    return maxIdx - minIdx
           + max(minIdx, N - 1 - maxIdx);
}

// Driver Code
int main()
{
    int arr[] = { 1, 4, 3, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << find_max_dist(arr, N) << endl;
    return 0;
}

Java

// Java program maximize the distance
// between smallest and largest array
// element by a single swap
import java.util.*;

class GFG{

// Function to maximize the distance
// between the smallest and largest
// array element by a single swap
static int find_max_dist(int arr[], int N)
{
    int minIdx = -1, maxIdx = -1;

    for(int i = 0; i < N; i++)
    {
       if (arr[i] == 1 || arr[i] == N)
       {
           if (minIdx == -1)
               minIdx = i;
           else
           {
               maxIdx = i;
               break;
           }
       }
    }
    return maxIdx - minIdx +
           Math.max(minIdx, N - 1 - maxIdx);
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 4, 3, 2 };
    int N = arr.length;

    System.out.print(find_max_dist(arr, N) + "\n");
}
}

// This code is contributed by Amit Katiyar

Python3

# Python3 program maximize the
# distance between smallest
# and largest array element
# by a single swap

# Function to maximize the distance
# between the smallest and largest
# array element by a single swap
def find_max_dist(arr, N):

    minIdx, maxIdx = -1, -1

    for i in range(N):
        if (arr[i] == 1 or arr[i] == N):
            if (minIdx == -1) :
                minIdx = i

            else :
                maxIdx = i
                break

    return (maxIdx - minIdx +
        max(minIdx, N - 1 - maxIdx))

# Driver code
arr = [ 1, 4, 3, 2 ]
N = len(arr)

print(find_max_dist(arr, N))

# This code is contributed by divyeshrabadiya07

C#

// C# program maximize the distance
// between smallest and largest array
// element by a single swap
using System;

class GFG{

// Function to maximize the distance
// between the smallest and largest
// array element by a single swap
static int find_max_dist(int []arr, int N)
{
    int minIdx = -1, maxIdx = -1;

    for(int i = 0; i < N; i++)
    {
       if (arr[i] == 1 || arr[i] == N)
       {
           if (minIdx == -1)
               minIdx = i;
           else
           {
               maxIdx = i;
               break;
           }
       }
    }
    return maxIdx - minIdx +
           Math.Max(minIdx, N - 1 - maxIdx);
}

// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 4, 3, 2 };
    int N = arr.Length;

    Console.Write(find_max_dist(arr, N) + "\n");
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>
// javascript program maximize the distance
// between smallest and largest array
// element by a single swap

// Function to maximize the distance
// between the smallest and largest
// array element by a single swap
function find_max_dist(arr , N)
{
    var minIdx = -1, maxIdx = -1;

    for(i = 0; i < N; i++)
    {
       if (arr[i] == 1 || arr[i] == N)
       {
           if (minIdx == -1)
               minIdx = i;
           else
           {
               maxIdx = i;
               break;
           }
       }
    }
    return maxIdx - minIdx +
           Math.max(minIdx, N - 1 - maxIdx);
}

// Driver Code
var arr = [ 1, 4, 3, 2 ];
var N = arr.length;

document.write(find_max_dist(arr, N) + "\n");

// This code is contributed by Amit Katiyar
</script>

Output:

3

Time Complexity: O(N)