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:
Swappinf 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.

    Hence, maxIdx – minIdx + max(minIdx, N – 1 – maxIdx) gives us the required answer.

    Illustration:
    arr[] = {2, 4, 6, 1, 3, 5}
    Smallest element = 1
    Largest element = 6
    minIdx = 2
    maxIdx = 3
    Required Answer = maxIdx – minIdx + max(minIdx, N – 1 – maxIdx)
    = 3 – 2 + max(2, 5 – 3)
    = 1 + 2 = 3

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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 =
                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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Output:

3

Time Complexity: O(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.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.