# 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 and arr maximizes the distance.

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

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

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 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++

 `// C++ program maximize the ` `// distance between smallest ` `// and largest array element ` `// by a single swap ` `#include ` `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); ` `    ``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 `

Output:

```3
```

Time Complexity: O(N)

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