通过一次交换最大和最小阵列元素之间的距离

给定arr []由[1，N]范围内的N个元素组成，任务是通过一次交换使最小和最大数组元素之间的距离最大化。
例子：

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. 
 

方法

1. 在数组中找到索引1和N。
2. 令minIdx和maxIdx分别为两个索引的最小值和最大值。
3. 现在，maxIdx – minIdx是两个元素之间的当前距离。可以通过以下两次交换将其最大化：
   • 将a [minIdx]替换为a [0]将距离增加minIdx 。
   • 将a [maxIdx]与a [N – 1]交换可使距离增加N – 1 – maxIdx 。

下面是上述方法的实现：

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[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

// 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

输出：

3

时间复杂度： O(N)