将相似元素排列在一起的最小相邻交换数

给定一个由 2 * N 个正整数组成的数组，其中每个数组元素位于 1 到 N 之间，并且在数组中恰好出现两次。任务是找到将所有相似的数组元素排列在一起所需的最小相邻交换数。
注意：没有必要对最终数组（执行交换后）进行排序。
例子：

Input: arr[] = { 1, 2, 3, 3, 1, 2 } 
Output: 5 
After first swapping, array will be arr[] = { 1, 2, 3, 1, 3, 2 }, 
after second arr[] = { 1, 2, 1, 3, 3, 2 }, after third arr[] = { 1, 1, 2, 3, 3, 2 }, 
after fourth arr[] = { 1, 1, 2, 3, 2, 3 }, after fifth arr[] = { 1, 1, 2, 2, 3, 3 }
Input: arr[] = { 1, 2, 1, 2 } 
Output: 1 
arr[2] can be swapped with arr[1] to get the required position. 

方法：这个问题可以使用贪心方法来解决。以下是步骤：

1. 保留一个数组visited[] ，如果没有对curr_ele 执行交换操作，它告诉visited[curr_ele] 为false。
2. 遍历原始数组，如果当前数组元素尚未被访问，即visited[arr[curr_ele]] = false ，将其设置为true 并从当前位置开始迭代另一个循环到数组末尾。
3. 初始化一个变量计数，它将确定将当前元素的伙伴放置在其正确位置所需的交换次数。
4. 在嵌套循环中，仅当visited[curr_ele] 为false 时才增加计数（因为如果为true，则表示curr_ele 已被放置在其正确位置）。
5. 如果在嵌套循环中找到当前元素的伙伴，则将 count 的值与总答案相加。

下面是上述方法的实现：

C++

// C++ Program to find the minimum number of
// adjacent swaps to arrange similar items together
 
#include 
using namespace std;
 
// Function to find minimum swaps
int findMinimumAdjacentSwaps(int arr[], int N)
{
    // visited array to check if value is seen already
    bool visited[N + 1];
 
    int minimumSwaps = 0;
    memset(visited, false, sizeof(visited));
 
    for (int i = 0; i < 2 * N; i++) {
 
        // If the arr[i] is seen first time
        if (visited[arr[i]] == false) {
            visited[arr[i]] = true;
 
            // stores the number of swaps required to
            // find the correct position of current
            // element's partner
            int count = 0;
 
            for (int j = i + 1; j < 2 * N; j++) {
 
                // Increment count only if the current
                // element has not been visited yet (if is
                // visited, means it has already been placed
                // at its correct position)
                if (visited[arr[j]] == false)
                    count++;
 
                // If current element's partner is found
                else if (arr[i] == arr[j])
                    minimumSwaps += count;
            }
        }
    }
    return minimumSwaps;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 3, 1, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
    N /= 2;
 
    cout << findMinimumAdjacentSwaps(arr, N) << endl;
    return 0;
}

Java

// Java Program to find the minimum number of
// adjacent swaps to arrange similar items together
import java.util.*;
 
class solution
{
 
// Function to find minimum swaps
static int findMinimumAdjacentSwaps(int arr[], int N)
{
    // visited array to check if value is seen already
    boolean[] visited = new boolean[N + 1];
 
    int minimumSwaps = 0;
    Arrays.fill(visited,false);
    
 
    for (int i = 0; i < 2 * N; i++) {
 
        // If the arr[i] is seen first time
        if (visited[arr[i]] == false) {
            visited[arr[i]] = true;
 
            // stores the number of swaps required to
            // find the correct position of current
            // element's partner
            int count = 0;
 
            for (int j = i + 1; j < 2 * N; j++) {
 
                // Increment count only if the current
                // element has not been visited yet (if is
                // visited, means it has already been placed
                // at its correct position)
                if (visited[arr[j]] == false)
                    count++;
 
                // If current element's partner is found
                else if (arr[i] == arr[j])
                    minimumSwaps += count;
            }
        }
    }
    return minimumSwaps;
}
 
// Driver Code
public static void main(String args[])
{
    int arr[] = { 1, 2, 3, 3, 1, 2 };
    int N = arr.length;
    N /= 2;
 
    System.out.println(findMinimumAdjacentSwaps(arr, N));
     
}
}
// This code is contributed by
// Sanjit_Prasad

Python3

# Python3 Program to find the minimum number of
# adjacent swaps to arrange similar items together
 
# Function to find minimum swaps
def findMinimumAdjacentSwaps(arr, N) :
     
    # visited array to check if value is seen already
    visited = [False] * (N + 1)
 
    minimumSwaps = 0
 
    for i in range(2 * N) :
 
        # If the arr[i] is seen first time
        if (visited[arr[i]] == False) :
            visited[arr[i]] = True
 
            # stores the number of swaps required to
            # find the correct position of current
            # element's partner
            count = 0
 
            for j in range( i + 1, 2 * N) :
 
                # Increment count only if the current
                # element has not been visited yet (if is
                # visited, means it has already been placed
                # at its correct position)
                if (visited[arr[j]] == False) :
                    count += 1
 
                # If current element's partner is found
                elif (arr[i] == arr[j]) :
                    minimumSwaps += count
         
    return minimumSwaps
 
 
# Driver Code
if __name__ == "__main__" :
 
    arr = [ 1, 2, 3, 3, 1, 2 ]
    N = len(arr)
    N //= 2
 
    print(findMinimumAdjacentSwaps(arr, N))
 
# This code is contributed by Ryuga

C#

// C# Program to find the minimum
// number of adjacent swaps to
// arrange similar items together
using System;
 
class GFG
{
 
    // Function to find minimum swaps
    static int findMinimumAdjacentSwaps(int []arr, int N)
    {
        // visited array to check
        // if value is seen already
        bool[] visited = new bool[N + 1];
 
        int minimumSwaps = 0;
 
 
        for (int i = 0; i < 2 * N; i++)
        {
 
            // If the arr[i] is seen first time
            if (visited[arr[i]] == false)
            {
                visited[arr[i]] = true;
 
                // stores the number of swaps required to
                // find the correct position of current
                // element's partner
                int count = 0;
 
                for (int j = i + 1; j < 2 * N; j++)
                {
 
                    // Increment count only if the current
                    // element has not been visited yet (if is
                    // visited, means it has already been placed
                    // at its correct position)
                    if (visited[arr[j]] == false)
                        count++;
 
                    // If current element's partner is found
                    else if (arr[i] == arr[j])
                        minimumSwaps += count;
                }
            }
        }
        return minimumSwaps;
    }
 
    // Driver Code
    public static void Main(String []args)
    {
        int []arr = { 1, 2, 3, 3, 1, 2 };
        int N = arr.Length;
        N /= 2;
 
        Console.WriteLine(findMinimumAdjacentSwaps(arr, N));
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript

 
// Javascript Program to find the minimum number of
// adjacent swaps to arrange similar items together
 
// Function to find minimum swaps
function findMinimumAdjacentSwaps(arr, N)
{
    // visited array to check if value is seen already
    let visited = Array(N + 1).fill(false);
   
    let minimumSwaps = 0;    
   
    for (let i = 0; i < 2 * N; i++) {
   
        // If the arr[i] is seen first time
        if (visited[arr[i]] == false) {
            visited[arr[i]] = true;
   
            // stores the number of swaps required to
            // find the correct position of current
            // element's partner
            let count = 0;
   
            for (let j = i + 1; j < 2 * N; j++) {
   
                // Increment count only if the current
                // element has not been visited yet (if is
                // visited, means it has already been placed
                // at its correct position)
                if (visited[arr[j]] == false)
                    count++;
   
                // If current element's partner is found
                else if (arr[i] == arr[j])
                    minimumSwaps += count;
            }
        }
    }
    return minimumSwaps;
}
 
// driver code
 
     let arr = [ 1, 2, 3, 3, 1, 2 ];
     let N = arr.length;
     N = Math.floor(N / 2);
   
    document.write(findMinimumAdjacentSwaps(arr, N));
   

输出：

5

时间复杂度： O(N ² )
辅助空间： O(N)

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。