Minimum number of adjacent swaps for arranging similar elements together

Given an array of 2 * N positive integers where each array element lies between 1 to N and appears exactly twice in the array. The task is to find the minimum number of adjacent swaps required to arrange all similar array elements together.

Note: It is not necessary that the final array (after performing swaps) should be sorted.

Examples:



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.

Approach : This problem can be solved using greedy approach. Following are the steps :

  1. Keep an array visited[] which tells that visited[curr_ele] is false if swap operation has not been performed on curr_ele.
  2. Traverse through the original array and if the current array element has not been visited yet i.e. visited[arr[curr_ele]] = false, set it to true and iterate over another loop starting from the current position to the end of array.
  3. Initialize a variable count which will determine the number of swaps required to place the current element’s partner at its correct position.
  4. In nested loop, increment count only if the visited[curr_ele] is false (since if it is true, means curr_ele has already been placed at its correct position).
  5. If the current element’s partner is found in the nested loop, add up the value of count to the total answer.

Below is the implementation of above approach:

C++

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// C++ Program to find the minimum number of
// adjacent swaps to arrange similar items together
  
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

5

Time Complexity: O(N2)



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