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 :

- Keep an array
*visited[]*which tells that visited[curr_ele] is false if swap operation has not been performed on curr_ele. - 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. - Initialize a variable count which will determine the number of swaps required to place the current element’s partner at its correct position.
- 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).
- 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++ 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; }

**Output:**

5

**Time Complexity:** O(N^{2})

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