Minimum number of swaps required to sort an array | Set 2

Given an array of N distinct elements, find the minimum number of swaps required to sort the array.

Note: The problem is not asking to sort the array by the minimum number of swaps. The problem is to find the minimum swaps in which the array can be sorted.

Examples:

Input: arr[] = {4, 3, 2, 1}
Output: 2
Explanation: Swap index 0 with 3 and 1 with
2 to get the sorted array {1, 2, 3, 4}.

Input: arr[] = { 3, 5, 2, 4, 6, 8}
Output: 3
Explanation: 
Swap 4 and 5 so array = 3, 4, 2, 5, 6, 8
Swap 2 and 3 so array = 2, 4, 3, 5, 6, 8
Swap 4 and 3 so array = 2, 3, 4, 5, 6, 8
So the array is sorted.

This problem is already discussed in the previous article using graph. In this article another approach to solve this problem is discussed which is slightly different from the cycle approach.

Approach:
The idea is to create a vector of pair in C++ with first element as array values and second element as array indices. The next step is to sort the vector of pair according to the first element of the pair. After that traverse the vector and check if the index mapped with the value is correct or not, if not then keep swapping until the element is placed correctly and keep counting the number of swaps.



Algorithm:

  1. Create a vector of pairs and traverse the array and for every element of the array insert a element-index pair in the vector
  2. Traverse the vector from start to the end (loop counter is i).
  3. For every element of the pair where the second element(index) is not equal to i. Swap the ith element of the vector with the second element(index) th element of the vector
  4. If the second element(index) is equal to i then skip the iteration of the loop.
  5. if after the swap the second element(index) is not equal to i then decrement i.
  6. Increment the counter.

Implementation:

CPP

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP program to find the minimum number 
// of swaps required to sort an array
// of distinct element
  
#include<bits/stdc++.h>
using namespace std;
  
// Function to find minimum swaps to 
// sort an array
int findMinSwap(int arr[] , int n)
{
    // Declare a vector of pair     
    vector<pair<int,int>> vec(n);
      
    for(int i=0;i<n;i++)
    {
        vec[i].first=arr[i];
        vec[i].second=i;
    }
  
    // Sort the vector w.r.t the first
    // element of pair
    sort(vec.begin(),vec.end());
  
    int ans=0,c=0,j;
  
    for(int i=0;i<n;i++)
    {   
        // If the element is already placed
        // correct, then continue
        if(vec[i].second==i) 
            continue;
        else
        {
            // swap with its respective index 
            swap(vec[i].first,vec[vec[i].second].first);
            swap(vec[i].second,vec[vec[i].second].second); 
        
          
        // swap until the correct 
        // index matches
        if(i!=vec[i].second)
            --i;
          
        // each swap makes one element
        // move to its correct index, 
        // so increment answer
        ans++;
    }
  
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = {1, 5, 4, 3, 2};
      
    int n = sizeof(arr)/sizeof(arr[0]);
      
    cout<<findMinSwap(arr,n);
      
    return 0;

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to find the minimum number
# of swaps required to sort an array
# of distinct element
  
# Function to find minimum swaps to
# sort an array
def findMinSwap(arr, n):
      
    # Declare a vector of pair
    vec = []
  
    for i in range(n):
        vec.append([arr[i], i])
  
    # Sort the vector w.r.t the first
    # element of pair
    vec = sorted(vec)
  
    ans, c, j = -1, 0, 0
  
    for i in range(n):
          
        # If the element is already placed
        # correct, then continue
        if(vec[i][1] == i):
            continue
        else:
            # swap with its respective index
            vec[i][0], vec[vec[i][1]][1] = \
                vec[vec[i][1]][1], vec[i][0]
            vec[i][1], vec[vec[i][1]][1] = \
                vec[vec[i][1]][1], vec[i][1]
  
        # swap until the correct
        # index matches
        if(i != vec[i][1]):
            i -= 1
  
        # each swap makes one element
        # move to its correct index,
        # so increment answer
        ans += 1
  
    return ans
  
# Driver code
arr = [1, 5, 4, 3, 2]
n = len(arr)
print(findMinSwap(arr,n))
  
# This code is contributed by mohit kumar 29

chevron_right


Output:

2

Complexity Analysis:

  • Time Complexity: O(n Log n).
    Time required to sort the array is n log n.
  • Auxiliary Space: O(n).
    An extra array or vector is created. So, the space complexity is O(n log n)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.