Minimum Swaps 2


Problem Statement :


You are given an unordered array consisting of consecutive integers  [1, 2, 3, ..., n] without any duplicates. You are allowed to swap any two elements. Find the minimum number of swaps required to sort the array in ascending order.

Example

arr =  [ 7 , 1 , 3 , 2, 4 , 5, 6 ]

Perform the following steps:

i   arr                         swap (indices)
0   [7, 1, 3, 2, 4, 5, 6]   swap (0,3)
1   [2, 1, 3, 7, 4, 5, 6]   swap (0,1)
2   [1, 2, 3, 7, 4, 5, 6]   swap (3,4)
3   [1, 2, 3, 4, 7, 5, 6]   swap (4,5)
4   [1, 2, 3, 4, 5, 7, 6]   swap (5,6)
5   [1, 2, 3, 4, 5, 6, 7]
It took 5 swaps to sort the array.

Function Description

Complete the function minimumSwaps in the editor below.

minimumSwaps has the following parameter(s):

int arr[n]: an unordered array of integers


Returns

int: the minimum number of swaps to sort the array



Input Format

The first line contains an integer, n, the size of arr .
The second line contains n space-separated integers arr[ i ] .


Constraints

1  <=   n  <=  10^5
1  <=   arr[i]  <=  n


Sample Input 0

4
4 3 1 2


Sample Output 0

3



Solution :



title-img




                        Solution in C++ :

In   C++ :






vector<int>v[100003];
bool visit[100003];


int dfs(int i)
{
    visit[i] = true;
    int z = 1;
    
    for(auto x: v[i])
        if(!visit[x])
            z += dfs(x);
            
    return z;  
}


int minimumSwaps(vector<int> A) {
    
    for(int i = 0; i < A.size(); ++i )
        v[i].push_back(A[i]-1), v[A[i]-1].push_back(i);
    
    int c = 0;

    for(int i = 0; i < A.size(); ++i)
    {
        if(!visit[i])
            c += dfs(i) - 1;
    }
    
    return c;
}
                    


                        Solution in Java :

In   Java  :





static int minimumSwaps(int[] a) {
        int swap=0;
        for(int i=0;i<a.length;i++){
            if(i+1!=a[i]){
                int t=i;
                while(a[t]!=i+1){
                    t++;  
                }
                int temp=a[t];
                a[t]=a[i];
                a[i]=temp;
                swap++;
            }
        }
        return swap;

    }
                    


                        Solution in Python : 
                            
In   Python3  :




def minimumSwaps(arr):
    ref_arr = sorted(arr)
    index_dict = {v: i for i,v in enumerate(arr)}
    swaps = 0
    
    for i,v in enumerate(arr):
        correct_value = ref_arr[i]
        if v != correct_value:
            to_swap_ix = index_dict[correct_value]
            arr[to_swap_ix],arr[i] = arr[i], arr[to_swap_ix]
            index_dict[v] = to_swap_ix
            index_dict[correct_value] = i
            swaps += 1
            
    return swaps
                    


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