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Minimum number of swaps required to sort an array

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Average time to solve is 10m

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Problem statement

You have been given an array 'ARR' of 'N' distinct elements.

Your task is to find the minimum no. of swaps required to sort the array.

For example:

For the given input array [4, 3, 2, 1], the minimum no. of swaps required to sort the array is 2, i.e. swap index 0 with 3 and 1 with 2 to form the sorted array [1, 2, 3, 4].

Detailed explanation ( Input/output format, Notes, Images )

Input Format:

The first line of input contains an integer ‘T’ representing the number of test cases. Then the test cases follow.

The first line of each test case contains an integer ‘N’ representing the size of the input array.

The second line of each test case contains the 'N' elements of the array separated by a single space.

Output Format:

For each test case, print a single line containing a single integer which represents the minimum no. of swaps required to sort the array.

The output for each test case is in a separate line.

Note:

You do not need to print anything; it has already been taken care of. Just implement the given function.

Constraints:

1 <= T <= 100
1 <= N <= 1000
0 <= ARR[i] <= 10 ^ 9

Where 'ARR[i]' is the value of the input array elements.

Time Limit: 1 sec

Sample Input 1:

Sample Output 1:

2
2

Explanation of Sample Input 1:

For the first test case, swap index 0 with 3 i.e. 4 -> 1 and 1 with 2 i.e. 3 -> 2 to form the sorted array {1, 2, 3, 4}.

For the second test case, swap index 1 with 4 i.e. 5 -> 2 and 2 with 3 i.e. 4 -> 3 to form the sorted array {1, 2, 3, 4, 5}.

Sample Input 2:

2
4
1 2 3 4
6
3 5 2 4 6 8

Sample Output 2:

0
3

Hint 1

Hint 2

Swap every element of the given array with its right index, if it is not at the right place.

Approaches (2)

Approach 1

Approach 2

Naive Approach

While iterating over the array, check the current element, and if not in the correct place, replace that element with the index of the element which should have come in this place.

Below is the algorithm:

Create a copy of the given input array and store it in temp.
Sort the temp array.
Iterate over the input array, and check whether the current element is at the right place or not by comparing it with temp[i].
1. If it is the right place, no need to do anything.
2. Else, increase the count by 1 and swap the current element with the right index. So that array will remain sorted till index i.

Time Complexity

O(N ^ 2), where ‘N’ is the size of the given array.

Since we are using indexOf() function to check the right index of every element of input array, having the worst case time complexity of O(N) for each array element. So, the overall complexity is O(N ^ 2).

Space Complexity

O(N), where ‘N’ is the size of the given array.

Since we are using a temp array to store the sorted form of a given input array, thus, the space complexity will be O(N).

(100% EXP penalty)

Minimum number of swaps required to sort an array

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