# Number of steps to sort the array by changing order of three elements in each step

Given an array **arr[]** of size **N** consisting of unique elements in the range **[0, N-1]**, the task is to find **K** which is the number of steps required to sort the given array by selecting three distinct elements and rearranging them. And also, print the indices selected in those K steps in K lines.

For example, in the array {5, 4, 3, 2, 1, 0}, one possible way to sort the given array by selecting three distinct elements is to select the numbers {2, 1, 0} and sort them as {0, 1, 2} thereby making the array {5, 4, 3, 0, 1, 2}. Similarly, the remaining operations are performed and the indices selected ({3, 4, 5} in the above case) are printed in separate lines.

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**Examples:**

Input:arr[] = {0, 5, 4, 3, 2, 1}Output:

2

1 2 5

2 5 4Explanation:

The above array can be sorted in 2 steps:

Step I: We change the order of elements at indices 1, 2, 5 then the array becomes {0, 1, 5, 3, 2, 4}.

Step II: We again change the order of elements at the indices 2, 5, 4 then the array becomes {0, 1, 2, 3, 4, 5} which is sorted.

Input:arr[] = {0, 3, 1, 6, 5, 2, 4}Output:-1Explanation:

The above array cannot be sorted in any number of steps.

Suppose we choose indices 1, 3, 2 then the array becomes {0, 1, 6, 3, 5, 2, 4}

After that, we choose indices 2, 6, 4 then the array becomes {0, 1, 5, 3, 4, 2, 6}.

Now only two elements are left unsorted so we cannot choose 3 elements so the above array cannot be sorted. We can try with any order of indices and we will always be left with 2 elements unsorted.

**Approach:** The idea is to first count the elements which are not sorted and insert them in an unordered set. If count is 0 then we don’t need any number of steps for sorting the array so we print 0 and exit. Else, we first erase all the elements from the set for which i = A[A[i]] then we perform the following operation till the set becomes empty:

- We select all the possible combination of indices(if available) such that minimum two elements will get sorted.
- Now, change the order of the elements and erase them from the set if i = A[i].
- Then, we are left with only those elements such that i = A[A[i]] and the count of those must be a multiple of 4 otherwise it is not possible to sort the elements.
- Then, we choose any two pair and perform changing the order of elements two times. Then all the four chosen elements will get sorted.
- We store all the indices which are involved in the changing of orders of elements in a vector and print it as the answer.

Let’s understand the above approach with an example. Let the array arr[] = {0, 8, 9, 10, 1, 7, 12, 4, 3, 2, 6, 5, 11}. Then:

- Initially, the set will contain all the 12 elements and there are no elements such that i = A[A[i]].
- Now, {11, 5, 7} are chosen and the order of the elements are changed. Then, arr[] = {0, 8, 9, 10, 1, 5, 12, 7, 3, 2, 6, 4, 11}.
- Now, {11, 4, 1} are chosen and the order of the elements are changed. Then, arr[] = {0, 1, 9, 10, 4, 5, 12, 7, 3, 2, 6, 8, 11}.
- Now, {11, 8, 3} are chosen and the order of the elements are changed. Then, arr[] = {0, 1, 9, 3, 4, 5, 12, 7, 8, 2, 6, 10, 11}.
- Now, {11, 10, 6} are chosen and the order of the elements are changed. Then, arr[] = {0, 1, 9, 3, 4, 5, 6, 7, 8, 2, 10, 12, 11}.
- After the above step, we are left with two pairs of unsorted elements such that i = A[A[i]].
- Finally, {2, 11, 9} and {11, 9, 5} are chosen and reordered. Then, arr[] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} which is sorted.

Below is the implementation of the above approach:

`// C++ program to sort the array` `// by changing the order of` `// three elements` ` ` `#include <bits/stdc++.h>` `using` `namespace` `std;` ` ` `// Function to change the order of` `// the elements having a temporary` `// vector and the required indices` `// as the arguments` `void` `cngorder(vector<` `int` `>& v, ` `int` `i,` ` ` `int` `j, ` `int` `k)` `{` ` ` `int` `temp = v[k];` ` ` `v[k] = v[j];` ` ` `v[j] = v[i];` ` ` `v[i] = temp;` `}` ` ` `// Function to sort the elements having` `// the given array and its size.` `void` `sortbyorder3(vector<` `int` `>& A, ` `int` `n)` `{` ` ` ` ` `// Flag to check whether the sorting` ` ` `// is possible or not` ` ` `bool` `flag = 0;` ` ` ` ` `int` `count = 0;` ` ` ` ` `// Set that will contains unsorted` ` ` `// elements` ` ` `unordered_set<` `int` `> s;` ` ` ` ` `// Iterating through the elements` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` ` ` `// Inserting the required elements` ` ` `// in the set` ` ` `if` `(i != A[i])` ` ` `count++, s.insert(i);` ` ` `}` ` ` ` ` `// When the given array is` ` ` `// already sorted` ` ` `if` `(count == 0)` ` ` `cout << ` `"0"` `<< endl;` ` ` ` ` `else` `{` ` ` ` ` `// Vector that will contain` ` ` `// the answer` ` ` `vector<vector<` `int` `> > ans;` ` ` ` ` `// Temporary vector to store` ` ` `// the indices` ` ` `vector<` `int` `> vv;` ` ` ` ` `int` `x, y, z;` ` ` ` ` `count = 0;` ` ` ` ` `// Loop that will execute till the` ` ` `// set becomes empty` ` ` `while` `(!s.empty()) {` ` ` `auto` `it = s.begin();` ` ` `int` `i = *it;` ` ` ` ` `// Check for the condition` ` ` `if` `(i == A[A[i]]) {` ` ` `s.erase(i);` ` ` `s.erase(A[i]);` ` ` `continue` `;` ` ` `}` ` ` ` ` `// Case when the minimum two` ` ` `// elements will get sorted` ` ` `else` `{` ` ` `x = A[i], y = A[A[i]], z = A[A[A[i]]];` ` ` `vv.push_back(x), vv.push_back(y),` ` ` `vv.push_back(z);` ` ` ` ` `// Changing the order of elements` ` ` `cngorder(A, x, y, z);` ` ` ` ` `// Pushing the indices to the` ` ` `// answer vector` ` ` `ans.push_back(vv);` ` ` ` ` `// If the third element also` ` ` `// gets sorted` ` ` `if` `(vv[0] == A[vv[0]])` ` ` `s.erase(vv[0]);` ` ` ` ` `// Erasing the two sorted elements` ` ` `// from the set` ` ` `s.erase(vv[1]), s.erase(vv[2]);` ` ` `vv.clear();` ` ` `}` ` ` `}` ` ` ` ` `count = 0;` ` ` ` ` `// The count of the remaining` ` ` `// unsorted elements` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `if` `(i != A[i])` ` ` `count++;` ` ` `}` ` ` ` ` `// If the count of the left` ` ` `// unsorted elements is not` ` ` `// a multiple of 4, then` ` ` `// sorting is not possible` ` ` `if` `(count % 4 != 0)` ` ` `flag = 1;` ` ` ` ` `// Only the elements such that` ` ` `// i = A[A[i]] are left` ` ` `// for sorting` ` ` `else` `{` ` ` ` ` `// Indices of any one element` ` ` `// from the two pairs that` ` ` `// will be sorted in 2 steps` ` ` `int` `i1 = -1, i2 = -1;` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` ` ` `// Index of any element of` ` ` `// the pair` ` ` `if` `(A[i] != i && i1 == -1) {` ` ` `i1 = i;` ` ` `}` ` ` ` ` `// When we find the second` ` ` `// pair and the index of` ` ` `// any one element is stored` ` ` `else` `if` `(A[i] != i && i1 != -1` ` ` `&& i2 == -1) {` ` ` `if` `(i1 == A[i])` ` ` `continue` `;` ` ` `else` ` ` `i2 = i;` ` ` `}` ` ` ` ` `// When we got both the pair` ` ` `// of elements` ` ` `if` `(i1 != -1 && i2 != -1) {` ` ` ` ` `// Remaining two indices` ` ` `// of the elements` ` ` `int` `i3 = A[i1], i4 = A[i2];` ` ` ` ` `// The first order of indices` ` ` `vv.push_back(i1),` ` ` `vv.push_back(i2),` ` ` `vv.push_back(A[i1]);` ` ` ` ` `// Pushing the indices to the` ` ` `// answer vector` ` ` `ans.push_back(vv);` ` ` `vv.clear();` ` ` ` ` `// The second order of indices` ` ` `vv.push_back(i2),` ` ` `vv.push_back(A[i1]),` ` ` `vv.push_back(A[i2]);` ` ` ` ` `// Pushing the indices to the` ` ` `// answer vector` ` ` `ans.push_back(vv);` ` ` `vv.clear();` ` ` ` ` `// Changing the order of the` ` ` `// first combination of` ` ` `// the indices` ` ` `cngorder(A, i1, i2, i3);` ` ` ` ` `// Changing the order of the` ` ` `// second combination of` ` ` `// the indices after which all` ` ` `// the 4 elements will be sorted` ` ` `cngorder(A, i2, i3, i4);` ` ` ` ` `i1 = -1, i2 = -1;` ` ` `}` ` ` `}` ` ` `}` ` ` ` ` `// If the flag value is 1` ` ` `// the sorting is not possible` ` ` `if` `(flag == 1)` ` ` `cout << ` `"-1"` `<< endl;` ` ` ` ` `else` `{` ` ` ` ` `// Printing the required number` ` ` `// of steps` ` ` `cout << ans.size() << endl;` ` ` ` ` `// Printing the indices involved` ` ` `// in the shifting` ` ` `for` `(` `int` `i = 0; i < ans.size(); i++) {` ` ` `cout << ans[i][0]` ` ` `<< ` `" "` `<< ans[i][1]` ` ` `<< ` `" "` `<< ans[i][2]` ` ` `<< endl;` ` ` `}` ` ` `}` ` ` `}` `}` ` ` `// Driver code` `int` `main()` `{` ` ` ` ` `int` `n;` ` ` `vector<` `int` `> A{ 0, 8, 9, 10, 1, 7, 12,` ` ` `4, 3, 2, 6, 5, 11 };` ` ` `n = A.size();` ` ` ` ` `// Calling the sorting function` ` ` `sortbyorder3(A, n);` ` ` ` ` `return` `0;` `}` |

**Output:**

6 11 5 7 11 4 1 11 8 3 11 10 6 2 11 9 11 9 12

**Time Complexity:** *O(N)*, where N is the size of the array.