Arrange the array such that upon performing given operations an increasing order is obtained

Given an array arr[] of size N, the task is to print the arrangement of the array such that upon performing following operations on this arrangement, an increasing order is obtained as the output:

1. Take the first (0th index) element, remove it from the array and print it.
2. If there are still elements left in the array, move the next top element to the end of the array.
3. Repeat the above steps until array is not empty.

Examples:

Input: arr = {1, 2, 3, 4, 5, 6, 7, 8}
Output: {1, 5, 2, 7, 3, 6, 4, 8}
Explanation:
Let initial array be {1, 5, 2, 7, 3, 6, 4, 8}, where 1 is the top of the array.
1 is printed, and 5 is moved to the end. The array is now {2, 7, 3, 6, 4, 8, 5}.
2 is printed, and 7 is moved to the end. The array is now {3, 6, 4, 8, 5, 7}.
3 is printed, and 6 is moved to the end. The array is now {4, 8, 5, 7, 6}.
4 is printed, and 8 is moved to the end. The array is now {5, 7, 6, 8}.
5 is printed, and 7 is moved to the end. The array is now {6, 8, 7}.
6 is printed, and 8 is moved to the end. The array is now {7, 8}.
7 is printed, and 8 is moved to the end. The array is now {8}.
8 is printed.
The printing order is 1, 2, 3, 4, 5, 6, 7, 8 which is increasing.

Input: arr = {3, 2, 25, 2, 3, 1, 2, 6, 5, 45, 4, 89, 5}
Output: {1, 45, 2, 5, 2, 25, 2, 5, 3, 89, 3, 6, 4}

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
The idea is to simulate the given process. For this a queue data structure is used.

1. The given array is sorted and the queue is prepared by adding array indexes.
2. Then the given array is traversed and for each element, the index from the front of the queue is popped and add the current array element is added at the popped index in the resultant array.
3. If the queue is still not empty, then the next index (in the queue front) is moved to the back of the queue.

Below is the implementation of the above approach:

C++

 `#include ` `#define mod 1000000007 ` `using` `namespace` `std; ` ` `  `// Function to print the arrangement ` `vector<``int``> arrangement(vector<``int``> arr) ` `{ ` `    ``// Sorting the list ` `    ``sort(arr.begin(),arr.end()); ` ` `  `    ``//Finding Length of the List ` `    ``int` `length = arr.size(); ` ` `  `    ``// Initializing the result array ` `    ``vector<``int``> ans(length,0); ` ` `  `    ``// Initializing the Queue ` `    ``deque <``int``> Q; ` `    ``for` `(``int` `i = 0; i < length; i++) ` `        ``Q.push_back(i); ` `         `  `    ``// Adding current array element to the ` `    ``// result at an index which is at the ` `    ``// front of the Q and then if still ` `    ``// elements are left then putting the next ` `    ``// top element the bottom of the array. ` `    ``for` `(``int` `i = 0; i < length; i++) ` `    ``{ ` `        ``int` `j = Q.front(); ` `        ``Q.pop_front(); ` `        ``ans[j] = arr[i]; ` `         `  `        ``if``(Q.size() != 0) ` `        ``{ ` `            ``j = Q.front(); ` `            ``Q.pop_front(); ` `            ``Q.push_back(j); ` `        ``} ` ` `  `        ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``vector<``int``> arr = {1, 2, 3, 4, 5, 6, 7, 8}; ` `     `  `    ``vector<``int``> answer = arrangement(arr); ` `     `  `    ``for``(``int` `i:answer) cout << i << ``" "``; ` `} ` ` `  `// This code is contributed by mohit kumar 29 `

Java

 `// Java implementation of the above approach ` ` `  `import` `java.util.*; ` ` `  `public` `class` `GfG { ` ` `  `    ``// Function to find the array ` `    ``// arrangement ` `    ``static` `public` `int``[] arrayIncreasing(``int``[] arr) ` `    ``{ ` ` `  `        ``// Sorting the array ` `        ``Arrays.sort(arr); ` ` `  `        ``// Finding size of array ` `        ``int` `length = arr.length; ` ` `  `        ``// Empty array to store resultant order ` `        ``int` `answer[] = ``new` `int``[length]; ` ` `  `        ``// Doubly Ended Queue to ` `        ``// simulate the process ` `        ``Deque dq = ``new` `LinkedList<>(); ` ` `  `        ``// Loop to initialize queue with indexes ` `        ``for` `(``int` `i = ``0``; i < length; i++) { ` `            ``dq.add(i); ` `        ``} ` ` `  `        ``// Adding current array element to the ` `        ``// result at an index which is at the ` `        ``// front of the queue and then if still ` `        ``// elements are left then putting the next ` `        ``// top element the bottom of the array. ` `        ``for` `(``int` `i = ``0``; i < length; i++) { ` ` `  `            ``answer[dq.pollFirst()] = arr[i]; ` ` `  `            ``if` `(!dq.isEmpty()) ` `                ``dq.addLast(dq.pollFirst()); ` `        ``} ` ` `  `        ``// Returning the resultant order ` `        ``return` `answer; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `A[] = { ``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7``, ``8` `}; ` ` `  `        ``// Calling the function ` `        ``int` `ans[] = arrayIncreasing(A); ` ` `  `        ``// Printing the obtained pattern ` `        ``for` `(``int` `i = ``0``; i < A.length; i++) ` `            ``System.out.print(ans[i] + ``" "``); ` `    ``} ` `} `

Python

 `# Python3 Code for the approach ` ` `  `# Importing Queue from Collections Module ` `from` `collections ``import` `deque ` ` `  `# Function to print the arrangement ` `def` `arrangement(arr): ` `    ``# Sorting the list ` `    ``arr.sort() ` `     `  `    ``# Finding Length of the List ` `    ``length ``=` `len``(arr) ` `     `  `    ``# Initializing the result array ` `    ``answer ``=` `[``0` `for` `x ``in` `range``(``len``(arr))] ` `     `  `    ``# Initializing the Queue ` `    ``queue ``=` `deque() ` `    ``for` `i ``in` `range``(length): ` `        ``queue.append(i) ` `     `  `    ``# Adding current array element to the ` `    ``# result at an index which is at the ` `    ``# front of the queue and then if still ` `    ``# elements are left then putting the next ` `    ``# top element the bottom of the array. ` `    ``for` `i ``in` `range``(length): ` `     `  `        ``answer[queue.popleft()] ``=` `arr[i] ` `     `  `        ``if` `len``(queue) !``=` `0``: ` `            ``queue.append(queue.popleft()) ` `    ``return` `answer ` ` `  `# Driver code ` `arr ``=` `[``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7``, ``8``] ` `answer ``=` `arrangement(arr) ` `# Printing the obtained result ` `print``(``*``answer, sep ``=` `' '``) `

C#

 `// C# implementation of the above approach ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GfG ` `{ ` ` `  `    ``// Function to find the array ` `    ``// arrangement ` `    ``static` `public` `int``[] arrayIncreasing(``int``[] arr) ` `    ``{ ` ` `  `        ``// Sorting the array ` `        ``Array.Sort(arr); ` ` `  `        ``// Finding size of array ` `        ``int` `length = arr.Length; ` ` `  `        ``// Empty array to store resultant order ` `        ``int` `[]answer = ``new` `int``[length]; ` ` `  `        ``// Doubly Ended Queue to ` `        ``// simulate the process ` `        ``List<``int``> dq = ``new` `List<``int``>(); ` ` `  `        ``// Loop to initialize queue with indexes ` `        ``for` `(``int` `i = 0; i < length; i++) ` `        ``{ ` `            ``dq.Add(i); ` `        ``} ` ` `  `        ``// Adding current array element to the ` `        ``// result at an index which is at the ` `        ``// front of the queue and then if still ` `        ``// elements are left then putting the next ` `        ``// top element the bottom of the array. ` `        ``for` `(``int` `i = 0; i < length; i++) ` `        ``{ ` ` `  `            ``answer[dq[0]] = arr[i]; ` `            ``dq.RemoveAt(0); ` `            ``if` `(dq.Count != 0) ` `            ``{ ` `                ``dq.Add(dq[0]); ` `                ``dq.RemoveAt(0); ` `            ``} ` `        ``} ` ` `  `        ``// Returning the resultant order ` `        ``return` `answer; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String []args) ` `    ``{ ` `        ``int` `[]A = { 1, 2, 3, 4, 5, 6, 7, 8 }; ` ` `  `        ``// Calling the function ` `        ``int` `[]ans = arrayIncreasing(A); ` ` `  `        ``// Printing the obtained pattern ` `        ``for` `(``int` `i = 0; i < A.Length; i++) ` `            ``Console.Write(ans[i] + ``" "``); ` `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```1 5 2 7 3 6 4 8
```

Time Complexity: O(NlogN)

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