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# Divide array into increasing and decreasing subsequence without changing the order

Given a merged sequence which consists of two sequences which got merged, one of them was strictly increasing and the other was strictly decreasing. Elements of increasing sequence were inserted between elements of the decreasing one without changing the order.

Sequences [1, 3, 4] and [10, 4, 2] can produce the following resulting sequences:
[10, 1, 3, 4, 2, 4], [1, 3, 4, 10, 4, 2].
The following sequence cannot be the result of these insertions:
[1, 10, 4, 4, 3, 2] because the order of elements in the increasing sequence was changed.

Given a merged sequence, the task is to find any two suitable initial sequences, one of them should be strictly increasing, and another should be strictly decreasing.
Note: An empty sequence and the sequence consisting of one element can be considered as increasing or decreasing.
Examples:

Input: arr[] = {5, 1, 3, 6, 8, 2, 9, 0, 10}
Output: [1, 3, 6, 8, 9, 10] [5, 2, 0]

Input: arr[] = {1, 2, 4, 0, 2}
Output: -1
No such sequences possible.

Method 1: We can modify Longest Increasing Sequence) and solve the required problem. It will take O(nlogn) time.

Method 2: We can also solve this problem only in a single traversal. The Idea used here is that maintain two sorted arrays.
For a new element x

• If it can be appended to only one of the arrays then append it.
• If it can be appended to neither, then the answer is -1.
• If it can be appended to both then check the next element y, if y > x then append x to the increasing one otherwise append x to the decreasing one.

Now when we encounter an element which can be included in both the sequence we need to decide based on the next element’s value. Let’s consider a situation where we need to iterate over the remaining value x,y,z where ( x < z < y) and we have already the last element of the increasing and decreasing sequence as inc and dec respectively from the visited portion of the array.

Case 1 : x<y and inc<x<dec

so we can include x in any sequence.

If we include it in decreasing sequence then dec will become x. And then for y we have only one choice i.e. to include it in increasing sequence as y>dec and inc becomes y. If we do this we cannot insert z in any sequence as z>dec and z<inc.

But if we include x to increasing sequence (inc becomes x) and y to decreasing sequence (dec becomes y) following the same logic then we can place z in any sequence and get an answer.

Case 2 : x>=y and inc<x<dec

it follows the same logic as above.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to print strictly increasing and``// strictly decreasing sequence if possible``void` `Find_Sequence(``int` `arr[], ``int` `n)``{``    ``// Arrays to store strictly increasing and``    ``// decreasing sequence``    ``vector<``int``> inc_arr, dec_arr;` `    ``// Initializing last element of both sequence``    ``int` `flag = 0;``    ``long` `inc = -1, dec = 1e7;` `    ``// Iterating through the array``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``// If current element can be appended``        ``// to both the sequences``        ``if` `(inc < arr[i] && arr[i] < dec)``        ``{``            ``// If next element is greater than``            ``// the current element``            ``// Then append it to the strictly``            ``// increasing array``            ``if` `(arr[i] < arr[i + 1])``            ``{``                ``inc = arr[i];``                ``inc_arr.emplace_back(arr[i]);``            ``}` `            ``// Otherwise append it to the``            ``// strictly decreasing array``            ``else``            ``{``                ``dec = arr[i];``                ``dec_arr.emplace_back(arr[i]);``            ``}``        ``}``        ` `        ``// If current element can be appended``        ``// to the increasing sequence only``        ``else` `if` `(inc < arr[i])``        ``{``            ``inc = arr[i];``            ``inc_arr.emplace_back(arr[i]);``        ``}``        ` `        ``// If current element can be appended``        ``// to the decreasing sequence only``        ``else` `if` `(dec > arr[i])``        ``{``            ``dec = arr[i];``            ``dec_arr.emplace_back(arr[i]);``        ``}``        ` `        ``// Else we can not make such sequences``        ``// from the given array``        ``else``        ``{``            ``cout << -1 << endl;``            ``flag = 1;``            ``break``;``        ``}``    ``}``    ` `    ``// Print the required sequences``    ``if` `(!flag)``    ``{``        ``for` `(``auto` `i = inc_arr.begin();``                  ``i != inc_arr.end(); i++)``            ``cout << *i << ``" "``;``        ``cout << endl;` `        ``for` `(``auto` `i = dec_arr.begin();``                  ``i != dec_arr.end(); i++)``            ``cout << *i << ``" "``;``        ``cout << endl;``    ``}``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 5, 1, 3, 6, 8, 2, 9, 0, 10 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``Find_Sequence(arr, n);``}` `// This code is contributed by sanjeev2552`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{` `    ``// Function to print strictly increasing and``    ``// strictly decreasing sequence if possible``    ``static` `void` `Find_Sequence(``int``[] arr, ``int` `n)``    ``{` `        ``// Arrays to store strictly increasing and``        ``// decreasing sequence``        ``Vector inc_arr = ``new` `Vector<>(),``                        ``dec_arr = ``new` `Vector<>();` `        ``// Initializing last element of both sequence``        ``int` `flag = ``0``;``        ``long` `inc = -``1``, dec = (``long``) 1e7;` `        ``// Iterating through the array``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{` `            ``// If current element can be appended``            ``// to both the sequences``            ``if` `(inc < arr[i] && arr[i] < dec)``            ``{` `                ``// If next element is greater than``                ``// the current element``                ``// Then append it to the strictly``                ``// increasing array``                ``if` `(arr[i] < arr[i + ``1``])``                ``{``                    ``inc = arr[i];``                    ``inc_arr.add(arr[i]);``                ``}` `                ``// Otherwise append it to the``                ``// strictly decreasing array``                ``else``                ``{``                    ``dec = arr[i];``                    ``dec_arr.add(arr[i]);``                ``}``            ``}` `            ``// If current element can be appended``            ``// to the increasing sequence only``            ``else` `if` `(inc < arr[i])``            ``{``                ``inc = arr[i];``                ``inc_arr.add(arr[i]);``            ``}` `            ``// If current element can be appended``            ``// to the decreasing sequence only``            ``else` `if` `(dec > arr[i])``            ``{``                ``dec = arr[i];``                ``dec_arr.add(arr[i]);``            ``}` `            ``// Else we can not make such sequences``            ``// from the given array``            ``else``            ``{``                ``System.out.println(-``1``);``                ``flag = ``1``;``                ``break``;``            ``}``        ``}` `        ``// Print the required sequences``        ``if` `(flag == ``0``)``        ``{``            ``for` `(``int` `i : inc_arr)``                ``System.out.print(i + ``" "``);``            ``System.out.println();` `            ``for` `(``int` `i : dec_arr)``                ``System.out.print(i + ``" "``);``            ``System.out.println();``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[] arr = { ``5``, ``1``, ``3``, ``6``, ``8``, ``2``, ``9``, ``0``, ``10` `};``        ``int` `n = arr.length;``        ``Find_Sequence(arr, n);``    ``}``}` `// This code is contributed by``// sanjeev2552`

## Python3

 `# Python3 implementation of the approach` `# Function to print strictly increasing and``# strictly decreasing sequence if possible``def` `Find_Sequence(array, n):` `    ``# Arrays to store strictly increasing and``    ``# decreasing sequence``    ``inc_arr, dec_arr ``=``[], []` `    ``# Initializing last element of both sequence``    ``inc, dec ``=` `-``1``, ``1e7` `    ``# Iterating through the array``    ``for` `i ``in` `range``(n):` `        ``# If current element can be appended``        ``# to both the sequences``        ``if` `inc < array[i] < dec:` `            ``# If next element is greater than``            ``# the current element``            ``# Then append it to the strictly``            ``# increasing array``            ``if` `array[i] < array[i ``+` `1``]:``                ``inc ``=` `array[i]``                ``inc_arr.append(array[i])` `            ``# Otherwise append it to the``            ``# strictly decreasing array``            ``else``:``                ``dec ``=` `array[i]``                ``dec_arr.append(array[i])` `        ``# If current element can be appended``        ``# to the increasing sequence only``        ``elif` `inc < array[i]:``            ``inc ``=` `array[i]``            ``inc_arr.append(array[i])` `        ``# If current element can be appended``        ``# to the decreasing sequence only``        ``elif` `dec > array[i]:``            ``dec ``=` `array[i]``            ``dec_arr.append(array[i])` `        ``# Else we can not make such sequences``        ``# from the given array``        ``else``:``            ``print``(``'-1'``)``            ``break` `    ``# Print the required sequences``    ``else``:``        ``print``(inc_arr, dec_arr)` `# Driver code``arr ``=` `[``5``, ``1``, ``3``, ``6``, ``8``, ``2``, ``9``, ``0``, ``10``]``n ``=` `len``(arr)``Find_Sequence(arr, n)`

## C#

 `// C# implementation of the approach``using` `System;``using` `System.Collections;``using` `System.Collections.Generic; ` `class` `GFG{``  ` `// Function to print strictly increasing and``// strictly decreasing sequence if possible``static` `void` `Find_Sequence(``int``[] arr, ``int` `n)``{` `    ``// Arrays to store strictly increasing and``    ``// decreasing sequence``    ``ArrayList inc_arr = ``new` `ArrayList();``    ``ArrayList dec_arr = ``new` `ArrayList();` `    ``// Initializing last element of both sequence``    ``int` `flag = 0;``    ``long` `inc = -1, dec = (``long``)1e7;` `    ``// Iterating through the array``    ``for``(``int` `i = 0; i < n; i++)``    ``{` `        ``// If current element can be appended``        ``// to both the sequences``        ``if` `(inc < arr[i] && arr[i] < dec)``        ``{` `            ``// If next element is greater than``            ``// the current element``            ``// Then append it to the strictly``            ``// increasing array``            ``if` `(arr[i] < arr[i + 1])``            ``{``                ``inc = arr[i];``                ``inc_arr.Add(arr[i]);``            ``}` `            ``// Otherwise append it to the``            ``// strictly decreasing array``            ``else``            ``{``                ``dec = arr[i];``                ``dec_arr.Add(arr[i]);``            ``}``        ``}` `        ``// If current element can be appended``        ``// to the increasing sequence only``        ``else` `if` `(inc < arr[i])``        ``{``            ``inc = arr[i];``            ``inc_arr.Add(arr[i]);``        ``}` `        ``// If current element can be appended``        ``// to the decreasing sequence only``        ``else` `if` `(dec > arr[i])``        ``{``            ``dec = arr[i];``            ``dec_arr.Add(arr[i]);``        ``}` `        ``// Else we can not make such sequences``        ``// from the given array``        ``else``        ``{``            ``Console.Write(-1);``            ``flag = 1;``            ``break``;``        ``}``    ``}` `    ``// Print the required sequences``    ``if` `(flag == 0)``    ``{``        ``foreach``(``int` `i ``in` `inc_arr)``            ``Console.Write(i + ``" "``);``            ` `        ``Console.Write(``'\n'``);` `        ``foreach``(``int` `i ``in` `dec_arr)``            ``Console.Write(i + ``" "``);``            ` `        ``Console.Write(``'\n'``);``    ``}``}` `// Driver Code``public` `static` `void` `Main(``string``[] args)``{``    ``int``[] arr = { 5, 1, 3, 6, 8,``                  ``2, 9, 0, 10 };``    ``int` `n = arr.Length;``    ` `    ``Find_Sequence(arr, n);``}``}` `// This code is contributed by rutvik_56`

## PHP

 ` ``\$arr``[``\$i``])``        ``{``            ``\$dec` `= ``\$arr``[``\$i``];``            ``array_push``(``\$dec_arr``, ``\$arr``[``\$i``]);``        ``}` `        ``// Else we can not make such sequences``        ``// from the given array``        ``else``        ``{``            ``echo` `'-1'``;``            ``break``;``        ``}``    ``}``    ` `    ``// Print the required sequences``    ``print_r(``\$inc_arr``);``    ``print_r(``\$dec_arr``);``}` `// Driver code``\$arr` `= ``array``(5, 1, 3, 6, 8, 2, 9, 0, 10);``\$n` `= ``count``(``\$arr``);``Find_Sequence(``\$arr``, ``\$n``);` `// This code is contributed by Ryuga``?>`

## Javascript

 ``

Output:

`[1, 3, 6, 8, 9, 10] [5, 2, 0]`

Time Complexity : O(n) ,where n is size of given array.

Space Complexity : O(n) ,extra space required to store strictly increasing and decreasing sequence.