Split the array elements into strictly increasing and decreasing sequence

Given an array of N elements. The task is to split the elements into two arrays say a1[] and a2[] such that one contains strictly increasing elements and the other contains strictly decreasing elements and a1.size() + a2.size() = a.size(). If it is not possible to do so, print -1 or else print both the arrays. 

Note: There can be multiple answers and the order of elements needs not to be the same in the child arrays. 

Examples:

Input: a[] = {7, 2, 7, 3, 3, 1, 4} 
Output: a1[] = {1, 2, 3, 4, 7} , a2[] = {7, 3} 

Input: a[] = {1, 2, 2, 1, 1} 
Output: -1 
It is not possible 

Approach: The following steps are followed to solve the above problem:  

• Initialize two vectors v1 and v2 which stores increasing and decreasing numbers.
• Use hashing to know the occurrence of the number in the array.
• If the number appears to come for the first time, then store it in v1.
• If the number appears to come for the second time, then store it in v2.
• If the number appears for more than 2 times, then it is not possible to store to create a strictly increasing and strictly decreasing array.
• At last, sort the first vector in increasing order and the second vector in decreasing order and print them.

Below is the implementation of the above approach:  

C++

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to print both the arrays
void PrintBothArrays(int a[], int n)
{

    // Store both arrays
    vector<int> v1, v2;

    // Used for hashing
    unordered_map<int, int> mpp;

    // Iterate for every element
    for (int i = 0; i < n; i++) {

        // Increase the count
        mpp[a[i]]++;

        // If first occurrence
        if (mpp[a[i]] == 1)
            v1.push_back(a[i]);

        // If second occurrence
        else if (mpp[a[i]] == 2)
            v2.push_back(a[i]);

        // If occurs more than 2 times
        else {
            cout << "Not possible";
            return;
        }
    }

    // Sort in increasing order
    sort(v1.begin(), v1.end());

    // Print the increasing array
    cout << "Strictly increasing array is:\n";
    for (auto it : v1)
        cout << it << " ";

    // Sort in reverse order
    sort(v2.begin(), v2.end(), greater<int>());

    // Print the decreasing array
    cout << "\nStrictly decreasing array is:\n";
    for (auto it : v2)
        cout << it << " ";
}

// Driver code
int main()
{

    int a[] = { 7, 2, 7, 3, 3, 1, 4 };
    int n = sizeof(a) / sizeof(a[0]);
    PrintBothArrays(a, n);

    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;

class GFG
{

// Function to print both the arrays
static void PrintBothArrays(int a[], int n)
{

    // Store both arrays
    Vector<Integer> v1 = new Vector<Integer>(),
                    v2 = new Vector<Integer>();

    // Used for hashing
    HashMap<Integer, Integer> mpp = new HashMap<>();

    // Iterate for every element
    for (int i = 0; i < n; i++)
    {

        // Increase the count
        mpp.put(a[i],(mpp.get(a[i]) == null?0:mpp.get(a[i]))+1);

        // If first occurrence
        if (mpp.get(a[i]) == 1)
            v1.add(a[i]);

        // If second occurrence
        else if (mpp.get(a[i]) == 2)
            v2.add(a[i]);

        // If occurs more than 2 times
        else
        {
            System.out.println( "Not possible");
            return;
        }
    }

    // Sort in increasing order
    Collections.sort(v1);

    // Print the increasing array
    System.out.println("Strictly increasing array is:");
    for (int i = 0; i < v1.size(); i++)
        System.out.print(v1.get(i) + " ");

    // Sort
    Collections.sort(v2);
    Collections.reverse(v2);

    // Print the decreasing array
    System.out.println("\nStrictly decreasing array is:");
    for (int i = 0; i < v2.size(); i++)
        System.out.print(v2.get(i) + " ");
}

// Driver code
public static void main(String args[])
{
    int a[] = { 7, 2, 7, 3, 3, 1, 4 };
    int n = a.length;
    PrintBothArrays(a, n);
}
}

// This code is contributed by Arnab Kundu

Python3

# Python3 program to implement
# the above approach

# Function to print both the arrays
def PrintBothArrays(a, n) :

    # Store both arrays
    v1, v2 = [], [];

    # Used for hashing
    mpp = dict.fromkeys(a, 0);

    # Iterate for every element
    for i in range(n) :

        # Increase the count
        mpp[a[i]] += 1;

        # If first occurrence
        if (mpp[a[i]] == 1) :
            v1.append(a[i]);

        # If second occurrence
        elif (mpp[a[i]] == 2) :
            v2.append(a[i]);

        # If occurs more than 2 times
        else :
            print("Not possible");
            return;

    # Sort in increasing order
    v1.sort();

    # Print the increasing array
    print("Strictly increasing array is:");
    for it in v1:
        print(it, end = " ");

    # Sort in reverse order
    v2.sort(reverse = True);

    # Print the decreasing array
    print("\nStrictly decreasing array is:");
    for it in v2 :
        print(it, end = " ")

# Driver code
if __name__ == "__main__" :

    a = [ 7, 2, 7, 3, 3, 1, 4 ];
    n = len(a);
    PrintBothArrays(a, n);

# This code is contributed by Ryuga

C#

// C# program to implement
// the above approach
using System;
using System.Collections;
using System.Collections.Generic;

class GFG
{

    // Function to print both the arrays
    static void PrintBothArrays(int [] a, int n)
    {

        // Store both arrays
        List<int> v1 = new List<int>();
        List<int> v2 = new List<int>();

        // Used for hashing
        Dictionary<int, int> mpp = new Dictionary<int, int>();

        // Iterate for every element
        for (int i = 0; i < n; i++)
        {

            // Increase the Count
            if(mpp.ContainsKey(a[i]))
                mpp[a[i]] = mpp[a[i]] + 1;
            else
                mpp[a[i]] = 1;

            // If first occurrence
            if (mpp[a[i]] == 1)
                v1.Add(a[i]);

            // If second occurrence
            else if (mpp[a[i]] == 2)
                v2.Add(a[i]);

            // If occurs more than 2 times
            else
            {
                Console.WriteLine( "Not possible");
                return;
            }
        }

        // Sort in increasing order
        v1.Sort();

        // Print the increasing array
        Console.WriteLine("Strictly increasing array is:");
        for (int i = 0; i < v1.Count; i++)
            Console.Write(v1[i] + " ");

        // Sort
        v2.Sort();
        v2.Reverse();

        // Print the decreasing array
        Console.WriteLine("\nStrictly decreasing array is:");
        for (int i = 0; i < v2.Count; i++)
            Console.Write(v2[i] + " ");
    }

    // Driver code
    public static void Main()
    {
        int [] a = { 7, 2, 7, 3, 3, 1, 4 };
        int n = a.Length;
        PrintBothArrays(a, n);
    }
}

// This code is contributed by ihritik

Javascript

<script>

    // Javascript implementation of the approach

// Function to print both the arrays
function PrletBothArrays(a, n)
{

    // Store both arrays
    let v1 = [], v2 = [];

    // Used for hashing
    let mpp = new Map();

    // Iterate for every element
    for (let i = 0; i < n; i++)
    {

        // Increase the count
        mpp.set(a[i],(mpp.get(a[i]) == null?0:mpp.get(a[i]))+1);

        // If first occurrence
        if (mpp.get(a[i]) == 1)
            v1.push(a[i]);

        // If second occurrence
        else if (mpp.get(a[i]) == 2)
            v2.push(a[i]);

        // If occurs more than 2 times
        else
        {
            document.write( "Not possible");
            return;
        }
    }

    // Sort in increasing order
    v1.sort();

    // Print the increasing array
    document.write("Strictly increasing array is:" + "<br/>");
    for (let i = 0; i < v1.length; i++)
        document.write(v1[i] + " ");

    // Sort
    v2.sort();
    v2.reverse();

    // Print the decreasing array
    document.write("<br/>" + "\nStrictly decreasing array is:" + "<br/>");
    for (let i = 0; i < v2.length; i++)
        document.write(v2[i] + " ");
}

    // Driver code

     let a = [ 7, 2, 7, 3, 3, 1, 4 ];
    let n = a.length;
    PrletBothArrays(a, n);

 // This code is contributed by sanjoy_62.
</script>

Output:

Strictly increasing array is:
1 2 3 4 7 
Strictly decreasing array is:
7 3

Time Complexity: O(N*logN), as in the worst case we will be using an inbuilt sort function to sort an array of size N. Where N is the number of elements in the array.

Auxiliary Space: O(N), as we are using extra space for the map and array v1 and v2. Where N is the length of the string.