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++ implementation of the approach #include <bits/stdc++.h> 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 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<Integer> 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 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# 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 // Php implementation of the approach // Function to print strictly increasing and // strictly decreasing sequence if possible function Find_Sequence( $arr , $n )
{ // Arrays to store strictly increasing and
// decreasing sequence
$inc_arr = array (); $dec_arr = array ();
// Initializing last element of both sequence
$inc = -1; $dec = 1e7;
// Iterating through the array
for ( $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 ];
array_push ( $inc_arr , $arr [ $i ]);
}
// Otherwise append it to the
// strictly decreasing array
else
{
$dec = $arr [ $i ];
array_push ( $dec_arr , $arr [ $i ]);
}
}
// If current element can be appended
// to the increasing sequence only
else if ( $inc < $arr [ $i ])
{
$inc = $arr [ $i ];
array_push ( $inc_arr , $arr [ $i ]);
}
// If current element can be appended
// to the decreasing sequence only
else if ( $dec > $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 ?> |
<script> // Javascript implementation of the approach // Function to print strictly increasing and
// strictly decreasing sequence if possible
function Find_Sequence(arr, n)
{
// Arrays to store strictly increasing and
// decreasing sequence
let inc_arr =[],
dec_arr = [];
// Initializing last element of both sequence
let flag = 0;
let inc = -1, dec = 1e7;
// Iterating through the array
for (let 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.push(arr[i]);
}
// Otherwise append it to the
// strictly decreasing array
else
{
dec = arr[i];
dec_arr.push(arr[i]);
}
}
// If current element can be appended
// to the increasing sequence only
else if (inc < arr[i])
{
inc = arr[i];
inc_arr.push(arr[i]);
}
// If current element can be appended
// to the decreasing sequence only
else if (dec > arr[i])
{
dec = arr[i];
dec_arr.push(arr[i]);
}
// Else we can not make such sequences
// from the given array
else
{
document.write(-1);
flag = 1;
break ;
}
}
// Print the required sequences
if (flag == 0)
{
document.write( "[" );
for (let i in inc_arr)
document.write( inc_arr[i] + " " );
document.write( "] " );
document.write( "[" );
for (let i in dec_arr)
document.write( dec_arr[i] + " " );
document.write( "]" );
}
}
// Driver Code let arr = [ 5, 1, 3, 6, 8, 2, 9, 0, 10 ];
let n = arr.length;
Find_Sequence(arr, n);
// This code is contributed by target_2. </script> |
[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.