Given an array arr[0 … n-1] containing n positive integers, a subsequence of arr[] is called Bitonic if it is first increasing, then decreasing. Write a function that takes an array as argument and returns the length of the longest bitonic subsequence.
A sequence, sorted in increasing order is considered Bitonic with the decreasing part as empty. Similarly, decreasing order sequence is considered Bitonic with the increasing part as empty.
Examples:
Input arr[] = {1, 11, 2, 10, 4, 5, 2, 1}; Output: 6 (A Longest Bitonic Subsequence of length 6 is 1, 2, 10, 4, 2, 1) Input arr[] = {12, 11, 40, 5, 3, 1} Output: 5 (A Longest Bitonic Subsequence of length 5 is 12, 11, 5, 3, 1) Input arr[] = {80, 60, 30, 40, 20, 10} Output: 5 (A Longest Bitonic Subsequence of length 5 is 80, 60, 30, 20, 10)
Source: Microsoft Interview Question
Solution
This problem is a variation of standard Longest Increasing Subsequence (LIS) problem. Let the input array be arr[] of length n. We need to construct two arrays lis[] and lds[] using Dynamic Programming solution of LIS problem. lis[i] stores the length of the Longest Increasing subsequence ending with arr[i]. lds[i] stores the length of the longest Decreasing subsequence starting from arr[i]. Finally, we need to return the max value of lis[i] + lds[i] – 1 where i is from 0 to n-1.
Following is the implementation of the above Dynamic Programming solution.
/* Dynamic Programming implementation of longest bitonic subsequence problem */ #include<stdio.h> #include<stdlib.h> /* lbs() returns the length of the Longest Bitonic Subsequence in arr[] of size n. The function mainly creates two temporary arrays
lis[] and lds[] and returns the maximum lis[i] + lds[i] - 1.
lis[i] ==> Longest Increasing subsequence ending with arr[i]
lds[i] ==> Longest decreasing subsequence starting with arr[i]
*/ int lbs( int arr[], int n )
{ int i, j;
/* Allocate memory for LIS[] and initialize LIS values as 1 for
all indexes */
int lis[n];
for (i = 0; i < n; i++)
lis[i] = 1;
/* Compute LIS values from left to right */
for (i = 1; i < n; i++)
for (j = 0; j < i; j++)
if (arr[i] > arr[j] && lis[i] < lis[j] + 1)
lis[i] = lis[j] + 1;
/* Allocate memory for lds and initialize LDS values for
all indexes */
int lds[n];
for (i = 0; i < n; i++)
lds[i] = 1;
/* Compute LDS values from right to left */
for (i = n-2; i >= 0; i--)
for (j = n-1; j > i; j--)
if (arr[i] > arr[j] && lds[i] < lds[j] + 1)
lds[i] = lds[j] + 1;
/* Return the maximum value of lis[i] + lds[i] - 1*/
int max = lis[0] + lds[0] - 1;
for (i = 1; i < n; i++)
if (lis[i] + lds[i] - 1 > max)
max = lis[i] + lds[i] - 1;
return max;
} /* Driver program to test above function */ int main()
{ int arr[] = {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5,
13, 3, 11, 7, 15};
int n = sizeof (arr)/ sizeof (arr[0]);
printf ( "Length of LBS is %d\n" , lbs( arr, n ) );
return 0;
} |
/* Dynamic Programming implementation in Java for longest bitonic subsequence problem */
import java.util.*;
import java.lang.*;
import java.io.*;
class LBS
{ /* lbs() returns the length of the Longest Bitonic Subsequence in
arr[] of size n. The function mainly creates two temporary arrays
lis[] and lds[] and returns the maximum lis[i] + lds[i] - 1.
lis[i] ==> Longest Increasing subsequence ending with arr[i]
lds[i] ==> Longest decreasing subsequence starting with arr[i]
*/
static int lbs( int arr[], int n )
{
int i, j;
/* Allocate memory for LIS[] and initialize LIS values as 1 for
all indexes */
int [] lis = new int [n];
for (i = 0 ; i < n; i++)
lis[i] = 1 ;
/* Compute LIS values from left to right */
for (i = 1 ; i < n; i++)
for (j = 0 ; j < i; j++)
if (arr[i] > arr[j] && lis[i] < lis[j] + 1 )
lis[i] = lis[j] + 1 ;
/* Allocate memory for lds and initialize LDS values for
all indexes */
int [] lds = new int [n];
for (i = 0 ; i < n; i++)
lds[i] = 1 ;
/* Compute LDS values from right to left */
for (i = n- 2 ; i >= 0 ; i--)
for (j = n- 1 ; j > i; j--)
if (arr[i] > arr[j] && lds[i] < lds[j] + 1 )
lds[i] = lds[j] + 1 ;
/* Return the maximum value of lis[i] + lds[i] - 1*/
int max = lis[ 0 ] + lds[ 0 ] - 1 ;
for (i = 1 ; i < n; i++)
if (lis[i] + lds[i] - 1 > max)
max = lis[i] + lds[i] - 1 ;
return max;
}
public static void main (String[] args)
{
int arr[] = { 0 , 8 , 4 , 12 , 2 , 10 , 6 , 14 , 1 , 9 , 5 ,
13 , 3 , 11 , 7 , 15 };
int n = arr.length;
System.out.println( "Length of LBS is " + lbs( arr, n ));
}
} |
# Dynamic Programming implementation of longest bitonic subsequence problem """ lbs() returns the length of the Longest Bitonic Subsequence in arr[] of size n. The function mainly creates two temporary arrays lis[] and lds[] and returns the maximum lis[i] + lds[i] - 1. lis[i] ==> Longest Increasing subsequence ending with arr[i] lds[i] ==> Longest decreasing subsequence starting with arr[i] """ def lbs(arr):
n = len (arr)
# allocate memory for LIS[] and initialize LIS values as 1
# for all indexes
lis = [ 1 for i in range (n + 1 )]
# Compute LIS values from left to right
for i in range ( 1 , n):
for j in range ( 0 , i):
if ((arr[i] > arr[j]) and (lis[i] < lis[j] + 1 )):
lis[i] = lis[j] + 1
# allocate memory for LDS and initialize LDS values for
# all indexes
lds = [ 1 for i in range (n + 1 )]
# Compute LDS values from right to left
for i in reversed ( range (n - 1 )): #loop from n-2 downto 0
for j in reversed ( range (i - 1 ,n)): #loop from n-1 downto i-1
if (arr[i] > arr[j] and lds[i] < lds[j] + 1 ):
lds[i] = lds[j] + 1
# Return the maximum value of (lis[i] + lds[i] - 1)
maximum = lis[ 0 ] + lds[ 0 ] - 1
for i in range ( 1 , n):
maximum = max ((lis[i] + lds[i] - 1 ), maximum)
return maximum
# Driver program to test the above function arr = [ 0 , 8 , 4 , 12 , 2 , 10 , 6 , 14 , 1 , 9 , 5 , 13 ,
3 , 11 , 7 , 15 ]
print ( "Length of LBS is" ,lbs(arr))
# This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
/* Dynamic Programming implementation in C# for longest bitonic subsequence problem */
using System;
class LBS {
/* lbs() returns the length of the Longest Bitonic Subsequence in
arr[] of size n. The function mainly creates two temporary arrays
lis[] and lds[] and returns the maximum lis[i] + lds[i] - 1.
lis[i] ==> Longest Increasing subsequence ending with arr[i]
lds[i] ==> Longest decreasing subsequence starting with arr[i]
*/
static int lbs( int [] arr, int n)
{
int i, j;
/* Allocate memory for LIS[] and initialize
LIS values as 1 for all indexes */
int [] lis = new int [n];
for (i = 0; i < n; i++)
lis[i] = 1;
/* Compute LIS values from left to right */
for (i = 1; i < n; i++)
for (j = 0; j < i; j++)
if (arr[i] > arr[j] && lis[i] < lis[j] + 1)
lis[i] = lis[j] + 1;
/* Allocate memory for lds and initialize LDS values for
all indexes */
int [] lds = new int [n];
for (i = 0; i < n; i++)
lds[i] = 1;
/* Compute LDS values from right to left */
for (i = n - 2; i >= 0; i--)
for (j = n - 1; j > i; j--)
if (arr[i] > arr[j] && lds[i] < lds[j] + 1)
lds[i] = lds[j] + 1;
/* Return the maximum value of lis[i] + lds[i] - 1*/
int max = lis[0] + lds[0] - 1;
for (i = 1; i < n; i++)
if (lis[i] + lds[i] - 1 > max)
max = lis[i] + lds[i] - 1;
return max;
}
// Driver code
public static void Main()
{
int [] arr = { 0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5,
13, 3, 11, 7, 15 };
int n = arr.Length;
Console.WriteLine( "Length of LBS is " + lbs(arr, n));
}
} // This code is contributed by vt_m. |
<?php // Dynamic Programming implementation // of longest bitonic subsequence problem /* lbs() returns the length of the Longest Bitonic Subsequence in arr[] of size n.
The function mainly creates two temporary
arrays lis[] and lds[] and returns the
maximum lis[i] + lds[i] - 1.
lis[i] ==> Longest Increasing subsequence
ending with arr[i]
lds[i] ==> Longest decreasing subsequence
starting with arr[i]
*/ function lbs(& $arr , $n )
{ /* Allocate memory for LIS[] and initialize
LIS values as 1 for all indexes */
$lis = array_fill (0, $n , NULL);
for ( $i = 0; $i < $n ; $i ++)
$lis [ $i ] = 1;
/* Compute LIS values from left to right */
for ( $i = 1; $i < $n ; $i ++)
for ( $j = 0; $j < $i ; $j ++)
if ( $arr [ $i ] > $arr [ $j ] &&
$lis [ $i ] < $lis [ $j ] + 1)
$lis [ $i ] = $lis [ $j ] + 1;
/* Allocate memory for lds and initialize
LDS values for all indexes */
$lds = array_fill (0, $n , NULL);
for ( $i = 0; $i < $n ; $i ++)
$lds [ $i ] = 1;
/* Compute LDS values from right to left */
for ( $i = $n - 2; $i >= 0; $i --)
for ( $j = $n - 1; $j > $i ; $j --)
if ( $arr [ $i ] > $arr [ $j ] &&
$lds [ $i ] < $lds [ $j ] + 1)
$lds [ $i ] = $lds [ $j ] + 1;
/* Return the maximum value of
lis[i] + lds[i] - 1*/
$max = $lis [0] + $lds [0] - 1;
for ( $i = 1; $i < $n ; $i ++)
if ( $lis [ $i ] + $lds [ $i ] - 1 > $max )
$max = $lis [ $i ] + $lds [ $i ] - 1;
return $max ;
} // Driver Code $arr = array (0, 8, 4, 12, 2, 10, 6, 14,
1, 9, 5, 13, 3, 11, 7, 15);
$n = sizeof( $arr );
echo "Length of LBS is " . lbs( $arr , $n );
// This code is contributed by ita_c ?> |
<script> /* Dynamic Programming implementation in JavaScript for longest bitonic subsequence problem */ /* lbs() returns the length of the Longest Bitonic Subsequence in
arr[] of size n. The function mainly creates two temporary arrays
lis[] and lds[] and returns the maximum lis[i] + lds[i] - 1.
lis[i] ==> Longest Increasing subsequence ending with arr[i]
lds[i] ==> Longest decreasing subsequence starting with arr[i]
*/
function lbs(arr,n)
{
let i, j;
/* Allocate memory for LIS[] and initialize LIS values as 1 for
all indexes */
let lis = new Array(n)
for (i = 0; i < n; i++)
lis[i] = 1;
/* Compute LIS values from left to right */
for (i = 1; i < n; i++)
for (j = 0; j < i; j++)
if (arr[i] > arr[j] && lis[i] < lis[j] + 1)
lis[i] = lis[j] + 1;
/* Allocate memory for lds and initialize LDS values for
all indexes */
let lds = new Array(n);
for (i = 0; i < n; i++)
lds[i] = 1;
/* Compute LDS values from right to left */
for (i = n-2; i >= 0; i--)
for (j = n-1; j > i; j--)
if (arr[i] > arr[j] && lds[i] < lds[j] + 1)
lds[i] = lds[j] + 1;
/* Return the maximum value of lis[i] + lds[i] - 1*/
let max = lis[0] + lds[0] - 1;
for (i = 1; i < n; i++)
if (lis[i] + lds[i] - 1 > max)
max = lis[i] + lds[i] - 1;
return max;
}
let arr=[0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]
let n = arr.length;
document.write( "Length of LBS is " + lbs( arr, n ));
// This code is contributed by avanitrachhadiya2155
</script> |
Length of LBS is 7
Time Complexity: O(n^2)
Auxiliary Space: O(n)