Dynamic Programming | Set 15 (Longest Bitonic Subsequence)

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 C++ implementation of the above Dynamic Programming solution.

C++

/* 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 = 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 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;
}

Java

/* 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 ));
    }
}

Python


# 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)



Output:
 Length of LBS is 7

Time Complexity: O(n^2)
Auxiliary Space: O(n)

Asked in: Microsoft

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

GATE CS Corner    Company Wise Coding Practice

Recommended Posts:







Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.