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Longest Bitonic Subsequence | DP-15

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  • Difficulty Level : Medium
  • Last Updated : 12 Jan, 2022

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. 
 

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 ));
    }
}

Python3




# 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)

C#




/* 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




<?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
?>

Javascript




<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>

Output: 

 Length of LBS is 7

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

 

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