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)
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.
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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;
}
Output :
Length of LBS is 7
Time Complexity: O(n^2)
Auxiliary Space: O(n)
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