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.

/* 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 ) );     getchar();   return 0; }

Output:

 Length of LBS is 7

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

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