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 ", lbs( arr, n ) ); return 0; } |
Output:
Length of LBS is 7
Time Complexity: O(n^2)
Auxiliary Space: O(n)
Please refer complete article on Longest Bitonic Subsequence | DP-15 for more details!
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