Length of longest strict bitonic subsequence

Given an array arr[] containing n integers. The problem is to find the length of the longest strict bitonic subsequence. A subsequence is called strict bitonic if it is first increasing and then decreasing with the condition that in both the increasing and decreasing parts the absolute difference between adjacents is 1 only. 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, 5, 2, 3, 4, 5, 3, 2}
Output : 6
The Longest Strict Bitonic Subsequence is:
{1, 2, 3, 4, 3, 2}.

Input : arr[] = {1, 2, 5, 3, 6, 7, 4, 6, 5}
Output : 5

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1: The problem could be solved using the concept of finding the longest bitonic subsequence. The only condition that needs to be maintained is that the adjacents should have a difference of 1 only. It has a time complexity of O(n2).

Method 2 (Efficient Approach): The idea is to create two hash maps inc and dcr having tuples in the form (ele, len), where len denotes the length of the longest increasing subsequence ending with the element ele in map inc and length of the longest decreasing subsequence starting with element ele in map dcr respectively. Also create two arrays len_inc[] and len_dcr[] where len_inc[i] represents the length of the largest increasing subsequence ending with element arr[i] and len_dcr[i] represents the length of the largest decreasing subsequence starting with element arr[i]. Now, for each element arr[i] we can find the length of the value (arr[i]-1) if it exists in the hash table inc. Let this value be v (initially v will be 0). Now, the length of longest increasing subsequence ending with arr[i] would be v+1. Update this length along with the element arr[i] in the hash table inc and in the array len_inc[] at respective index i. Now, traversing the array from right to left we can similarly fill the hash table dcr and array len_dcr[] for longest decreasing subsequence. Finally, for each element arr[i] we calculate (len_inc[i] + len_dcr[i] – 1) and return the maximum value.

Note: Here increasing and decreasing subsequences only mean that the difference between adjacent elements is 1 only.

C++

 // C++ implementation to find length of longest  // strict bitonic subsequence  #include using namespace std;       // function to find length of longest  // strict bitonic subsequence  int longLenStrictBitonicSub(int arr[], int n) {     // hash table to map the array element with the     // length of the longest subsequence of which     // it is a part of and is the last/first element of     // that subsequence     unordered_map inc, dcr;            // arrays to store the length of increasing and     // decreasing subsequences which end at them     // or start from them       int len_inc[n], len_dcr[n];            // to store the length of longest strict      // bitonic subsequence     int longLen = 0;            // traverse the array elements     // from left to right     for (int i=0; i=0; i--)     {         // initialize current length          // for element arr[i] as 0         int len = 0;                      // if 'arr[i]-1' is in 'dcr'          if (dcr.find(arr[i]-1) != dcr.end())             len = dcr[arr[i]-1];                      // update arr[i] subsequence length in 'dcr'         // and in len_dcr[]             dcr[arr[i]] = len_dcr[i] = len + 1;      }            // calculating the length of all the strict      // bitonic subsequence     for (int i=0; i

Java

 // Java implementation to find length of longest  // strict bitonic subsequence  import java.util.*;    class GfG  {        // function to find length of longest  // strict bitonic subsequence  static int longLenStrictBitonicSub(int arr[], int n)  {      // hash table to map the array element with the      // length of the longest subsequence of which      // it is a part of and is the last/first element of      // that subsequence      HashMap inc = new HashMap ();     HashMap dcr = new HashMap ();             // arrays to store the length of increasing and      // decreasing subsequences which end at them      // or start from them      int len_inc[] = new int[n];     int len_dcr[] = new int[n];             // to store the length of longest strict      // bitonic subsequence      int longLen = 0;             // traverse the array elements      // from left to right      for (int i = 0; i < n; i++)      {          // initialize current length          // for element arr[i] as 0          int len = 0;                         // if 'arr[i]-1' is in 'inc'          if (inc.containsKey(arr[i] - 1))              len = inc.get(arr[i] - 1);                         // update arr[i] subsequence length in 'inc'              // and in len_inc[]          len_inc[i] = len + 1;         inc.put(arr[i], len_inc[i]);     }             // traverse the array elements      // from right to left      for (int i = n - 1; i >= 0; i--)      {          // initialize current length          // for element arr[i] as 0          int len = 0;                         // if 'arr[i]-1' is in 'dcr'          if (dcr.containsKey(arr[i] - 1))              len = dcr.get(arr[i] - 1);                         // update arr[i] subsequence length in 'dcr'          // and in len_dcr[]          len_dcr[i] = len + 1;         dcr.put(arr[i], len_dcr[i]);      }             // calculating the length of all the strict      // bitonic subsequence      for (int i = 0; i < n; i++)          if (longLen < (len_inc[i] + len_dcr[i] - 1))              longLen = len_inc[i] + len_dcr[i] - 1;                 // required longest length strict      // bitonic subsequence      return longLen;      }         // Driver code  public static void main(String[] args)  {      int arr[] = {1, 5, 2, 3, 4, 5, 3, 2};      int n = arr.length;      System.out.println("Longest length strict " +                              "bitonic subsequence = " +                              longLenStrictBitonicSub(arr, n));  } }     // This code is contributed by  // prerna saini

Python3

 # Python3 implementation to find length of  # longest strict bitonic subsequence    # function to find length of longest # strict bitonic subsequence def longLenStrictBitonicSub(arr, n):        # hash table to map the array element      # with the length of the longest subsequence      # of which it is a part of and is the      # last/first element of that subsequence     inc, dcr = dict(), dict()        # arrays to store the length of increasing     # and decreasing subsequences which end at      # them or start from them     len_inc, len_dcr =  * n,  * n        # to store the length of longest strict     # bitonic subsequence     longLen = 0        # traverse the array elements     # from left to right     for i in range(n):            # initialize current length         # for element arr[i] as 0         len = 0            # if 'arr[i]-1' is in 'inc'         if inc.get(arr[i] - 1) in inc.values():             len = inc.get(arr[i] - 1)                    # update arr[i] subsequence length in 'inc'              # and in len_inc[]          inc[arr[i]] = len_inc[i] = len + 1            # traverse the array elements     # from right to left     for i in range(n - 1, -1, -1):            # initialize current length         # for element arr[i] as 0         len = 0            # if 'arr[i]-1' is in 'dcr'         if dcr.get(arr[i] - 1) in dcr.values():             len = dcr.get(arr[i] - 1)                    # update arr[i] subsequence length           # in 'dcr' and in len_dcr[]          dcr[arr[i]] = len_dcr[i] = len + 1            # calculating the length of      # all the strict bitonic subsequence      for i in range(n):         if longLen < (len_inc[i] + len_dcr[i] - 1):             longLen = len_inc[i] + len_dcr[i] - 1            # required longest length strict      # bitonic subsequence      return longLen    # Driver Code if __name__ == "__main__":     arr = [1, 5, 2, 3, 4, 5, 3, 2]     n = len(arr)     print("Longest length strict bitonic subsequence =",            longLenStrictBitonicSub(arr, n))    # This code is contributed by sanjeev2552

C#

 // C# implementation to find length of longest  // strict bitonic subsequence  using System; using System.Collections.Generic;    class GfG  {         // function to find length of longest      // strict bitonic subsequence      static int longLenStrictBitonicSub(int []arr, int n)      {          // hash table to map the array          // element with the length of          // the longest subsequence of          // which it is a part of and          // is the last/first element of          // that subsequence          Dictionary inc = new Dictionary ();          Dictionary dcr = new Dictionary ();             // arrays to store the length          // of increasing and decreasing          // subsequences which end at them          // or start from them          int []len_inc = new int[n];          int []len_dcr = new int[n];             // to store the length of longest strict          // bitonic subsequence          int longLen = 0;             // traverse the array elements          // from left to right          for (int i = 0; i < n; i++)          {              // initialize current length              // for element arr[i] as 0              int len = 0;                 // if 'arr[i]-1' is in 'inc'              if (inc.ContainsKey(arr[i] - 1))                  len = inc[arr[i] - 1];                 // update arr[i] subsequence length                   // in 'inc' and in len_inc[]              len_inc[i] = len + 1;              if (inc.ContainsKey(arr[i]))             {                 inc.Remove(arr[i]);                 inc.Add(arr[i], len_inc[i]);              }             else                 inc.Add(arr[i], len_inc[i]);         }             // traverse the array elements          // from right to left          for (int i = n - 1; i >= 0; i--)          {              // initialize current length              // for element arr[i] as 0              int len = 0;                 // if 'arr[i]-1' is in 'dcr'              if (dcr.ContainsKey(arr[i] - 1))                  len = dcr[arr[i] - 1];                 // update arr[i] subsequence length in 'dcr'              // and in len_dcr[]              len_dcr[i] = len + 1;              if (dcr.ContainsKey(arr[i]))             {                 dcr.Remove(arr[i]);                 dcr.Add(arr[i], len_dcr[i]);              }             else                 dcr.Add(arr[i], len_dcr[i]);          }             // calculating the length of all the strict          // bitonic subsequence          for (int i = 0; i < n; i++)              if (longLen < (len_inc[i] + len_dcr[i] - 1))                  longLen = len_inc[i] + len_dcr[i] - 1;             // required longest length strict          // bitonic subsequence          return longLen;      }         // Driver code      public static void Main(String[] args)      {          int []arr = {1, 5, 2, 3, 4, 5, 3, 2};          int n = arr.Length;          Console.WriteLine("Longest length strict " +                              "bitonic subsequence = " +                              longLenStrictBitonicSub(arr, n));      }  }     // This code is contributed by Rajput-Ji

Output:

Longest length strict bitonic subsequence = 6

Time Complexity: O(n).
Auxiliary Space: O(n).

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