Maximum length subsequence with difference between adjacent elements as either 0 or 1 | Set 2

Given an array of n integers. The problem is to find maximum length of the subsequence with difference between adjacent elements in the subsequence as either 0 or 1. Time Complexity of O(n) is required.

Examples:

Input : arr[] = {2, 5, 6, 3, 7, 6, 5, 8}
Output : 5
The subsequence is {5, 6, 7, 6, 5}.

Input : arr[] = {-2, -1, 5, -1, 4, 0, 3}
Output : 4
The subsequence is {-2, -1, -1, 0}.

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

Method 1: Previously an approach having time complexity of O(n2) have been discussed in this post.

Method 2 (Efficient Approach): The idea is to create a hash map having tuples in the form (ele, len), where len denotes the length of the longest subsequence ending with the element ele. Now, for each element arr[i] we can find the length of the values arr[i]-1, arr[i] and arr[i]+1 in the hash table and consider the maximum among them. Let this maximum value be max. Now, the length of longest subsequence ending with arr[i] would be max+1. Update this length along with the element arr[i] in the hash table. Finally, the element having the maximum length in the hash table gives the maximum length subsequence.

C++

 // C++ implementation to find maximum length // subsequence with difference between adjacent  // elements as either 0 or 1 #include using namespace std;     // function to find maximum length subsequence  // with difference between adjacent elements as // either 0 or 1 int maxLenSub(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 element of     // that subsequence     unordered_map um;            // to store the maximum length subsequence     int maxLen = 0;            // traverse the array elements     for (int i=0; i

Java

 // Java implementation to find maximum length // subsequence with difference between adjacent  // elements as either 0 or 1 import java.util.HashMap;    class GFG {            // function to find maximum length subsequence      // with difference between adjacent elements as     // either 0 or 1     public static int maxLengthSub(int[] arr)     {                    // to store the maximum length subsequence         int max_val = 0;         int start = 0;                    // 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 element of         // that subsequence         HashMap map = new HashMap<>();            // traverse the array elements         for (int i = 0; i < arr.length; i++)          {                            // initialize current length              // for element arr[i] as 0             int temp = 0;             if (map.containsKey(arr[i] - 1))             {                 temp = map.get(arr[i] - 1);             }                if (map.containsKey(arr[i]))             {                 temp = Math.max(temp, map.get(arr[i]));             }                            if (map.containsKey(arr[i] + 1))             {                 temp = Math.max(temp, map.get(arr[i] + 1));             }             temp++;                            // update maximum length             if (temp > max_val)              {                 max_val = temp;             }             map.put(arr[i], temp);         }                    // required maximum length subsequence         return max_val;     }            // Driver Code     public static void main(String[] args)     {         int arr[] = {2, 5, 6, 3, 7, 6, 5, 8};         System.out.println(maxLengthSub(arr));     } }    // This code is contributed  // by tushar jajodia

Python3

 # Python3 implementation to find maximum  # length subsequence with difference between  # adjacent elements as either 0 or 1  from collections import defaultdict    # Function to find maximum length subsequence with  # difference between adjacent elements as either 0 or 1  def maxLenSub(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 element of that subsequence      um = defaultdict(lambda:0)            # to store the maximum length subsequence      maxLen = 0            # traverse the array elements      for i in range(0, n):                 # initialize current length          # for element arr[i] as 0          length = 0                    # if 'arr[i]-1' is in 'um' and its length of          # subsequence is greater than 'len'          if (arr[i]-1) in um and length < um[arr[i]-1]:             length = um[arr[i]-1]                     # if 'arr[i]' is in 'um' and its length of          # subsequence is greater than 'len'          if arr[i] in um and length < um[arr[i]]:              length = um[arr[i]]                         # if 'arr[i]+1' is in 'um' and its length of          # subsequence is greater than 'len'              if (arr[i]+1) in um and length < um[arr[i]+1]:              length = um[arr[i]+1]                     # update arr[i] subsequence length in 'um'          um[arr[i]] = length + 1                    # update maximum length          if maxLen < um[arr[i]]:              maxLen = um[arr[i]]             # required maximum length subsequence      return maxLen    # Driver program to test above  if __name__ == "__main__":         arr = [2, 5, 6, 3, 7, 6, 5, 8]      n = len(arr)      print("Maximum length subsequence =", maxLenSub(arr, n))        # This code is contributed by Rituraj Jain

Output:

Maximum length subsequence = 5

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

Thanks to Neeraj for suggesting the above solution in the comments of this post.

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