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}.

**Method 1:** Previously an approach having time complexity of O(n^{2}) 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 <bits/stdc++.h> ` `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<` `int` `, ` `int` `> um; ` ` ` ` ` `// to store the maximum length subsequence ` ` ` `int` `maxLen = 0; ` ` ` ` ` `// traverse the array elements ` ` ` `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 'um' and its length of ` ` ` `// subsequence is greater than 'len' ` ` ` `if` `(um.find(arr[i]-1) != um.end() && len < um[arr[i]-1]) ` ` ` `len = um[arr[i]-1]; ` ` ` ` ` `// if 'arr[i]' is in 'um' and its length of ` ` ` `// subsequence is greater than 'len' ` ` ` `if` `(um.find(arr[i]) != um.end() && len < um[arr[i]]) ` ` ` `len = um[arr[i]]; ` ` ` ` ` `// if 'arr[i]+1' is in 'um' and its length of ` ` ` `// subsequence is greater than 'len' ` ` ` `if` `(um.find(arr[i]+1) != um.end() && len < um[arr[i]+1]) ` ` ` `len = um[arr[i]+1]; ` ` ` ` ` `// update arr[i] subsequence length in 'um' ` ` ` `um[arr[i]] = len + 1; ` ` ` ` ` `// update maximum length ` ` ` `if` `(maxLen < um[arr[i]]) ` ` ` `maxLen = um[arr[i]]; ` ` ` `} ` ` ` ` ` `// required maximum length subsequence ` ` ` `return` `maxLen; ` `} ` ` ` `// Driver program to test above ` `int` `main() ` `{ ` ` ` `int` `arr[] = {2, 5, 6, 3, 7, 6, 5, 8}; ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` `cout << ` `"Maximum length subsequence = "` ` ` `<< maxLenSub(arr, n); ` ` ` `return` `0; ` `} ` |

## 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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

## leave a comment

## 0 Comments