# 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}.

**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; ` `} ` |

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## 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<Integer, Integer> 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 ` |

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## 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 ` |

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**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.

This article is contributed by **Ayush Jauhari**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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