# Longest Increasing consecutive subsequence

Given N elements, write a program that prints the length of the longest increasing subsequence whose adjacent element difference is one.

Examples:

Input : a[] = {3, 10, 3, 11, 4, 5, 6, 7, 8, 12}
Output : 6
Explanation: 3, 4, 5, 6, 7, 8 is the longest increasing subsequence whose adjacent element differs by one.

Input : a[] = {6, 7, 8, 3, 4, 5, 9, 10}
Output : 5
Explanation: 6, 7, 8, 9, 10 is the longest increasing subsequence

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

Naive Approach: A normal approach will be to iterate for every element and find out the longest increasing subsequence. For any particular element, find the length of the subsequence starting from that element. Print the longest length of the subsequence thus formed. The time complexity of this approach will be O(n2).

Dynamic Programming Approach: Let DP[i] store the length of the longest subsequence which ends with A[i]. For every A[i], if A[i]-1 is present in the array before i-th index, then A[i] will add to the increasing subsequence which has A[i]-1. Hence DP[i] = DP[ index(A[i]-1) ] + 1. If A[i]-1 is not present in the array before i-th index, then DP[i]=1 since the A[i] element forms a subsequence which starts with A[i]. Hence the relation for DP[i] is:

If A[i]-1 is present before i-th index:

• DP[i] = DP[ index(A[i]-1) ] + 1
• else:

• DP[i] = 1
• Given below is the illustration of the above approach:

## C++

 `// CPP program to find length of the ` `// longest increasing subsequence ` `// whose adjacent element differ by 1 ` `#include ` `using` `namespace` `std; ` ` `  `// function that returns the length of the ` `// longest increasing subsequence ` `// whose adjacent element differ by 1 ` `int` `longestSubsequence(``int` `a[], ``int` `n) ` `{ ` `    ``// stores the index of elements ` `    ``unordered_map<``int``, ``int``> mp; ` ` `  `    ``// stores the length of the longest ` `    ``// subsequence that ends with a[i] ` `    ``int` `dp[n]; ` `    ``memset``(dp, 0, ``sizeof``(dp)); ` ` `  `    ``int` `maximum = INT_MIN; ` ` `  `    ``// iterate for all element ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// if a[i]-1 is present before i-th index ` `        ``if` `(mp.find(a[i] - 1) != mp.end()) { ` ` `  `            ``// last index of a[i]-1 ` `            ``int` `lastIndex = mp[a[i] - 1] - 1; ` ` `  `            ``// relation ` `            ``dp[i] = 1 + dp[lastIndex]; ` `        ``} ` `        ``else` `            ``dp[i] = 1; ` ` `  `        ``// stores the index as 1-index as we need to ` `        ``// check for occurrence, hence 0-th index ` `        ``// will not be possible to check ` `        ``mp[a[i]] = i + 1; ` ` `  `        ``// stores the longest length ` `        ``maximum = max(maximum, dp[i]); ` `    ``} ` ` `  `    ``return` `maximum; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `a[] = { 3, 10, 3, 11, 4, 5, 6, 7, 8, 12 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` `    ``cout << longestSubsequence(a, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find length of the ` `// longest increasing subsequence ` `// whose adjacent element differ by 1 ` ` `  `import` `java.util.*; ` `class` `lics { ` `    ``static` `int` `LongIncrConseqSubseq(``int` `arr[], ``int` `n) ` `    ``{ ` `        ``// create hashmap to save latest consequent  ` `        ``// number as "key" and its length as "value" ` `        ``HashMap map = ``new` `HashMap<>(); ` `        `  `        ``// put first element as "key" and its length as "value" ` `        ``map.put(arr[``0``], ``1``); ` `        ``for` `(``int` `i = ``1``; i < n; i++) { ` `        `  `            ``// check if last consequent of arr[i] exist or not ` `            ``if` `(map.containsKey(arr[i] - ``1``)) { ` `        `  `                ``// put the updated consequent number  ` `                ``// and increment its value(length) ` `                ``map.put(arr[i], map.get(arr[i] - ``1``) + ``1``); ` `           `  `                ``// remove the last consequent number ` `                ``map.remove(arr[i] - ``1``); ` `            ``} ` ` `  `            ``// if their is no last consequent of ` `            ``// arr[i] then put arr[i] ` `            ``else` `{ ` `                ``map.put(arr[i], ``1``); ` `            ``} ` `        ``} ` `        ``return` `Collections.max(map.values()); ` `    ``} ` ` `  `    ``// driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``// Take input from user ` `        ``Scanner sc = ``new` `Scanner(System.in); ` `        ``int` `n = sc.nextInt(); ` `        ``int` `arr[] = ``new` `int``[n]; ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `            ``arr[i] = sc.nextInt(); ` `        ``System.out.println(LongIncrConseqSubseq(arr, n)); ` `    ``} ` `} ` `// This code is contributed by CrappyDoctor `

## Python3

 `# python program to find length of the  ` `# longest increasing subsequence  ` `# whose adjacent element differ by 1  ` ` `  `from` `collections ``import` `defaultdict ` `import` `sys ` ` `  `# function that returns the length of the  ` `# longest increasing subsequence  ` `# whose adjacent element differ by 1  ` ` `  `def` `longestSubsequence(a, n): ` `    ``mp ``=` `defaultdict(``lambda``:``0``) ` ` `  `    ``# stores the length of the longest  ` `    ``# subsequence that ends with a[i]  ` `    ``dp ``=` `[``0` `for` `i ``in` `range``(n)] ` `    ``maximum ``=` `-``sys.maxsize ` ` `  `    ``# iterate for all element  ` `    ``for` `i ``in` `range``(n): ` ` `  `        ``# if a[i]-1 is present before i-th index  ` `        ``if` `a[i] ``-` `1` `in` `mp: ` ` `  `            ``# last index of a[i]-1  ` `            ``lastIndex ``=` `mp[a[i] ``-` `1``] ``-` `1` ` `  `            ``# relation  ` `            ``dp[i] ``=` `1` `+` `dp[lastIndex] ` `        ``else``: ` `            ``dp[i] ``=` `1` ` `  `            ``# stores the index as 1-index as we need to  ` `            ``# check for occurrence, hence 0-th index  ` `            ``# will not be possible to check  ` `        ``mp[a[i]] ``=` `i ``+` `1` ` `  `        ``# stores the longest length  ` `        ``maximum ``=` `max``(maximum, dp[i]) ` `    ``return` `maximum ` ` `  ` `  `# Driver Code  ` `a ``=` `[``3``, ``10``, ``3``, ``11``, ``4``, ``5``, ``6``, ``7``, ``8``, ``12``] ` `n ``=` `len``(a) ` `print``(longestSubsequence(a, n)) ` ` `  `# This code is contributed by Shrikant13 `

Output:

```6
```

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

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Improved By : shrikanth13, CrappyDoctor