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# 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 :
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 :
Explanation: 6, 7, 8, 9, 10 is the longest increasing subsequence

Naive Approach: For any particular element, find the length of the subsequence starting from the first element. Print the longest length of the subsequence thus formed. The time complexity of this approach will be O(n).

## C++

 `#include ` `using` `namespace` `std;` `int` `LongestSubsequence(``int` `a[], ``int` `n)` `{` `    ``// initialize x to the first element of the array;` `    ``int` `x = a;` `    ``// initialize count to Zero` `    ``int` `count = 0, y = 0;` `    ``// iterate for all elements` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``if` `(a[i] == (x + y)) {` `            ``count++;` `            ``y++;` `        ``}` `    ``}` `    ``return` `count;` `}` `int` `main()` `{` `    ``int` `a[] = { 6, 7, 8, 3, 4, 5, 9, 10 };` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a);` `    ``cout << LongestSubsequence(a, n);`   `    ``return` `0;` `}`

## Java

 `import` `java.util.Scanner;`   `public` `class` `Main {`   `  ``public` `static` `int` `LongestSubsequence(``int` `a[], ``int` `n)` `  ``{`   `    ``// initialize x to the first element of the array;` `    ``int` `x = a[``0``];`   `    ``// initialize count to Zero` `    ``int` `count = ``0``, y = ``0``;`   `    ``// iterate for all elements` `    ``for` `(``int` `i = ``0``; i < n; i++) {` `      ``if` `(a[i] == (x + y)) {` `        ``count++;` `        ``y++;` `      ``}` `    ``}` `    ``return` `count;` `  ``}` `  ``public` `static` `void` `main(String[] args) {` `    ``int``[] a = {``6``, ``7``, ``8``, ``3``, ``4``, ``5``, ``9``, ``10``};` `    ``int` `n = a.length;` `    ``System.out.println(LongestSubsequence(a, n));` `  ``}` `}`   `// This code contributed by Ajax`

## Python3

 `def` `LongestSubsequence(a, n):` `    ``# initialize x to the first element of the array;` `    ``x ``=` `a[``0``]` `    `  `    ``# initialize count to Zero` `    ``count ``=` `0` `    ``y ``=` `0` `    `  `    ``# iterate for all elements` `    ``for` `i ``in` `range``(n):` `        ``if` `a[i] ``=``=` `(x ``+` `y):` `            ``count ``+``=` `1` `            ``y ``+``=` `1` `    ``return` `count`   `a ``=` `[``6``, ``7``, ``8``, ``3``, ``4``, ``5``, ``9``, ``10``]` `n ``=` `len``(a)` `print``(LongestSubsequence(a, n))`   `# This code is contributed by laxmisinde5t82.`

## C#

 `using` `System;`   `public` `class` `GFG {`   `    ``public` `static` `int` `LongestSubsequence(``int``[] a, ``int` `n)` `    ``{`   `        ``// initialize x to the first element of the array;` `        ``int` `x = a;`   `        ``// initialize count to Zero` `        ``int` `count = 0, y = 0;`   `        ``// iterate for all elements` `        ``for` `(``int` `i = 0; i < n; i++) {` `            ``if` `(a[i] == (x + y)) {` `                ``count++;` `                ``y++;` `            ``}` `        ``}` `        ``return` `count;` `    ``}` `    ``public` `static` `void` `Main(``string``[] args)` `    ``{` `        ``int``[] a = { 6, 7, 8, 3, 4, 5, 9, 10 };` `        ``int` `n = a.Length;` `        ``Console.WriteLine(LongestSubsequence(a, n));` `    ``}` `}`   `// This code contributed by Karandeep1234`

## Javascript

 `function` `LongestSubsequence(a, n)` `{`   `    ``// initialize x to the first element of the array;` `    ``var` `x = a;` `    `  `    ``// initialize count to Zero` `    ``var` `count = 0, y = 0;` `    `  `    ``// iterate for all elements` `    ``for` `(``var` `i = 0; i < n; i++) {` `        ``if` `(a[i] == (x + y)) {` `            ``count++;` `            ``y++;` `        ``}` `    ``}` `    ``return` `count;` `}`   `var` `a = [6, 7, 8, 3, 4, 5, 9, 10];` `var` `n = a.length;` `console.log(LongestSubsequence(a, n));`   `// This code contributed by Prasad Kandekar(prasad264)`

Output

`5`

Time Complexity: O(n)
Auxiliary Space: O(1)

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 there 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`

## C#

 `// C# program to find length of the` `// longest increasing subsequence` `// whose adjacent element differ by 1` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG{` `    `  `static` `int` `longIncrConseqSubseq(``int` `[]arr, ` `                                ``int` `n)` `{` `  ``// Create hashmap to save ` `  ``// latest consequent number ` `  ``// as "key" and its length ` `  ``// as "value"` `  ``Dictionary<``int``, ` `             ``int``> map = ``new` `Dictionary<``int``, ` `                                       ``int``>();`   `  ``// Put first element as "key" ` `  ``// and its length as "value"` `  ``map.Add(arr, 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.Add(arr[i], map[arr[i] - 1] + 1);`   `      ``// Remove the last consequent number` `      ``map.Remove(arr[i] - 1);` `    ``}`   `    ``// If there is no last consequent of` `    ``// arr[i] then put arr[i]` `    ``else` `    ``{` `      ``if``(!map.ContainsKey(arr[i]))` `        ``map.Add(arr[i], 1);` `    ``}` `  ``}` `  `  `  ``int` `max = ``int``.MinValue;` `  ``foreach``(KeyValuePair<``int``, ` `                       ``int``> entry ``in` `map)` `  ``{` `    ``if``(entry.Value > max)` `    ``{` `      ``max = entry.Value;` `    ``}` `  ``}` `  ``return` `max;` `}`   `// Driver code` `public` `static` `void` `Main(String []args)` `{` `  ``// Take input from user` `  ``int` `[]arr = {3, 10, 3, 11, ` `               ``4, 5, 6, 7, 8, 12};` `  ``int` `n = arr.Length;` `  ``Console.WriteLine(longIncrConseqSubseq(arr, n));` `}` `}`   `// This code is contributed by gauravrajput1`

## Javascript

 ``

Output

`6`

Complexity Analysis:

• Time Complexity: O(N), as we are using a loop to traverse N times.
• Auxiliary Space: O(N), as we are using extra space for dp and map m.

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