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# Largest increasing subsequence of consecutive integers

• Difficulty Level : Medium
• Last Updated : 22 Jun, 2021

Given an array of n positive integers. We need to find the largest increasing sequence of consecutive positive integers.
Examples:

```Input : arr[] = {5, 7, 6, 7, 8}
Output : Size of LIS = 4
LIS = 5, 6, 7, 8

Input : arr[] = {5, 7, 8, 7, 5}
Output : Size of LIS = 2
LIS = 7, 8```

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This problem can be solved easily by the concept of LIS where each next greater element differ from earlier one by 1. But this will take O(n^2) time complexity.
With the use of hashing we can finding the size of longest increasing sequence with consecutive integers in time complexity of O(n).
We create a hash table.. Now for each element arr[i], we perform hash[arr[i]] = hash[arr[i] – 1] + 1. So, for every element we know longest consecutive increasing subsequence ending with it. Finally we return maximum value from hash table.

## C++

 `// C++ implementation of longest continuous increasing``// subsequence``#include ``using` `namespace` `std;` `// Function for LIS``int` `findLIS(``int` `A[], ``int` `n)``{``    ``unordered_map<``int``, ``int``> hash;` `    ``// Initialize result``    ``int` `LIS_size = 1;``    ``int` `LIS_index = 0;` `    ``hash[A] = 1;` `    ``// iterate through array and find``    ``// end index of LIS and its Size``    ``for` `(``int` `i = 1; i < n; i++) {``        ``hash[A[i]] = hash[A[i] - 1] + 1;``        ``if` `(LIS_size < hash[A[i]]) {``            ``LIS_size = hash[A[i]];``            ``LIS_index = A[i];``        ``}``    ``}` `    ``// print LIS size``    ``cout << ``"LIS_size = "` `<< LIS_size << ``"\n"``;` `    ``// print LIS after setting start element``    ``cout << ``"LIS : "``;``    ``int` `start = LIS_index - LIS_size + 1;``    ``while` `(start <= LIS_index) {``        ``cout << start << ``" "``;``        ``start++;``    ``}``}` `// driver``int` `main()``{``    ``int` `A[] = { 2, 5, 3, 7, 4, 8, 5, 13, 6 };``    ``int` `n = ``sizeof``(A) / ``sizeof``(A);``    ``findLIS(A, n);``    ``return` `0;``}`

## Java

 `// Java implementation of longest continuous increasing``// subsequence``import` `java.util.*;` `class` `GFG``{` `// Function for LIS``static` `void` `findLIS(``int` `A[], ``int` `n)``{``    ``Map hash = ``new` `HashMap();` `    ``// Initialize result``    ``int` `LIS_size = ``1``;``    ``int` `LIS_index = ``0``;` `    ``hash.put(A[``0``], ``1``);``    ``// iterate through array and find``    ``// end index of LIS and its Size``    ``for` `(``int` `i = ``1``; i < n; i++)``    ``{``        ``hash.put(A[i], hash.get(A[i] - ``1``)==``null``? ``1``:hash.get(A[i] - ``1``)+``1``);``        ``if` `(LIS_size < hash.get(A[i]))``        ``{``            ``LIS_size = hash.get(A[i]);``            ``LIS_index = A[i];``        ``}``    ``}` `    ``// print LIS size``    ``System.out.println(``"LIS_size = "` `+ LIS_size);` `    ``// print LIS after setting start element``    ``System.out.print(``"LIS : "``);``    ``int` `start = LIS_index - LIS_size + ``1``;``    ``while` `(start <= LIS_index)``    ``{``        ``System.out.print(start + ``" "``);``        ``start++;``    ``}``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `A[] = { ``2``, ``5``, ``3``, ``7``, ``4``, ``8``, ``5``, ``13``, ``6` `};``    ``int` `n = A.length;``    ``findLIS(A, n);``}``}` `// This code is contributed by Princi Singh`

## Python3

 `# Python3 implementation of longest``# continuous increasing subsequence` `# Function for LIS``def` `findLIS(A, n):``    ``hash` `=` `dict``()` `    ``# Initialize result``    ``LIS_size, LIS_index ``=` `1``, ``0` `    ``hash``[A[``0``]] ``=` `1` `    ``# iterate through array and find``    ``# end index of LIS and its Size``    ``for` `i ``in` `range``(``1``, n):` `        ``# If the desired key is not present``        ``# in dictionary, it will throw key error,``        ``# to avoid this error this is necessary``        ``if` `A[i] ``-` `1` `not` `in` `hash``:``            ``hash``[A[i] ``-` `1``] ``=` `0` `        ``hash``[A[i]] ``=` `hash``[A[i] ``-` `1``] ``+` `1``        ``if` `LIS_size < ``hash``[A[i]]:``            ``LIS_size ``=` `hash``[A[i]]``            ``LIS_index ``=` `A[i]``    ` `    ``# print LIS size``    ``print``(``"LIS_size ="``, LIS_size)` `    ``# print LIS after setting start element``    ``print``(``"LIS : "``, end ``=` `"")` `    ``start ``=` `LIS_index ``-` `LIS_size ``+` `1``    ``while` `start <``=` `LIS_index:``        ``print``(start, end ``=` `" "``)``        ``start ``+``=` `1` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ``A ``=` `[ ``2``, ``5``, ``3``, ``7``, ``4``, ``8``, ``5``, ``13``, ``6` `]``    ``n ``=` `len``(A)``    ``findLIS(A, n)` `# This code is contributed by sanjeev2552`

## C#

 `// C# implementation of longest continuous increasing``// subsequence``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `// Function for LIS``static` `void` `findLIS(``int` `[]A, ``int` `n)``{``    ``Dictionary<``int``,``int``> hash = ``new` `Dictionary<``int``,``int``>();` `    ``// Initialize result``    ``int` `LIS_size = 1;``    ``int` `LIS_index = 0;` `    ``hash.Add(A, 1);``    ` `    ``// iterate through array and find``    ``// end index of LIS and its Size``    ``for` `(``int` `i = 1; i < n; i++)``    ``{``        ``if``(hash.ContainsKey(A[i]-1))``        ``{``            ``var` `val = hash[A[i]-1];``            ``hash.Remove(A[i]);``            ``hash.Add(A[i], val + 1);``        ``}``        ``else``        ``{``            ``hash.Add(A[i], 1);``        ``}``        ``if` `(LIS_size < hash[A[i]])``        ``{``            ``LIS_size = hash[A[i]];``            ``LIS_index = A[i];``        ``}``    ``}` `    ``// print LIS size``    ``Console.WriteLine(``"LIS_size = "` `+ LIS_size);` `    ``// print LIS after setting start element``    ``Console.Write(``"LIS : "``);``    ``int` `start = LIS_index - LIS_size + 1;``    ``while` `(start <= LIS_index)``    ``{``        ``Console.Write(start + ``" "``);``        ``start++;``    ``}``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]A = { 2, 5, 3, 7, 4, 8, 5, 13, 6 };``    ``int` `n = A.Length;``    ``findLIS(A, n);``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``
Output:
```LIS_size = 5
LIS : 2 3 4 5 6 ```

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