# Longest Consecuetive Subsequence when only one insert operation is allowed

Given a sequence of positive integers of length N. The only operation allowed is to insert a single integer of any value at any position in the sequence. The task is to find the sub-sequence of maximum length that contains consecutive values in increasing order.

Examples:

Input: arr[] = {2, 1, 4, 5}
Output: 4
Insert element with value 3 at the 3rd position.(1 based indexing)
The new sequence becomes {2, 1, 3, 4, 5}
Longest consecutive sub-sequence would be {2, 3, 4, 5}

Input: arr[] = {2, 1, 2, 3, 5, 7}
Output: 5

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

Approach: The idea is to use Dynamic Programming.
Let dp[val] be the length of required subsequence that ends in an element equal to val and the element is not inserted yet. Let dp[val] be the length of required subsequence that ends in an element equal to val and some element has been inserted already.
Now break the problem into its subproblems as follows:

To calculate dp[val], as no element in inserted, the length of the subsequence will increase by 1 from its previous value
dp[val] = 1 + dp[val – 1].

To calculate dp[val], consider these two cases:

1. When the element is already inserted for (val-1), then there would be an increment of length 1 from dp[ val-1 ][ 1 ]
2. When the element has not been inserted yet, then the element with value (val-1) can be inserted . Hence there would be an increment of length 2 from dp[ val-2 ][ 0 ].

Take maximum of both the above cases.
dp[val] = max(1 + dp[val – 1], 2 + dp[val – 2]).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the length of longest ` `// consecuetive subsequence after inserting an element ` `int` `LongestConsSeq(``int` `arr[], ``int` `N) ` `{ ` ` `  `    ``// Variable to find maximum value of the array ` `    ``int` `maxval = 1; ` ` `  `    ``// Calulating maximum value of the array ` `    ``for` `(``int` `i = 0; i < N; i += 1) { ` ` `  `        ``maxval = max(maxval, arr[i]); ` `    ``} ` ` `  `    ``// Declaring the DP table ` `    ``int` `dp[maxval + 1] = { 0 }; ` ` `  `    ``// Variable to store the maximum length ` `    ``int` `ans = 1; ` ` `  `    ``// Iterating for every value present in the array ` `    ``for` `(``int` `i = 0; i < N; i += 1) { ` ` `  `        ``// Recurrence for dp[val] ` `        ``dp[arr[i]] = (1 + dp[arr[i] - 1]); ` ` `  `        ``// No value can be inserted before 1, ` `        ``// hence the element value should be ` `        ``// greater than 1 for this recurrance relation ` `        ``if` `(arr[i] >= 2) ` ` `  `            ``// Recurrence for dp[val] ` `            ``dp[arr[i]] = max(1 + dp[arr[i] - 1], ` `                                ``2 + dp[arr[i] - 2]); ` `        ``else` ` `  `            ``// Maximum length of consecutive sequence ` `            ``// ending at 1 is equal to 1 ` `            ``dp[arr[i]] = 1; ` ` `  `        ``// Update the ans variable with ` `        ``// the new maximum length possible ` `        ``ans = max(ans, dp[arr[i]]); ` `    ``} ` ` `  `    ``// Return the ans ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``// Input array ` `    ``int` `arr[] = { 2, 1, 4, 5 }; ` ` `  `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << LongestConsSeq(arr, N); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of above approach ` ` `  `class` `GFG ` `{  ` `    ``// Function to return the length of longest ` `    ``// consecuetive subsequence after inserting an element ` `    ``static` `int` `LongestConsSeq(``int` `[] arr, ``int` `N) ` `    ``{ ` `     `  `        ``// Variable to find maximum value of the array ` `        ``int` `maxval = ``1``; ` `     `  `        ``// Calulating maximum value of the array ` `        ``for` `(``int` `i = ``0``; i < N; i += ``1``) ` `        ``{ ` `            ``maxval = Math. max(maxval, arr[i]); ` `        ``} ` `     `  `        ``// Declaring the DP table ` `        ``int` `[][] dp = ``new` `int``[maxval + ``1``][``2``]; ` `     `  `        ``// Variable to store the maximum length ` `        ``int` `ans = ``1``; ` `     `  `        ``// Iterating for every value present in the array ` `        ``for` `(``int` `i = ``0``; i < N; i += ``1``)  ` `        ``{ ` `     `  `            ``// Recurrence for dp[val] ` `            ``dp[arr[i]][``0``] = (``1` `+ dp[arr[i] - ``1``][``0``]); ` `     `  `            ``// No value can be inserted before 1, ` `            ``// hence the element value should be ` `            ``// greater than 1 for this recurrance relation ` `            ``if` `(arr[i] >= ``2``) ` `     `  `                ``// Recurrence for dp[val] ` `                ``dp[arr[i]][``1``] = Math.max(``1` `+ dp[arr[i] - ``1``][``1``], ` `                                    ``2` `+ dp[arr[i] - ``2``][``0``]); ` `            ``else` `     `  `                ``// Maximum length of consecutive sequence ` `                ``// ending at 1 is equal to 1 ` `                ``dp[arr[i]][``1``] = ``1``; ` `     `  `            ``// Update the ans variable with ` `            ``// the new maximum length possible ` `            ``ans = Math.max(ans, dp[arr[i]][``1``]); ` `        ``} ` `     `  `        ``// Return the ans ` `        ``return` `ans; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `         `  `        ``// Input array ` `        ``int` `[] arr = { ``2``, ``1``, ``4``, ``5` `}; ` `     `  `        ``int` `N = arr.length; ` `     `  `        ``System.out.println(LongestConsSeq(arr, N)); ` `    ``} ` `} ` ` `  `// This code is contributed by ihritik `

## Python3

 `# Python3 implementation of above approach ` ` `  `# Function to return the length of longest ` `# consecuetive subsequence after inserting an element ` `def` `LongestConsSeq(arr, N): ` ` `  `    ``# Variable to find maximum value of the array ` `    ``maxval ``=` `1` ` `  `    ``# Calulating maximum value of the array ` `    ``for` `i ``in` `range``(N): ` ` `  `        ``maxval ``=` `max``(maxval, arr[i]) ` `     `  ` `  `    ``# Declaring the DP table ` `    ``dp``=``[[ ``0` `for` `i ``in` `range``(``2``)] ``for` `i ``in` `range``(maxval ``+` `1``)] ` ` `  `    ``# Variable to store the maximum length ` `    ``ans ``=` `1` ` `  `    ``# Iterating for every value present in the array ` `    ``for` `i ``in` `range``(N): ` ` `  `        ``# Recurrence for dp[val] ` `        ``dp[arr[i]][``0``] ``=` `1` `+` `dp[arr[i] ``-` `1``][``0``] ` ` `  `        ``# No value can be inserted before 1, ` `        ``# hence the element value should be ` `        ``# greater than 1 for this recurrance relation ` `        ``if` `(arr[i] >``=` `2``): ` ` `  `            ``# Recurrence for dp[val] ` `            ``dp[arr[i]][``1``] ``=` `max``(``1` `+` `dp[arr[i] ``-` `1``][``1``], ` `                                ``2` `+` `dp[arr[i] ``-` `2``][``0``]) ` `        ``else``: ` ` `  `            ``# Maximum length of consecutive sequence ` `            ``# ending at 1 is equal to 1 ` `            ``dp[arr[i]][``1``] ``=` `1` ` `  `        ``# Update the ans variable with ` `        ``# the new maximum length possible ` `        ``ans ``=` `max``(ans, dp[arr[i]][``1``]) ` `     `  ` `  `    ``# Return the ans ` `    ``return` `ans ` ` `  `# Driver code ` ` `  `arr``=``[``2``, ``1``, ``4``, ``5``] ` ` `  `N ``=` `len``(arr) ` ` `  `print``(LongestConsSeq(arr, N)) ` ` `  `# This code is contributed by mohit kumar 29 `

## C#

 `// C# implementation of above approach ` `using` `System; ` ` `  `class` `GFG ` `{  ` `    ``// Function to return the length of longest ` `    ``// consecuetive subsequence after inserting an element ` `    ``static` `int` `LongestConsSeq(``int` `[] arr, ``int` `N) ` `    ``{ ` `     `  `        ``// Variable to find maximum value of the array ` `        ``int` `maxval = 1; ` `     `  `        ``// Calulating maximum value of the array ` `        ``for` `(``int` `i = 0; i < N; i += 1)  ` `        ``{ ` `     `  `            ``maxval =Math.Max(maxval, arr[i]); ` `        ``} ` `     `  `        ``// Declaring the DP table ` `        ``int` `[ , ] dp = ``new` `int``[maxval + 1, 2]; ` `     `  `        ``// Variable to store the maximum length ` `        ``int` `ans = 1; ` `     `  `        ``// Iterating for every value present in the array ` `        ``for` `(``int` `i = 0; i < N; i += 1)  ` `        ``{ ` `     `  `            ``// Recurrence for dp[val] ` `            ``dp[arr[i], 0] = (1 + dp[arr[i] - 1, 0]); ` `     `  `            ``// No value can be inserted before 1, ` `            ``// hence the element value should be ` `            ``// greater than 1 for this recurrance relation ` `            ``if` `(arr[i] >= 2) ` `     `  `                ``// Recurrence for dp[val] ` `                ``dp[arr[i], 1] = Math.Max(1 + dp[arr[i] - 1, 1], ` `                                    ``2 + dp[arr[i] - 2, 0]); ` `            ``else` `     `  `                ``// Maximum length of consecutive sequence ` `                ``// ending at 1 is equal to 1 ` `                ``dp[arr[i], 1] = 1; ` `     `  `            ``// Update the ans variable with ` `            ``// the new maximum length possible ` `            ``ans = Math.Max(ans, dp[arr[i], 1]); ` `        ``} ` `     `  `        ``// Return the ans ` `        ``return` `ans; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main ()  ` `    ``{ ` `         `  `        ``// Input array ` `        ``int` `[] arr = ``new` `int` `[] { 2, 1, 4, 5 }; ` `     `  `        ``int` `N = arr.Length; ` `     `  `        ``Console.WriteLine(LongestConsSeq(arr, N)); ` `    ``} ` `} ` ` `  `// This code is contributed by ihritik `

Output:

```4
```

Time Complexity: O(N)
Space Complexity: O(MaxValue) where MaxValue is the maximum value present in the array.

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Improved By : ihritik, mohit kumar 29