# Longest subsequence having greater corner values

Given an array arr[] containing a random permutation of first N natural numbers, the task is to find the longest sub-sequence having the property that the first and the last elements are greater than all the other sub-sequence elements.

Examples:

Input: arr[] = {3, 1, 5, 2, 4}
Output: 4
The sub-sequence is {3, 1, 2, 4}. The corner elements of this subsequence are greater than all other elements.

Input: arr[] = {1, 2, 3, 4, 5}
Output: 2
We cannot make a subsequence of size greater than 2.

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

Approach: If we fix the leftmost and the rightmost elements of a sub-sequence, we are interested in counting how many elements between them have a smaller value than both. A straight forward implementation of this idea has a complexity of O(N3).
In order to reduce the complexity, we approach the problem differently. Instead of fixing the ends of the sub-sequence, we fix the elements in between. The idea is that for a given X (1 ≤ X ≤ N), we want to find two elements greater or equal to X, that have between them as many elements as possible less than X. For a fixed X it’s optimal to choose the leftmost and rightmost elements ≤ X. Now we have a better O(N2) solution.
As X increases, the leftmost element can only increase, while the rightmost one can only decrease. We can use a pointer for each of them to get an amortised complexity of O(N).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `#define MAXN 100005 ` ` `  `// Function to return the length of the ` `// longest required sub-sequence ` `int` `longestSubSeq(``int` `n, ``int` `arr []) ` `{ ` `    ``int` `max_length = 0; ` ` `  `    ``// Create a position array to find ` `    ``// where an element is present ` `    ``int` `pos[MAXN]; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``pos[arr[i] - 1] = i; ` ` `  `    ``int` `left = n, right = 0; ` ` `  `    ``for` `(``int` `i = n - 1, num = 1; i >= 0;  ` `                       ``i -= 1, num += 1)  ` `    ``{ ` ` `  `        ``// Store the minimum position  ` `        ``// to the left ` `        ``left = min(left, pos[i]); ` ` `  `        ``// Store the maximum position to  ` `        ``// the right ` `        ``right = max(right, pos[i]); ` ` `  `        ``// Recompute current maximum ` `        ``max_length = max(max_length, ` `                         ``right - left - num + 3); ` `    ``} ` ` `  `    ``// Edge case when there is a single ` `    ``// element in the sequence ` `    ``if` `(n == 1) ` `        ``max_length = 1; ` ` `  `    ``return` `max_length; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 3, 4, 5 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``cout << longestSubSeq(n, arr); ` `} ` ` `  `// This code is contributed by ihritik `

## Java

 `// Java implementation of the approach ` `class` `GFG { ` ` `  `    ``static` `int` `MAXN = (``int``)1e5 + ``5``; ` ` `  `    ``// Function to return the length of the ` `    ``// longest required sub-sequence ` `    ``static` `int` `longestSubSeq(``int` `n, ``int``[] arr) ` `    ``{ ` `        ``int` `max_length = ``0``; ` ` `  `        ``// Create a position array to find ` `        ``// where an element is present ` `        ``int``[] pos = ``new` `int``[MAXN]; ` ` `  `        ``for` `(``int` `i = ``0``; i < n; i++) ` `            ``pos[arr[i] - ``1``] = i; ` ` `  `        ``int` `left = n, right = ``0``; ` ` `  `        ``for` `(``int` `i = n - ``1``, num = ``1``; i >= ``0``;  ` `                         ``i -= ``1``, num += ``1``) { ` ` `  `            ``// Store the minimum position  ` `            ``// to the left ` `            ``left = Math.min(left, pos[i]); ` ` `  `            ``// Store the maximum position to  ` `            ``// the right ` `            ``right = Math.max(right, pos[i]); ` ` `  `            ``// Recompute current maximum ` `            ``max_length = Math.max(max_length, ` `                      ``right - left - num + ``3``); ` `        ``} ` ` `  `        ``// Edge case when there is a single ` `        ``// element in the sequence ` `        ``if` `(n == ``1``) ` `            ``max_length = ``1``; ` ` `  `        ``return` `max_length; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `arr[] = { ``1``, ``2``, ``3``, ``4``, ``5` `}; ` `        ``int` `n = arr.length; ` `        ``System.out.println(longestSubSeq(n, arr)); ` `    ``} ` `} `

## Python3

 `# Python3 implementation of the approach ` ` `  `MAXN ``=` `100005` ` `  `# Function to return the length of the ` `# longest required sub-sequence ` `def` `longestSubSeq(n, arr): ` ` `  `    ``max_length ``=` `0` ` `  `    ``# Create a position array to find ` `    ``# where an element is present ` `    ``pos ``=` `[``0``] ``*` `MAXN ` ` `  `    ``for` `i ``in` `range` `(``0``, n): ` `        ``pos[arr[i] ``-` `1``] ``=` `i ` ` `  `    ``left ``=` `n ` `    ``right ``=` `0` `    ``num ``=` `1` `     `  `    ``for` `i ``in` `range` `(n ``-` `1``, ``-``1``, ``-``1``) :  ` ` `  `        ``# Store the minimum position  ` `        ``# to the left ` `        ``left ``=` `min``(left, pos[i]) ` ` `  `        ``# Store the maximum position to  ` `        ``# the right ` `        ``right ``=` `max``(right, pos[i]) ` ` `  `        ``# Recompute current maximum ` `        ``max_length ``=` `max``(max_length, ` `                ``right ``-` `left ``-` `num ``+` `3``) ` `     `  `        ``num ``=` `num ``+` `1` `         `  `    ``# Edge case when there is a single ` `    ``# element in the sequence ` `    ``if` `(n ``=``=` `1``) : ` `        ``max_length ``=` `1` ` `  `    ``return` `max_length ` ` `  `# Driver code ` `arr ``=` `[ ``1``, ``2``, ``3``, ``4``, ``5` `] ` `n ``=` `len``(arr) ` `print``(longestSubSeq(n, arr)) ` ` `  `# This code is contributed by ihritik `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` ` `  `    ``static` `int` `MAXN = (``int``)1e5 + 5;  ` ` `  `    ``// Function to return the length of the  ` `    ``// longest required sub-sequence  ` `    ``static` `int` `longestSubSeq(``int` `n, ``int``[] arr)  ` `    ``{  ` `        ``int` `max_length = 0;  ` ` `  `        ``// Create a position array to find  ` `        ``// where an element is present  ` `        ``int``[] pos = ``new` `int``[MAXN];  ` ` `  `        ``for` `(``int` `i = 0; i < n; i++)  ` `            ``pos[arr[i] - 1] = i;  ` ` `  `        ``int` `left = n, right = 0;  ` ` `  `        ``for` `(``int` `i = n - 1, num = 1; i >= 0;  ` `                        ``i -= 1, num += 1)  ` `        ``{  ` ` `  `            ``// Store the minimum position  ` `            ``// to the left  ` `            ``left = Math.Min(left, pos[i]);  ` ` `  `            ``// Store the maximum position to  ` `            ``// the right  ` `            ``right = Math.Max(right, pos[i]);  ` ` `  `            ``// Recompute current maximum  ` `            ``max_length = Math.Max(max_length,  ` `                    ``right - left - num + 3);  ` `        ``}  ` ` `  `        ``// Edge case when there is a single  ` `        ``// element in the sequence  ` `        ``if` `(n == 1)  ` `            ``max_length = 1;  ` ` `  `        ``return` `max_length;  ` `    ``}  ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `        ``int` `[]arr = { 1, 2, 3, 4, 5 };  ` `        ``int` `n = arr.Length;  ` `        ``Console.WriteLine(longestSubSeq(n, arr));  ` `    ``}  ` `}  ` ` `  `// This code is contributed by Ryuga `

## PHP

 `= 0 ; ``\$i``--, ``\$num``++)  ` `    ``{ ` `         `  `        ``// Store the minimum position  ` `        ``// to the left ` `        ``\$left` `= min(``\$left``, ``\$pos``[``\$i``]); ` ` `  `        ``// Store the maximum position to  ` `        ``// the right ` `        ``\$right` `= max(``\$right``, ``\$pos``[``\$i``]); ` ` `  `        ``// Recompute current maximum ` `        ``\$max_length` `= max(``\$max_length``, ` `                          ``\$right` `- ``\$left` `- ``\$num` `+ 3); ` `    ``} ` `     `  `    ``// Edge case when there is a single ` `    ``// element in the sequence ` `    ``if` `(``\$n` `== 1)  ` `        ``\$max_length` `= 1; ` ` `  `    ``return` `\$max_length``; ` `} ` ` `  `// Driver code ` `\$arr` `= ``array``(1, 2, 3, 4, 5); ` `\$n` `= sizeof(``\$arr``); ` `echo` `longestSubSeq(``\$n``, ``\$arr``); ` ` `  `// This code is contributed by ihritik ` `?> `

Output:

```2
```

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