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.

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[0]);
cout << longestSubSeq(n, arr); } // This code is contributed by ihritik [tabby title = "Java"]

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// 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));
    }
}

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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#

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// 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

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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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Improved By : AnkitRai01, ihritik



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