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.


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:





// 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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