Subsequence with longest Increment

I. Problem Description

Set the sequence L = <A1, A2, A3 ,..., an> is a sequence whose length is N. An ascending sequence of L is described as: <aI1, ai2 ,..., AIK>, where the subscript sequence <I1, I2 ,..., ik> Yes

The sub-sequence <aI1, ai2,..., AIK> is also incremental. The length of this incremental sequence is K.

Solution 1: Converting to LCS

First, sort the sequence L in ascending order to get another sequence s, and then obtain the longest common subsequence of L and S.

Three solutions 2: Dynamic Planning

In addition, Len [I] indicates the length of the longest incrementing sub-sequence ending with element I. Finally, obtain max {Len [I ]}.

Instance <1, 3, 4, 2, 7>

Len [1] = 1. The longest incrementing subsequence ending with 1 is <1> and the length is 1.

Len [2] = 2. The longest incrementing subsequence ending with 3 is <1, 3> and the length is 2.

Len [3] = 3. The longest incrementing subsequence ending with 4 is <1, 3, 4>, and the length is 3.

Len [4] = 2. The longest incrementing sub-sequence ending with 2 is <1, 2> and the length is 2.

So how can we find Len [5? Into the following subproblem:

If 5th elements are equal to 7, find the element a [I] whose Sn is before a [5] and earlier than a [5]. the longest incrementing subsequence ending with a [I] plus a [5] to form a new longest incrementing subsequence.

The length is 1 longer than the original one. There are multiple (up to four) elements smaller than a [5], and the resulting incremental subsequence also has multiple, the longest one is equal to the longest incrementing subsequence ending with a [5.

In this example, a [3] is smaller than a [5], and the longest incrementing subsequence ending with a [3] is the longest, therefore, the length of the incremental sub-sequence ending with a [5] is incremented by 1 to 4.

Len [5] = 4. The longest incrementing sub-sequence ending with 5 is <,> and the length is 4.

Solution 4 3: Improvement of solution 2

In solution 2, when Len [I] is obtained, find all the elements smaller than a [I] in a [I-1], and a [1],... A [I-1] Is unordered. The search speed is slow.

F [k] is the final element in the ascending subsequence with the length equal to K. The longer the length is, the larger the end element is. Therefore, F [k] is incremental.

Instance <1, 3, 4, 2, 7>

Len [1] = 1, F [1] = A [1] = 1

Len [2] = 2, F [2] = A [2] = 3

Len [3] = 3, F [3] = A [3] = 4

Len [4] = 2, F [2] = A [4] = 2 (updated f [2])

So how can we find Len [5?

F [1] = 1, F [2] = 2, F [3] = 4, a [5] = 7,

Find the ascending subsequence whose end element is smaller than a [5] and has the longest length. In this example,

F [3] = 4 indicates a child sequence with a length equal to 3. The end element of the Child sequence is 4, and the length of the Child sequence is the longest.

A [5] adds the length of the new sequence formed by this sequence to 4, and then updates f [4] = A [5] = 7 to indicate the length.

Child sequence equal to 4, whose end element is equal to 7

In the above step, find the child sequence with the smallest element and the longest length than a [I ].

For f [1],... f [K], looking forward from the back, the first element smaller than a [I] is the longest subsequence. When searching

You can use the binary search method. Therefore, it is faster than solution 1.