Problem:
Score : 500 points

Problem Statement

You are given a sequence with N integers: A = \{ A_1, A_2, \cdots, A_N \}.
For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:

If A_i and A_j (i < j) are painted with the same color, A_i < A_j.

Find the minimum number of colors required to satisfy the condition.

Constraints

1 \leq N \leq 10^5

0 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print the minimum number of colors required to satisfy the condition.

Sample Input 1

5
2
1
4
5
3

Sample Output 1

2

We can satisfy the condition with two colors by, for example, painting 2 and 3 red and painting 1, 4, and 5 blue.

Sample Input 2

4
0
0
0
0

Sample Output 2

4

We have to paint all the integers with distinct colors.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int A[] = new int[N];
        TreeMap<Integer, ArrayList<Integer>> m = new TreeMap<>();
        for (int i = 0; i < N; ++i) A[i] = sc.nextInt();
        int dup = 0;
        int currentKey = A[0];
        for (int i = 0; i < N; ++i) {
            if (currentKey < A[i]) {
                ArrayList<Integer> current = m.get(currentKey);
                m.remove(currentKey);
                current.add(A[i]);
                m.put(A[i], current);
                currentKey = A[i];
            } else {
                Integer lower = m.lowerKey(A[i]);
                if (lower == null) {
                    if (m.containsKey(A[i])) ++dup;
                    ArrayList<Integer> a = new ArrayList<>();
                    a.add(A[i]);
                    m.put(A[i], a);
                } else {
                    ArrayList<Integer> current = m.get(lower);
                    m.remove(lower);
                    current.add(A[i]);
                    m.put(A[i], current);
                }
                currentKey = A[i];
            }
        }
        System.out.println(m.size() + dup);
    }
}