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: Accepted)

import java.util.*;

class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = Integer.parseInt(sc.next());
		Integer[] A = new Integer[N];
		for (int i=0;i<N;i++) {
			A[i] = Integer.parseInt(sc.next());
		}
		List<Integer> A_list = Arrays.asList(A);
		Collections.reverse(A_list);
		Integer[] A_2 = (Integer[])A_list.toArray(new Integer[N]);
		System.out.println(LIS(A_2));
	}

	static int LIS (Integer[] A) {
		int N = A.length;
		Integer[] dp = new Integer[N];
		for (int i=0;i<N;i++) {
			dp[i] = Integer.MAX_VALUE;
		}
		for (int i=0;i<N;i++) {
			int index = Arrays.binarySearch(dp, A[i], (x,y)->x.compareTo(y)>0?1:-1);
			if (index<0) {index = -index-2;}
			dp[index+1] = A[i];
		}
		int ans = 0;
		for (int i=0;i<N;i++) {
			if (dp[i] != Integer.MAX_VALUE) {
				ans++;
			}
		}
		return ans;
	}
}