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.Arrays;
import java.util.Scanner;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        int a[] = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }
        int pos = 0;
        int ans[] = new int[n];
        Arrays.fill(ans, Integer.MAX_VALUE);
        int min = Integer.MAX_VALUE;
        for (int i = 0; i < n; i++) {
            int add = a[i];
            if (min >= add) {
                min = add;
                ans[pos] = add;
                pos++;
                continue;
            }
            // int max = -1;
            // int maxpos = -1;
            for (int j = 0; j < pos; j++) {
                if (ans[j] < add) {
                    // max = ans[j];
                    // maxpos = j;
                    ans[j] = add;
                    if (j == pos - 1) {
                        min = add;
                    }
                    break;
                }
            }
            /*
             * if (min > add) { min = add; } if (maxpos != -1) { ans[maxpos] = add; } else {
             * ans[pos] = add; pos++; }
             */
        }
        System.out.println(pos);
        sc.close();
    }
}