Problem:
Score : 300 points

Problem Statement

In a flower bed, there are N flowers, numbered 1,2,......,N. Initially, the heights of all flowers are 0.
You are given a sequence h=\{h_1,h_2,h_3,......\} as input. You would like to change the height of Flower k to h_k for all k (1 \leq k \leq N), by repeating the following "watering" operation:

Specify integers l and r. Increase the height of Flower x by 1 for all x such that l \leq x \leq r.

Find the minimum number of watering operations required to satisfy the condition.

Constraints

1 \leq N \leq 100

0 \leq h_i \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
h_1 h_2 h_3 ...... h_N

Output

Print the minimum number of watering operations required to satisfy the condition.

Sample Input 1

4
1 2 2 1

Sample Output 1

2

The minimum number of watering operations required is 2.
One way to achieve it is:

Perform the operation with (l,r)=(1,3).

Perform the operation with (l,r)=(2,4).

Sample Input 2

5
3 1 2 3 1

Sample Output 2

5

Sample Input 3

8
4 23 75 0 23 96 50 100

Sample Output 3

221

Solution in Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int[] h = new int[n];
    for(int i = 0; i < n; i++) {
      h[i] = sc.nextInt();
    }
    System.out.println(dfs(h));
  }

  public static int dfs(int[] hana) {
    int m = 100;
    for(int i = 0; i < hana.length; i++) {
      m = Math.min(m, hana[i]);
    }
    int ans = m;
    for(int i = 0; i < hana.length; i++) {
      hana[i] -= m;
    }
    int p = -1;
    int q = -1;
    int t = 0;
    for(int i = 0; i < hana.length; i++) {
      if(hana[i] > 0) {
        if(t == 0) {
          p = i;
          t = 1;
        } else {

        }
      } else {
        if(t == 1) {
          q = i - 1;
          t = 0;
          int[] hana2 = new int[q - p + 1];
          for(int j = 0; j < q - p + 1; j++) { 
            hana2[j] = hana[j + p];
          }
          ans += dfs(hana2);
          p = -1;
        }
      }
      if((i == hana.length - 1) && (p != -1)) {
        q = hana.length - 1;
        int[] hana2 = new int[q - p + 1];
        for(int j = 0; j < q - p + 1; j++) { 
          hana2[j] = hana[j + p];
        }
        ans += dfs(hana2);
      }
    }
    return ans;
  }
}