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 Scala (2.11.7) (judge verdict: Accepted)

import java.util.Scanner
import scala.util.control.TailCalls._

object Main {
  def main(args: Array[String]): Unit = {
    val sc = new Scanner(System.in)
    val n = sc.nextInt()
    val h = Vector.fill(n)(sc.nextInt)

    def count(vector: Vector[Int], acc: Int): TailRec[Int] = {
      if (vector.isEmpty) done(acc)
      else if (vector.forall(_ > 0)) tailcall(count(vector.map(_ - 1), acc + 1))
      else if (vector.forall(_ == 0)) done(acc)
      else {
        val idx = vector.indexOf(0)
        if (idx == 0) tailcall(count(vector.tail, acc))
        else for {
          x <- tailcall(count(vector.take(idx), 0))
          y <- tailcall(count(vector.drop(idx + 1), acc))
        } yield x + y
      }
    }

    println(count(h, 0).result)
  }
}