Problem:
Score : 300 points

Problem Statement

N persons are standing in a row. The height of the i-th person from the front is A_i.

We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person:

Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool.

Find the minimum total height of the stools needed to meet this goal.

Constraints

1 \leq N \leq 2\times 10^5

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 \ldots A_N

Output

Print the minimum total height of the stools needed to meet the goal.

Sample Input 1

5
2 1 5 4 3

Sample Output 1

4

If the persons stand on stools of heights 0, 1, 0, 1, and 2, respectively, their heights will be 2, 2, 5, 5, and 5, satisfying the condition.

We cannot meet the goal with a smaller total height of the stools.

Sample Input 2

5
3 3 3 3 3

Sample Output 2

0

Giving a stool of height 0 to everyone will work.

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

import java.io.*;
import java.util.Arrays;
import java.util.StringTokenizer;

import static java.lang.Long.max;

public class Main {
  static class Pair {
    int num;
    int cut;

    Pair(int num, int cut) {
      this.num = num;
      this.cut = cut;
    }

    double getVal() {
      return -num / (double) cut;
    }
  }

  public static void main(String[] args) throws IOException {
    sc = new MScanner(System.in);
    pw = new PrintWriter(System.out);
    int N = sc.nextInt();
    int[] as = sc.intArr(N);
    long m = 0;
    long total = 0;
    for(int h: as) {
        total += max(0, m - h);
        m = max(h, m);
    }


    pw.println(total);

    // pw.println("end");
    // pw.println();

    pw.flush();
    // System.out.println(cnt);
  }

  /**
   * https://www.slideshare.net/hcpc_hokudai/binary-indexed-tree A space efficient version segment
   * tree. Usually for range sum. Or the required metric can be calulated in that:
   *
   * <p>right_metric = parant_metric - left_metric Only then, we can stop memoizing right_metric
   *
   * <p>The internal structure is a 1-base array, to simplify calculation of parent node's idx. lsb
   * = 3 & -3 = 2 (0x0011 & 0x1101 = 0x0001) lastParentOf3 = 3 - lsb = 2 curParentOf3 = 3 + lsb = 4
   */
  static class FenwickTree {
    private final long[] state;
    private final int size;

    FenwickTree(int size) {
      this.size = size;
      state = new long[size + 1];
    }

    @Override
    public String toString() {
      return Arrays.toString(state);
    }

    long queryRangeSumFromZeroTo(int idx) {
      long sum = 0;
      while (idx != 0) {
        sum += state[idx];
        idx -= (idx & -idx);
      }
      return sum;
    }

    /** @param val can be negative */
    void updateElement(int idx, long val) {
      while (idx <= size) {
        state[idx] += val;
        idx += (idx & -idx);
      }
    }
  }

  // sc.close();
  // pw.println(cnt);
  // pw.flush();

  static PrintWriter pw;
  static MScanner sc;

  static class MScanner {
    StringTokenizer st;
    BufferedReader br;

    public MScanner(InputStream system) {
      br = new BufferedReader(new InputStreamReader(system));
    }

    public MScanner(String file) throws Exception {
      br = new BufferedReader(new FileReader(file));
    }

    public String next() throws IOException {
      while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
      return st.nextToken();
    }

    public int[] intArr(int n) throws IOException {
      int[] in = new int[n];
      for (int i = 0; i < n; i++) in[i] = nextInt();
      return in;
    }

    public long[] longArr(int n) throws IOException {
      long[] in = new long[n];
      for (int i = 0; i < n; i++) in[i] = nextLong();
      return in;
    }

    public int[] intSortedArr(int n) throws IOException {
      int[] in = new int[n];
      for (int i = 0; i < n; i++) in[i] = nextInt();
      shuffle(in);
      Arrays.sort(in);
      return in;
    }

    public long[] longSortedArr(int n) throws IOException {
      long[] in = new long[n];
      for (int i = 0; i < n; i++) in[i] = nextLong();
      shuffle(in);
      Arrays.sort(in);
      return in;
    }

    public Integer[] IntegerArr(int n) throws IOException {
      Integer[] in = new Integer[n];
      for (int i = 0; i < n; i++) in[i] = nextInt();
      return in;
    }

    public Long[] LongArr(int n) throws IOException {
      Long[] in = new Long[n];
      for (int i = 0; i < n; i++) in[i] = nextLong();
      return in;
    }

    public String nextLine() throws IOException {
      return br.readLine();
    }

    public int nextInt() throws IOException {
      return Integer.parseInt(next());
    }

    public double nextDouble() throws IOException {
      return Double.parseDouble(next());
    }

    public char nextChar() throws IOException {
      return next().charAt(0);
    }

    public long nextLong() throws IOException {
      return Long.parseLong(next());
    }

    public boolean ready() throws IOException {
      return br.ready();
    }

    public void waitForInput() throws InterruptedException {
      Thread.sleep(3000);
    }
  }

  static void shuffle(int[] in) {
    for (int i = 0; i < in.length; i++) {
      int idx = (int) (Math.random() * in.length);
      int tmp = in[i];
      in[i] = in[idx];
      in[idx] = tmp;
    }
  }

  static void shuffle(long[] in) {
    for (int i = 0; i < in.length; i++) {
      int idx = (int) (Math.random() * in.length);
      long tmp = in[i];
      in[i] = in[idx];
      in[idx] = tmp;
    }
  }
}