// Efficient Method : Time Complexity : O(n), Auxilliary Space : O(n)

import java.util.*;
import java.io.*;

class GFG 
{ 
    static int getWater(int arr[], int n)
    {
    	int res = 0;

    	int lMax[] = new int[n];
    	int rMax[] = new int[n];

    	lMax[0] = arr[0];
    	for(int i = 1; i < n; i++)
    		lMax[i] = Math.max(arr[i], lMax[i - 1]);


    	rMax[n - 1] = arr[n - 1];
    	for(int i = n - 2; i >= 0; i--)
    		rMax[i] = Math.max(arr[i], rMax[i + 1]);

    	for(int i = 1; i < n - 1; i++)
    		res = res + (Math.min(lMax[i], rMax[i]) - arr[i]);
    	
    	return res;
    }


    public static void main(String args[]) 
    { 
       int arr[] = {5, 0, 6, 2, 3}, n = 5;

      System.out.println( getWater(arr, n)); // Output : 6
    } 
}


// Naive Method : Time Complexity : O(n^2)

    static int getWater(int arr[], int n)
    {
    	int res = 0;

    	for(int i = 1; i < n - 1; i++)
    	{
    		int lMax = arr[i];

    		for(int j = 0; j < i; j++)
    			lMax = Math.max(lMax, arr[j]);

    		int rMax = arr[i];

    		for(int j = i + 1; j < n; j++)
    			rMax = Math.max(rMax, arr[j]);

    		res = res + (Math.min(lMax, rMax) - arr[i]);
    	}
    
    	return res; // Output : 6
    }