// Source : https://oj.leetcode.com/problems/trapping-rain-water/
// Author : Hao Chen
// Date   : 2014-07-02

/********************************************************************************** 
* 
* Given n non-negative integers representing an elevation map where the width of each bar is 1, 
* compute how much water it is able to trap after raining. 
* 
* For example, 
* Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.
* 
* The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 
* 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!
*               
**********************************************************************************/

#include <stdio.h>

int trap(int a[], int n) {
    int result = 0;

    //find the highest value/position
    int maxHigh = 0;
    int maxIdx = 0;
    for(int i=0; i<n; i++){
        if (a[i]>maxHigh){
            maxHigh = a[i];
            maxIdx = i;
        }
    }

    //from the left to the highest postion
    int prevHigh = 0;
    for(int i=0; i<maxIdx; i++){
        if(a[i]>prevHigh){
            prevHigh = a[i];
        }
        result += (prevHigh - a[i]);
    }

    //from the right to the highest postion
    prevHigh=0;
    for(int i=n-1; i>maxIdx; i--){
        if(a[i]>prevHigh){
            prevHigh = a[i];
        }
        result += (prevHigh - a[i]);
    }

    return result;
}

#define TEST(a) printf("%d\n", trap(a, sizeof(a)/sizeof(int)))
int main()
{
    int a[]={2,0,2};
    TEST(a);
    int b[]={0,1,0,2,1,0,1,3,2,1,2,1};
    TEST(b);
    return 0;
}