42. Trapping Rain Water *HARD*

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.

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;
}

* The idea is:
* 1) find the highest bar.
* 2) traverse the bar from left the highest bar.
* becasue we have the highest bar in right, so, any bar higher than its right bar(s) can contain the water.
* 3) traverse the bar from right the highest bar.
* becasue we have the highest bar in left, so, any bar higher than its left bar(s) can contain the water.