Stack - Question 2

Leetcode 42. Trapping Rain Water

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining.

Constraints:

n == height.length

1 <= n <= 2 * 10^4

0 <= height[i] <= 10^5

Analysis:

This a classic question for interview.

If the water can be trapped at one position, then it requires there are two higher integers on its both side. So the question becomes how to decide the two sides.

We can use two vectors to record the largest integer so far: one is vector<int> left and the other is vector<int> right,

1. left[i] represents the largest elements so far from the left side;

2. right[i] represents the largest elements so far from the right side;

(Note: this is a very common trick used in dynamic programming.)

After having this information, we can do a third scan: for each element, just need to pick up the difference between nums[i] and the min(left[i], right[i]).

The time complexity is O(3N), and the space complexity is O(2N).

See the code below:

class Solution {
public:
    int trap(vector<int>& height) {
        int res = 0, n = height.size();
        vector<int> left(n+1, 0), right(n+1, 0);
        for(int i=0; i<n; ++i) {
            left[i+1] = max(height[i], left[i]);
        }
        for(int i=n-1; i>=0; --i) {
            right[i] = max(height[i], right[i+1]);
        }
        for(int i=0; i<n; ++i) {
            res += max(min(left[i+1], right[i]) - height[i], 0);
        }
        return res;
    }
};

The space and number of scanning can be further optimized. For example, we can find the maximum number first, which can then divide the array into two parts: one is on the left, the other is on the right side. 

The largest integer is always the "right wall" for the numbers on its left side, and the "left wall" for the numbers on its right side.

So we just need to find the other wall for each numbers.

The time complexity is O(2N) (yes, there are 3 for loops, but only 2N), and the space complexity is O(1).

See the code below:

class Solution {
public:
    int trap(vector<int>& height) {
        int res = 0, n = height.size(), id = 0, left = 0, right = n-1;
        for(int i=1; i<n; ++i) {
            if(height[i] > height[id]) id = i;
        }
        for(int i=1; i<id; ++i) {
            if(height[i] > height[left]) left = i;
            res += height[left] - height[i];
        }
        for(int i=n-1; i>id; --i) {
            if(height[i] > height[right]) right = i;
            res += height[right] - height[i];
        }
        return res;
    }
};

If use a stack, we just need to scan the array once. 

There are some key observations:

1. If all the numbers on its left side are smaller, then there is no way for one number to contribute rain water;

2. If we keep a decreasing order for the number in a stack, once we meet with a larger element, then we can determine the trapped water just between its closest left and right walls easily;

3. The rain water trapped due to other possible left and right walls can still be calculated, since we keep the index in the stack. When calculate the contribution, we use the formula: height diff * width.

In the above two method, we calculate the contribution from each number "vertically", which means all the rain water on top of one number; while using a stack, we calculate the contribution from each number "horizontally", which means we slice the water trapped into different rows, and add them together when popping out the corresponding walls from the stack.

The time complexity is O(N) (yes, only one scan), and the space complexity is O(N) as well.

See the code below:

class Solution {
public:
    int trap(vector<int>& height) {
        int res = 0, n = height.size();
        stack<int> sk;
        for(int i=0; i<n; ++i) {
            while(sk.size() && height[sk.top()] < height[i]) {
                auto t = sk.top();
                sk.pop();
                if(sk.size()) {
                    // height * width
                    res += (min(height[i], height[sk.top()]) - height[t]) * (i - sk.top() - 1);
                }
            }
            sk.push(i);
        }
        return res;
    }
};

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