LeetCode "Largest Rectangle in Histogram"

I got a DP solution first which is O(n^2). Apparently it is not a optimized one - think about: it is linear value space. There must be O(n) solution.

And yes there is: http://fisherlei.blogspot.com/2012/12/leetcode-largest-rectangle-in-histogram.html

It reminds me the mono-queue optimization upon DP - it is said to be even out-of-scope of IOI.. Precisely that is not a typical mono-queue style used. The mono-queue maintanence is tricky: if new height is less than top, we pop out all anti-mono elements and calculate corresponding areas - to make sure the queue is monotonous.

But how to conduct yourself to the wonderland you are longing for? Like this:

O(n^3) solution is direct -> Has duplicated calculation? Use DP -> O(n^2) -> is it linear space? think about MonoQ -> figure out how to maintanent MonoQ -> O(n)

Please keep the animated pictures of the whole procedure in mind. That helps.

I put comments:

class Solution {
public:
    int largestRectangleArea(vector<int> &height) {
        int cnt = height.size();
        if(cnt == 0) return 0;
        if(cnt == 1) return height[0];

        int max = std::numeric_limits<int>::min();

        stack<int> stk; // store indices
        height.push_back(0); // as sentinel for wrapping up everything
        for(int i = 0; i < height.size(); i ++)
        {
            if (stk.empty() || height[i] > height[stk.top()])
            {
                stk.push(i); // keep growing the non-descending mono-queue
            }
            else
            {
                int lastInx = stk.top();
                stk.pop();
                int lastH = height[lastInx];
                max = std::max(max, lastH * 
                            (stk.empty() ? i - (-1) - 1 : i - stk.top() - 1)); // expands to the left as much as possible
                i --; // going back for all heights that are higher than height[i]
            }
        }

        return max;
    }
};

TO BE REVISED LATER