LeetCode

Largest Rectangle in Histogram

2014.2.26 05:06

Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.

Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].

The largest rectangle is shown in the shaded area, which has area = 10 unit.

For example,
Given height = [2,1,5,6,2,3],
return 10.

Solution1:

　　In this problem, you'll find the largest rectangle that is completely contained in the histogram.

　　The brute-force solution is O(n^2). For each grid, you scan them all through to determine the left and right borders, calculate the area and update the result. Surely it won't get you an acceptable outcome.

　　In the brute-force solution, we start from the current grid and scan in both directions one by one. If not one by one, it would be more efficient. The solution below will use an extra array to do dymamic programming. The very few differences in the while loop is the key to this efficient DP solution.

　　Total time and space complexities are both O(n).

Accepted code:

// 3TLE, 1AC, DP solution
class Solution {
public:
    int largestRectangleArea(vector<int> &height) {
        int i;
        int n;
        
        n = (int)height.size();
        if (n == 0) {
            return 0;
        }
        
        vector<int> ll, rr;
        int max_area = 0;
        int area = 0;
        
        ll.resize(n);
        rr.resize(n);
        for (i = 0; i < n; ++i) {
            ll[i] = i;
            rr[i] = i;
        }
        for (i = 0; i <= n - 1; ++i) {
            while (ll[i] - 1 >= 0 && height[ll[i] - 1] >= height[i]) {
                ll[i] = ll[ll[i] - 1];
            }
        }
        for (i = n - 1; i >= 0; --i) {
            while (rr[i] + 1 <= n - 1 && height[rr[i] + 1] >= height[i]) {
                rr[i] = rr[rr[i] + 1];
            }
        }
        
        for (i = 0; i < n; ++i) {
            area = (rr[i] - ll[i] + 1) * height[i];
            if (area > max_area) {
                max_area = area;
            }
        }
        
        ll.clear();
        rr.clear();
        return max_area;
    }
};

Solution2:

　　This solution is quite difficult to figure out. Though I would say I can understand how it works, I'd say the original author of this algorithm is just magician to be so creative.

　　Here is where my reference came from, for your information: Largest Rectangular Area in Histogram.

　　Although the time and space are on the same level, but the constant coefficients are better indeed.

　　This algorithm is better than the last one, but I'd still prefer the last one because it's easier to write and understand, as well as efficient enough to solve the problem at hand. Tricky code is a double-edged sword, use it wisely.

Accepted code:

// 1WA, 1AC, the more optimized O(n) solution with stack.
#include <stack>
#include <vector>
using namespace std;

class Solution {
public:
    int largestRectangleArea(vector<int> &height) {
        int i;
        int n;
        
        n = (int)height.size();
        if (n == 0) {
            return 0;
        }
        
        int max_area = 0;
        int area;
        stack<int> st;
        int top;
        
        for (i = 0; i < n; ++i) {
            if (st.empty() || height[st.top()] <= height[i]) {
                st.push(i);
            } else {
                while (!st.empty() && height[st.top()] > height[i]) {
                    top = st.top();
                    st.pop();
                    if (st.empty()) {
                        area = i * height[top];
                    } else {
                        area = (i - st.top() - 1) * height[top];
                    }
                    if (area > max_area) {
                        max_area = area;
                    }
                }
                st.push(i);
            }
        }
        while (!st.empty()) {
            top = st.top();
            st.pop();
            if (st.empty()) {
                area = i * height[top];
            } else {
                area = (i - st.top() - 1) * height[top];
            }
            if (area > max_area) {
                max_area = area;
            }
        }
        
        return max_area;
    }
};