ajinkyajawale14499

public static int largestRectangleArea(int[] height) {
    if (height == null || height.length == 0) {
        return 0;
    }
    int[] lessFromLeft = new int[height.length]; // idx of the first bar the left that is lower than current
    int[] lessFromRight = new int[height.length]; // idx of the first bar the right that is lower than current
    lessFromRight[height.length - 1] = height.length;
    lessFromLeft[0] = -1;

    for (int i = 1; i < height.length; i++) {

ajinkyajawale14499

public class Solution {
    public int largestRectangleArea(int[] height) {
        int len = height.length;
        Stack<Integer> s = new Stack<Integer>();
        int maxArea = 0;
        for(int i = 0; i <= len; i++){
            int h = (i == len ? 0 : height[i]);
            if(s.isEmpty() || h >= height[s.peek()]){
                s.push(i);
            }else{

ajinkyajawale14499

def largestRectangleArea(self, height):
    height.append(0)
    stack = [-1]
    ans = 0
    for i in xrange(len(height)):
        while height[i] < height[stack[-1]]:
            h = height[stack.pop()]
            w = i - stack[-1] - 1
            ans = max(ans, h * w)
        stack.append(i)

ajinkyajawale14499

class Solution {
public:
    int largestRectangleArea(vector<int>& height) {
        height.insert(height.begin(),0);                        // dummy "0" added to make sure stack s will never be empty
        height.push_back(0);                                    // dummy "0" added to clear the stack at the end
        int len = height.size();
        int i, res = 0, idx;
        stack<int> s;                                          // stack to save "height" index
        s.push(0);                                             // index to the first dummy "0"
        for(i=1;i<len;i++)