Largest Rectangle in Histogram – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
Given an array of integers heights representing the histogram’s bar height where the width of each bar is 1, return the area of the largest rectangle in the histogram.
Example 1:
Input: heights = [2,1,5,6,2,3] Output: 10 Explanation: The above is a histogram where width of each bar is 1. The largest rectangle is shown in the red area, which has an area = 10 units.
Example 2:
Input: heights = [2,4] Output: 4
Constraints:
1 <= heights.length <= 1050 <= heights[i] <= 104
C++ Largest Rectangle in Histogram LeetCode Solution
class Solution {
public:
int largestRectangleArea(vector<int> &height) {
int ret = 0;
height.push_back(0);
vector<int> index;
for(int i = 0; i < height.size(); i++)
{
while(index.size() > 0 && height[index.back()] >= height[i])
{
int h = height[index.back()];
index.pop_back();
int sidx = index.size() > 0 ? index.back() : -1;
if(h * (i-sidx-1) > ret)
ret = h * (i-sidx-1);
}
index.push_back(i);
}
return ret;
}
};
Java Largest Rectangle in Histogram LeetCode Solution
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++) {
int p = i - 1;
while (p >= 0 && height[p] >= height[i]) {
p = lessFromLeft[p];
}
lessFromLeft[i] = p;
}
for (int i = height.length - 2; i >= 0; i--) {
int p = i + 1;
while (p < height.length && height[p] >= height[i]) {
p = lessFromRight[p];
}
lessFromRight[i] = p;
}
int maxArea = 0;
for (int i = 0; i < height.length; i++) {
maxArea = Math.max(maxArea, height[i] * (lessFromRight[i] - lessFromLeft[i] - 1));
}
return maxArea;
}
Python 3 Largest Rectangle in Histogram LeetCode Solution
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)
height.pop()
return ans
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