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 <= 105
0 <= 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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