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
Largest Rectangle in Histogram Solutions
✅Time: O(n)
✅Space: O(n)
C++
Will be updated Soonclass Solution {
public:
int largestRectangleArea(vector<int>& heights) {
int ans = 0;
stack<int> stack;
for (int i = 0; i <= heights.size(); ++i) {
while (!stack.empty() &&
(i == heights.size() || heights[stack.top()] > heights[i])) {
const int h = heights[stack.top()];
stack.pop();
const int w = stack.empty() ? i : i - stack.top() - 1;
ans = max(ans, h * w);
}
stack.push(i);
}
return ans;
}
};
Will be updated Soon
Java
class Solution {
public int largestRectangleArea(int[] heights) {
int ans = 0;
Deque<Integer> stack = new ArrayDeque<>();
for (int i = 0; i <= heights.length; ++i) {
while (!stack.isEmpty() && (i == heights.length || heights[stack.peek()] > heights[i])) {
final int h = heights[stack.pop()];
final int w = stack.isEmpty() ? i : i - stack.peek() - 1;
ans = Math.max(ans, h * w);
}
stack.push(i);
}
return ans;
}
}
Will be updated Soon
Python
Will be updated Soonclass Solution:
def largestRectangleArea(self, heights: List[int]) -> int:
ans = 0
stack = []
for i in range(len(heights) + 1):
while stack and (i == len(heights) or heights[stack[-1]] > heights[i]):
h = heights[stack.pop()]
w = i - stack[-1] - 1 if stack else i
ans = max(ans, h * w)
stack.append(i)
return ans
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