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In this post, you will find the solution for the **Largest Rectangle in Histogram** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Largest Rectangle in Histogram– LeetCode Problem

Largest Rectangle in Histogram– LeetCode Problem

**Problem:**

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:10Explanation: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 <= 10`

^{5}`0 <= heights[i] <= 10`

^{4}

Largest Rectangle in Histogram– LeetCode Solutions

Largest Rectangle in Histogram in C++:

class Solution { public: int largestRectangleArea(vector& heights) { int ans = 0; stack 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; } };

Largest Rectangle in Histogram in Java:

class Solution { public int largestRectangleArea(int[] heights) { int ans = 0; Stackstack = new Stack<>(); 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; } }

Largest Rectangle in Histogram in Python:

class 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