Hard
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
To solve the “Largest Rectangle in Histogram” problem in Java with the Solution class, follow these steps:
largestRectangleArea
in the Solution
class that takes an array of integers heights
as input and returns the area of the largest rectangle in the histogram.Here’s the implementation of the largestRectangleArea
method in Java:
import java.util.Stack;
class Solution {
public int largestRectangleArea(int[] heights) {
Stack<Integer> stack = new Stack<>();
int maxArea = 0;
int i = 0;
while (i < heights.length) {
if (stack.isEmpty() || heights[i] >= heights[stack.peek()]) {
stack.push(i++);
} else {
int top = stack.pop();
int width = stack.isEmpty() ? i : i - stack.peek() - 1;
maxArea = Math.max(maxArea, heights[top] * width);
}
}
while (!stack.isEmpty()) {
int top = stack.pop();
int width = stack.isEmpty() ? i : i - stack.peek() - 1;
maxArea = Math.max(maxArea, heights[top] * width);
}
return maxArea;
}
}
This implementation uses a stack-based approach to find the largest rectangle in the histogram, with a time complexity of O(N), where N is the number of bars in the histogram.