Largest Rectangle in Histogram & Stack using Queues

Histogram Calculation

Finds left and right smaller boundaries for each bar using monotonic stack in $O(N)$ time.

Production Code Example:

import java.util.Stack;
public class HistogramDemo {
    public int largestRectangleArea(int[] heights) {
        Stack<Integer> st = new Stack<>();
        int maxArea = 0, n = heights.length;
        for (int i = 0; i <= n; i++) {
            int h = (i == n) ? 0 : heights[i];
            while (!st.isEmpty() && heights[st.peek()] >= h) {
                int height = heights[st.pop()];
                int width = st.isEmpty() ? i : i - st.peek() - 1;
                maxArea = Math.max(maxArea, height * width);
            }
            st.push(i);
        }
        return maxArea;
    }
}

Key Complexity & Algorithmic Takeaways:

When implementing Largest Rectangle in Histogram & Stack using Queues in coding interviews and production applications, keep these core guidelines in mind:

• Time Complexity Analysis: Always evaluate best-case, average-case, and worst-case time complexities ($O(1)$, $O(\log n)$, $O(n)$, $O(n \log n)$, $O(n^2)$).
• Space Complexity & Memory Bounds: Account for auxiliary memory usage, call stack frame recursion overhead, and heap allocations.
• Edge Cases & Validation: Test empty inputs, null pointers, single-element collections, duplicate values, and integer overflow bounds.
• Optimal vs Naive Solutions: Start with a clear brute-force solution, then optimize using techniques like Hashing, Two Pointers, Windowing, or Dynamic Programming.

Summary Takeaway:

Mastering Largest Rectangle in Histogram & Stack using Queues provides the foundational problem-solving skills needed to pass technical coding interviews at top tech companies and write ultra-performant software systems.