LeetCode #84 — HARD

Largest Rectangle in Histogram

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Stack

Example 1

[2,1,5,6,2,3]

Example 2

[2,4]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

import java.util.*;

class Solution {
    public int largestRectangleArea(int[] heights) {
        Deque<Integer> stack = new ArrayDeque<>();
        int best = 0;

        for (int i = 0; i <= heights.length; i++) {
            int h = (i == heights.length) ? 0 : heights[i];
            while (!stack.isEmpty() && heights[stack.peek()] > h) {
                int height = heights[stack.pop()];
                int left = stack.isEmpty() ? -1 : stack.peek();
                int width = i - left - 1;
                best = Math.max(best, height * width);
            }
            stack.push(i);
        }

        return best;
    }
}

func largestRectangleArea(heights []int) int {
    stack := []int{}
    best := 0

    for i := 0; i <= len(heights); i++ {
        h := 0
        if i < len(heights) {
            h = heights[i]
        }
        for len(stack) > 0 && heights[stack[len(stack)-1]] > h {
            idx := stack[len(stack)-1]
            stack = stack[:len(stack)-1]
            left := -1
            if len(stack) > 0 {
                left = stack[len(stack)-1]
            }
            width := i - left - 1
            area := heights[idx] * width
            if area > best {
                best = area
            }
        }
        stack = append(stack, i)
    }
    return best
}

class Solution:
    def largestRectangleArea(self, heights: List[int]) -> int:
        stack = []
        best = 0

        for i in range(len(heights) + 1):
            h = 0 if i == len(heights) else heights[i]
            while stack and heights[stack[-1]] > h:
                idx = stack.pop()
                left = stack[-1] if stack else -1
                width = i - left - 1
                best = max(best, heights[idx] * width)
            stack.append(i)

        return best

impl Solution {
    pub fn largest_rectangle_area(heights: Vec<i32>) -> i32 {
        let mut stack: Vec<usize> = Vec::new();
        let mut best = 0;

        for i in 0..=heights.len() {
            let h = if i == heights.len() { 0 } else { heights[i] };
            while let Some(&top) = stack.last() {
                if heights[top] <= h {
                    break;
                }
                let idx = stack.pop().unwrap();
                let left = stack.last().copied().map(|x| x as i32).unwrap_or(-1);
                let width = i as i32 - left - 1;
                best = best.max(heights[idx] * width);
            }
            stack.push(i);
        }

        best
    }
}

function largestRectangleArea(heights: number[]): number {
  const stack: number[] = [];
  let best = 0;

  for (let i = 0; i <= heights.length; i++) {
    const h = i === heights.length ? 0 : heights[i];
    while (stack.length && heights[stack[stack.length - 1]] > h) {
      const idx = stack.pop()!;
      const left = stack.length ? stack[stack.length - 1] : -1;
      const width = i - left - 1;
      best = Math.max(best, heights[idx] * width);
    }
    stack.push(i);
  }
  return best;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.