Question

84. Largest Rectangle in Histogram

中文文档

Description

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

Solutions

Solution 1: Monotonic Stack

We can enumerate the height $h$ of each bar as the height of the rectangle. Using a monotonic stack, we find the index $left_i$, $right_i$ of the first bar with a height less than $h$ to the left and right. The area of the rectangle at this time is $h \times (right_i-left_i-1)$. We can find the maximum value.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ represents the length of $heights$.

Common model of monotonic stack: Find the nearest number to the left/right of each number that is larger/smaller than it. Template:

stk = []
for i in range(n):
  while stk and check(stk[-1], i):
    stk.pop()
  stk.append(i)

class Solution:
  def largestRectangleArea(self, heights: List[int]) -> int:
    n = len(heights)
    stk = []
    left = [-1] * n
    right = [n] * n
    for i, h in enumerate(heights):
      while stk and heights[stk[-1]] >= h:
        right[stk[-1]] = i
        stk.pop()
      if stk:
        left[i] = stk[-1]
      stk.append(i)
    return max(h * (right[i] - left[i] - 1) for i, h in enumerate(heights))

class Solution {
  public int largestRectangleArea(int[] heights) {
    int res = 0, n = heights.length;
    Deque<Integer> stk = new ArrayDeque<>();
    int[] left = new int[n];
    int[] right = new int[n];
    Arrays.fill(right, n);
    for (int i = 0; i < n; ++i) {
      while (!stk.isEmpty() && heights[stk.peek()] >= heights[i]) {
        right[stk.pop()] = i;
      }
      left[i] = stk.isEmpty() ? -1 : stk.peek();
      stk.push(i);
    }
    for (int i = 0; i < n; ++i) {
      res = Math.max(res, heights[i] * (right[i] - left[i] - 1));
    }
    return res;
  }
}

class Solution {
public:
  int largestRectangleArea(vector<int>& heights) {
    int res = 0, n = heights.size();
    stack<int> stk;
    vector<int> left(n, -1);
    vector<int> right(n, n);
    for (int i = 0; i < n; ++i) {
      while (!stk.empty() && heights[stk.top()] >= heights[i]) {
        right[stk.top()] = i;
        stk.pop();
      }
      if (!stk.empty()) left[i] = stk.top();
      stk.push(i);
    }
    for (int i = 0; i < n; ++i)
      res = max(res, heights[i] * (right[i] - left[i] - 1));
    return res;
  }
};

func largestRectangleArea(heights []int) int {
  res, n := 0, len(heights)
  var stk []int
  left, right := make([]int, n), make([]int, n)
  for i := range right {
    right[i] = n
  }
  for i, h := range heights {
    for len(stk) > 0 && heights[stk[len(stk)-1]] >= h {
      right[stk[len(stk)-1]] = i
      stk = stk[:len(stk)-1]
    }
    if len(stk) > 0 {
      left[i] = stk[len(stk)-1]
    } else {
      left[i] = -1
    }
    stk = append(stk, i)
  }
  for i, h := range heights {
    res = max(res, h*(right[i]-left[i]-1))
  }
  return res
}

impl Solution {
  #[allow(dead_code)]
  pub fn largest_rectangle_area(heights: Vec<i32>) -> i32 {
    let n = heights.len();
    let mut left = vec![-1; n];
    let mut right = vec![-1; n];
    let mut stack: Vec<(usize, i32)> = Vec::new();
    let mut ret = -1;

    // Build left vector
    for (i, h) in heights.iter().enumerate() {
      while !stack.is_empty() && stack.last().unwrap().1 >= *h {
        stack.pop();
      }
      if stack.is_empty() {
        left[i] = -1;
      } else {
        left[i] = stack.last().unwrap().0 as i32;
      }
      stack.push((i, *h));
    }

    stack.clear();

    // Build right vector
    for (i, h) in heights.iter().enumerate().rev() {
      while !stack.is_empty() && stack.last().unwrap().1 >= *h {
        stack.pop();
      }
      if stack.is_empty() {
        right[i] = n as i32;
      } else {
        right[i] = stack.last().unwrap().0 as i32;
      }
      stack.push((i, *h));
    }

    // Calculate the max area
    for (i, h) in heights.iter().enumerate() {
      ret = std::cmp::max(ret, (right[i] - left[i] - 1) * *h);
    }

    ret
  }
}

using System;
using System.Collections.Generic;
using System.Linq;

public class Solution {
  public int LargestRectangleArea(int[] height) {
    var stack = new Stack<int>();
    var result = 0;
    var i = 0;
    while (i < height.Length || stack.Any())
    {
      if (!stack.Any() || (i < height.Length && height[stack.Peek()] < height[i]))
      {
        stack.Push(i);
        ++i;
      }
      else
      {
        var previousIndex = stack.Pop();
        var area = height[previousIndex] * (stack.Any() ? (i - stack.Peek() - 1) : i);
        result = Math.Max(result, area);
      }
    }

    return result;
  }
}

Solution 2

class Solution:
  def largestRectangleArea(self, heights: List[int]) -> int:
    n = len(heights)
    stk = []
    left = [-1] * n
    right = [n] * n
    for i, h in enumerate(heights):
      while stk and heights[stk[-1]] >= h:
        stk.pop()
      if stk:
        left[i] = stk[-1]
      stk.append(i)
    stk = []
    for i in range(n - 1, -1, -1):
      h = heights[i]
      while stk and heights[stk[-1]] >= h:
        stk.pop()
      if stk:
        right[i] = stk[-1]
      stk.append(i)
    return max(h * (right[i] - left[i] - 1) for i, h in enumerate(heights))