Largest Rectangle in Histogram

Largest Rectangle in Histogram - 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.

A rectangle in a histogram is formed by selecting a contiguous range of bars. The height of the rectangle is determined by the shortest bar in that range, and the width is the number of bars in the range.

Your goal is to find the maximum possible area among all such rectangles.

Input & Output

Example 1 — Standard Histogram

$ Input: heights = [2,1,5,6,2,3]

› Output: 10

💡 Note: The largest rectangle has height 5 and width 2 (indices 2-3), giving area = 5 × 2 = 10

Example 2 — Single Bar

$ Input: heights = [2,4]

› Output: 4

💡 Note: The largest rectangle is the single bar with height 4 and width 1, area = 4 × 1 = 4

Example 3 — Decreasing Heights

$ Input: heights = [6,5,4,3,2,1]

› Output: 12

💡 Note: Best rectangle uses height 3 spanning 4 bars (indices 0-3), area = 3 × 4 = 12

Constraints

1 ≤ heights.length ≤ 10⁵
0 ≤ heights[i] ≤ 10⁴

Visualization

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Asked in

G Google 45 a Amazon 38 M Microsoft 32 f Facebook 28

The key insight is that each bar can serve as the minimum height of some rectangle. Use a monotonic stack to efficiently find how far each bar can extend left and right. Best approach is the stack solution with Time: O(n), Space: O(n).

Common Approaches

✓ Brute Force - Check All Subarrays

⏱️ Time: O(n³) Space: O(1)

For each starting position, extend to all possible ending positions. For each subarray, find the minimum height and calculate the rectangle area.

Monotonic Stack

⏱️ Time: O(n) Space: O(n)

Maintain a stack of indices with increasing heights. When we find a smaller height, pop from stack and calculate areas using popped heights as the minimum height of rectangles.

Brute Force - Check All Subarrays — Algorithm Steps

For each starting index i
For each ending index j >= i
Find minimum height in range [i,j]
Calculate area = min_height × (j-i+1)
Track maximum area

Visualization

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Step-by-Step Walkthrough

Input Array

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

Check Subarrays

Try all i,j pairs and find min height

Maximum Area

Track the largest area found: 10

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* heights, int n) {
    int maxArea = 0;
    
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            int minHeight = heights[i];
            for (int k = i; k <= j; k++) {
                if (heights[k] < minHeight) {
                    minHeight = heights[k];
                }
            }
            int area = minHeight * (j - i + 1);
            if (area > maxArea) {
                maxArea = area;
            }
        }
    }
    
    return maxArea;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int heights[1000];
    int n = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        heights[n++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(heights, n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Triple nested loops: n positions × n extensions × n to find minimum

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables to track current and maximum area

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check All Subarrays — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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