LeetCode-in-All

84. Largest Rectangle in Histogram

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

Solution

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

int maxOfThreeNums(int a, int b, int c) {
    return (a > b ? (a > c ? a : c) : (b > c ? b : c));
}

int findMinInArray(int* a, int start, int limit) {
    int min = __INT_MAX__;
    int minIndex = -1;
    for (int index = start; index < limit; index++) {
        if (a[index] < min) {
            min = a[index];
            minIndex = index;
        }
    }
    return minIndex;
}

bool checkIfSorted(int* a, int start, int limit) {
    for (int i = start + 1; i < limit; i++) {
        if (a[i] < a[i - 1]) {
            return false;
        }
    }
    return true;
}

int largestArea(int* a, int start, int limit) {
    if (a == NULL || limit == 0) {
        return 0;
    }
    if (start == limit) {
        return 0;
    }
    if (limit - start == 1) {
        return a[start];
    }
    if (limit - start == 2) {
        int maxOfTwoBars = (a[start] > a[start + 1]) ? a[start] : a[start + 1];
        int areaFromTwo = (a[start] < a[start + 1] ? a[start] : a[start + 1]) * 2;
        return (maxOfTwoBars > areaFromTwo) ? maxOfTwoBars : areaFromTwo;
    }
    if (checkIfSorted(a, start, limit)) {
        int maxWhenSorted = 0;
        for (int i = start; i < limit; i++) {
            int area = a[i] * (limit - i);
            if (area > maxWhenSorted) {
                maxWhenSorted = area;
            }
        }
        return maxWhenSorted;
    } else {
        int minInd = findMinInArray(a, start, limit);
        return maxOfThreeNums(
            largestArea(a, start, minInd),
            a[minInd] * (limit - start),
            largestArea(a, minInd + 1, limit)
        );
    }
}

int largestRectangleArea(int* heights, int length) {
    return largestArea(heights, 0, length);
}

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