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 <= 1050 <= heights[i] <= 104import kotlin.math.max
class Solution {
fun largestRectangleArea(heights: IntArray): Int {
return largestArea(heights, 0, heights.size)
}
private fun largestArea(a: IntArray, start: Int, limit: Int): Int {
if (a.isEmpty()) {
return 0
}
if (start == limit) {
return 0
}
if (limit - start == 1) {
return a[start]
}
if (limit - start == 2) {
val maxOfTwoBars = Math.max(a[start], a[start + 1])
val areaFromTwo = Math.min(a[start], a[start + 1]) * 2
return Math.max(maxOfTwoBars, areaFromTwo)
}
return if (checkIfSorted(a, start, limit)) {
var maxWhenSorted = 0
for (i in start until limit) {
if (a[i] * (limit - i) > maxWhenSorted) {
maxWhenSorted = a[i] * (limit - i)
}
}
maxWhenSorted
} else {
val minInd = findMinInArray(a, start, limit)
maxOfThreeNums(
largestArea(a, start, minInd),
a[minInd] * (limit - start),
largestArea(a, minInd + 1, limit),
)
}
}
private fun findMinInArray(a: IntArray, start: Int, limit: Int): Int {
var min = Int.MAX_VALUE
var minIndex = -1
for (index in start until limit) {
if (a[index] < min) {
min = a[index]
minIndex = index
}
}
return minIndex
}
private fun checkIfSorted(a: IntArray, start: Int, limit: Int): Boolean {
for (i in start + 1 until limit) {
if (a[i] < a[i - 1]) {
return false
}
}
return true
}
private fun maxOfThreeNums(a: Int, b: Int, c: Int): Int {
return max(max(a, b), c)
}
}