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
import 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)
}
}