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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 <= 1050 <= heights[i] <= 104
Solution
object Solution {
def largestRectangleArea(heights: Array[Int]): Int = {
largestArea(heights, 0, heights.length)
}
private def largestArea(a: Array[Int], start: Int, limit: Int): Int = {
if (a == null || 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)
}
if (checkIfSorted(a, start, limit)) {
var maxWhenSorted = 0
for (i <- start until limit) {
if (a(i) * (limit - i) > maxWhenSorted) {
maxWhenSorted = a(i) * (limit - i)
}
}
return maxWhenSorted
} else {
val minInd = findMinInArray(a, start, limit)
maxOfThreeNums(
largestArea(a, start, minInd),
a(minInd) * (limit - start),
largestArea(a, minInd + 1, limit)
)
}
}
private def findMinInArray(a: Array[Int], start: Int, limit: Int): Int = {
var min = Int.MaxValue
var minIndex = -1
for (index <- start until limit) {
if (a(index) < min) {
min = a(index)
minIndex = index
}
}
minIndex
}
private def checkIfSorted(a: Array[Int], start: Int, limit: Int): Boolean = {
for (i <- start + 1 until limit) {
if (a(i) < a(i - 1)) {
return false
}
}
true
}
private def maxOfThreeNums(a: Int, b: Int, c: Int): Int = {
math.max(math.max(a, b), c)
}
}