LeetCode in Kotlin
2444. Count Subarrays With Fixed Bounds
Hard
You are given an integer array nums and two integers minK and maxK.
A fixed-bound subarray of nums is a subarray that satisfies the following conditions:
- The minimum value in the subarray is equal to
minK. - The maximum value in the subarray is equal to
maxK.
Return the number of fixed-bound subarrays.
A subarray is a contiguous part of an array.
Example 1:
Input: nums = [1,3,5,2,7,5], minK = 1, maxK = 5
Output: 2
Explanation: The fixed-bound subarrays are [1,3,5] and [1,3,5,2].
Example 2:
Input: nums = [1,1,1,1], minK = 1, maxK = 1
Output: 10
Explanation: Every subarray of nums is a fixed-bound subarray. There are 10 possible subarrays.
Constraints:
2 <= nums.length <= 1051 <= nums[i], minK, maxK <= 106
Solution
class Solution {
fun countSubarrays(nums: IntArray, minK: Int, maxK: Int): Long {
var ans: Long = 0
var i = 0
while (i < nums.size) {
if (nums[i] in minK..maxK) {
var a = i
var b = i
var mini = 0
var maxi = 0
while (i != nums.size && nums[i] >= minK && nums[i] <= maxK) {
i++
}
while (true) {
while (b != i && (mini == 0 || maxi == 0)) {
if (nums[b] == minK) {
mini++
}
if (nums[b] == maxK) {
maxi++
}
b++
}
if (mini == 0 || maxi == 0) {
break
}
while (mini != 0 && maxi != 0) {
if (nums[a] == minK) {
mini--
}
if (nums[a] == maxK) {
maxi--
}
ans += (1 + (i - b)).toLong()
a++
}
}
}
i++
}
return ans
}
}