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CountSubArrayWhereMaxAppearsAtleastKtimes.kt
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CountSubArrayWhereMaxAppearsAtleastKtimes.kt
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/*
* Count Subarrays Where Max Element Appears at Least K Times
You are given an integer array nums and a positive integer k.
Return the number of subarrays where the maximum element of nums appears at least k times in that subarray.
A subarray is a contiguous sequence of elements within an array.
Example 1:
Input: nums = [1,3,2,3,3], k = 2
Output: 6
Explanation: The subarrays that contain the element 3 at least 2 times are: [1,3,2,3], [1,3,2,3,3], [3,2,3], [3,2,3,3], [2,3,3] and [3,3].
Example 2:
Input: nums = [1,4,2,1], k = 3
Output: 0
Explanation: No subarray contains the element 4 at least 3 times.
Constraints:
1 <= nums.length <= 105
1 <= nums[i] <= 106
1 <= k <= 105
*/
fun main() {
print(countSubArrays(nums = intArrayOf(1, 3, 2, 3, 3), k = 2)) // 6
}
// The sub-arrays that contain the element 3 at least 2 times are:
// [1,3,2,3], [1,3,2,3,3], [3,2,3], [3,2,3,3], [2,3,3] and [3,3].
fun countSubarrays(nums: IntArray, k: Int): Long {
val max = nums.max()
var counter = 0
var result = 0L
var left = 0
var right = 0
while (right < nums.size) {
if (nums[right] == max) {
counter++
while (left <= right && counter >= k) {
result += (nums.size - right)
if (nums[left] == max)
counter--
left++
}
}
right++
}
return result
}
class Solution {
fun countSubarrays(nums: IntArray, k: Int): Long {
var i = 0;
var j = 0
var count = 0L
val maxNum = nums.max()
var maxNumCount = 0
while (j < nums.size) {
if (nums[j] == maxNum) maxNumCount++
while (maxNumCount >= k) {
if (nums[i] == maxNum) maxNumCount--
count += nums.size - j
i++
}
j++
}
return count
}
}
class Solution {
fun countSubarrays(nums: IntArray, k: Int): Long {
var result = 0L
var max = nums.max() ?: 0
var left = 0
var right = 0
var count = 0
while (right < nums.size) {
if (nums[right] == max)
count++
while (count >= k) {
if (nums[left++] == max)
count--
result += (nums.size - right)
}
right++
}
return result
}
}
class Solution {
fun countSubarrays(nums: IntArray, k: Int): Long {
val n = nums.size
val totalCnt = (n.toLong() * (n.toLong() + 1)) / 2
var lessCnt: Long = 0
val maxE = nums.maxOrNull() ?: 0
var j = 0
var cnt = 0
for (i in nums.indices) {
if (nums[i] == maxE)
cnt++
while (j < n && cnt >= k) {
if (nums[j] == maxE)
cnt--
j++
}
lessCnt += (i - j + 1).toLong()
}
return totalCnt - lessCnt
}
}
fun countSubarrays(nums: IntArray, k: Int): Long {
val n = nums.size.toLong()
val m = nums.max()
var ck = 0
var j = 0
return n * (n + 1) / 2 +
nums.indices.sumOf { i ->
if (nums[i] == m)
ck++
while (ck >= k)
if (nums[j++] == m)
ck--
-(i - j + 1).toLong()
}
}
class Solution {
fun countSubarrays(nums: IntArray, k: Int): Long {
val maxElement = nums.max()
var solutionCount = 0L
var maxElementCount = 0
var left = 0
for (right in nums.indices) {
if (nums[right] == maxElement)
maxElementCount++
while (maxElementCount >= k) {
maxElementCount -= if (nums[left++] == maxElement) 1 else 0
}
solutionCount += left
}
return solutionCount
}
}