LeetCode in Kotlin
2790. Maximum Number of Groups With Increasing Length
Hard
You are given a 0-indexed array usageLimits of length n.
Your task is to create groups using numbers from 0 to n - 1, ensuring that each number, i, is used no more than usageLimits[i] times in total across all groups. You must also satisfy the following conditions:
- Each group must consist of distinct numbers, meaning that no duplicate numbers are allowed within a single group.
- Each group (except the first one) must have a length strictly greater than the previous group.
Return an integer denoting the maximum number of groups you can create while satisfying these conditions.
Example 1:
Input: usageLimits = [1,2,5]
Output: 3
Explanation:
In this example, we can use 0 at most once, 1 at most twice, and 2 at most five times.
One way of creating the maximum number of groups while satisfying the conditions is:
Group 1 contains the number [2].
Group 2 contains the numbers [1,2].
Group 3 contains the numbers [0,1,2].
It can be shown that the maximum number of groups is 3.
So, the output is 3.
Example 2:
Input: usageLimits = [2,1,2]
Output: 2
Explanation:
In this example, we can use 0 at most twice, 1 at most once, and 2 at most twice.
One way of creating the maximum number of groups while satisfying the conditions is:
Group 1 contains the number [0].
Group 2 contains the numbers [1,2].
It can be shown that the maximum number of groups is 2.
So, the output is 2.
Example 3:
Input: usageLimits = [1,1]
Output: 1
Explanation:
In this example, we can use both 0 and 1 at most once.
One way of creating the maximum number of groups while satisfying the conditions is:
Group 1 contains the number [0].
It can be shown that the maximum number of groups is 1.
So, the output is 1.
Constraints:
1 <= usageLimits.length <= 1051 <= usageLimits[i] <= 109
Solution
import kotlin.math.min
class Solution {
fun maxIncreasingGroups(usageLimits: List<Int>): Int {
val n: Int = usageLimits.size
var total: Long = 0
var k: Long = 0
val count = IntArray(n + 1)
count.fill(0)
for (a in usageLimits) {
val localA = min(a.toDouble(), n.toDouble()).toInt()
count[localA]++
}
for (i in 0..n) {
for (j in 0 until count[i]) {
total += i.toLong()
if (total >= (k + 1) * (k + 2) / 2) {
k++
}
}
}
return k.toInt()
}
}