LeetCode-in-Java
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 java.util.Arrays;
import java.util.List;
public class Solution {
public int maxIncreasingGroups(List<Integer> usageLimits) {
int n = usageLimits.size();
long total = 0;
long k = 0;
int[] count = new int[n + 1];
Arrays.fill(count, 0);
for (int a : usageLimits) {
int localA = (int) Math.min(a, (double) n);
count[localA]++;
}
for (int i = 0; i <= n; i++) {
for (int j = 0; j < count[i]; j++) {
total += i;
if (total >= (k + 1) * (k + 2) / 2) {
k++;
}
}
}
return (int) k;
}
}