Class Solution
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public final class Solution2790 - Maximum Number of Groups With Increasing Length\.
Hard
You are given a 0-indexed array
usageLimitsof lengthn.Your task is to create groups using numbers from
0ton - 1, ensuring that each number,i, is used no more thanusageLimits[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= 1,2,5Output: 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= 2,1,2Output: 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= 1,1Output: 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:
<code>1 <= usageLimits.length <= 10<sup>5</sup></code>
<code>1 <= usageLimitsi<= 10<sup>9</sup></code>
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Constructor Summary
Constructors Constructor Description Solution()
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Method Summary
Modifier and Type Method Description final IntegermaxIncreasingGroups(List<Integer> usageLimits)-
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Method Detail
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maxIncreasingGroups
final Integer maxIncreasingGroups(List<Integer> usageLimits)
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