Maximum Number of Groups With Increasing Length

Problem

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.

Examples

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

• $1 <= usageLimits.length <= 10^5$
• $1 <= usageLimits[i] <= 10^9$

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Approach

To solve the problem, we need to understand the nature of the allowed moves:

1. Sort the Usage Limits:

   • Start by sorting the 'usageLimits' array. This helps in efficiently forming groups by using the smallest available numbers first, which maximizes the number of groups we can form.

2. Initialize Variables:

   • 'groups': Keeps track of the number of groups formed.

   • 'currentGroupSize': Represents the required size for the next group to be formed, starting at 1.

   • 'availableNumbers': Tracks the total available slots from all numbers seen so far.

3. Form Groups:

   • Iterate through the sorted 'usageLimits'.

   • For each limit, add it to 'availableNumbers'.

   • Check if 'availableNumbers' is at least as large as 'currentGroupSize'.

   • If true, it means we can form a new group of size 'currentGroupSize'.

   • Increment the 'groups' counter, decrement 'availableNumbers' by 'currentGroupSize', and 'increase' currentGroupSize by 1 for the next group.

Solution for Maximum Number of Groups With Increasing Length

• The core idea is to maximize the number of groups by ensuring each group has a distinct set of numbers and each subsequent group is larger than the previous one.

• By sorting the 'usageLimits', we can efficiently use the smallest available counts first to form valid groups.

• This approach ensures that we can form as many groups as possible without exceeding the usage limits.

Code in Java

import java.util.Collections;
import java.util.List;

class Solution {
    public int maxIncreasingGroups(List<Integer> usageLimits) {
        // Sort the usageLimits in ascending order
        Collections.sort(usageLimits);

        // Initialize variables
        int groups = 0;
        int currentGroupSize = 1; // the size of the next group we want to form
        long availableNumbers = 0; // to track the number of available slots across all numbers

        // Traverse the sorted usageLimits
        for (int limit : usageLimits) {
            availableNumbers += limit; // add the current limit to available slots

            // Check if we can form the next group
            if (availableNumbers >= currentGroupSize) {
                groups++; // form a new group
                availableNumbers -= currentGroupSize; // reduce the used slots
                currentGroupSize++; // increase the size requirement for the next group
            }
        }

        return groups;
    }
}    

Complexity Analysis

Time Complexity: $O(nlogn)$

Reason: Time Complexity is $O(nlogn)$. Sorting the 'usageLimits' takes $O(nlogn)$ time. Iterating through the sorted list takes $O(n)$ time. Therefore, the overall time complexity is $O(nlogn)$.

Space Complexity: $O(1)$

Reason: The space complexity is $O(1)$, Because the additional space (ignoring the space used by the input list and the space needed for sorting, which is $O(n)$).

References

• LeetCode Problem: Maximum Number of Groups With Increasing Length
• Solution Link: Maximum Number of Groups With Increasing Length Solution on LeetCode
• Authors LeetCode Profile: Vivek Vardhan