#2790

Hard

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Maximum Number of Groups With Increasing Length

Maximize the number of groups that can be formed with given usage limits, leveraging binary search for optimal solutions.

array math binary search greedy sorting

Problem Statement

You are given a 0-indexed array usageLimits of length n. Your goal is to form the maximum number of groups such that each group contains integers from 0 to n - 1, where each integer, i, can be used no more than usageLimits[i] times in total across all groups.

Return the maximum number of groups you can create while satisfying these conditions. The solution involves finding the maximum possible value for the number of groups that can be formed using binary search and a greedy strategy to ensure each integer is used appropriately.

Examples

Example 1

Input: usageLimits = [1,2,5]

Output: 3

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

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

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 <= 105
1 <= usageLimits[i] <= 109

Solution Approach

Binary Search Over Valid Answer Space

The core approach for solving this problem is to apply binary search over the possible number of groups. This binary search is done over the range from 0 to n, where n is the length of the usageLimits array. At each step, the binary search checks if it is possible to form that many groups by leveraging a greedy approach to check if the usage limits can accommodate the groups.

Greedy Group Formation

A greedy approach is used to form groups incrementally, ensuring that each integer is used no more than its corresponding limit from the usageLimits array. This greedy method helps in determining whether the current mid-point number of groups in binary search is achievable under the given constraints.

Validating Group Forming with Sorting

Sorting the usageLimits array can aid in efficiently managing the group formation. Once sorted, the greedy approach becomes more manageable, as the numbers with lower usage limits will be placed in earlier groups, ensuring that we do not exceed any usage limit while maximizing the number of groups.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends primarily on the binary search, which operates in O(log n) time, and the group validation step, which requires iterating over the usageLimits array in O(n) time. Therefore, the overall time complexity is O(n log n). The space complexity is O(n) due to the space required for sorting and storing intermediate results during the greedy process.

What Interviewers Usually Probe

Can the candidate efficiently apply binary search over the possible answer space?
Is the candidate comfortable using a greedy approach to validate group formation?
Does the candidate optimize the solution using sorting for easier greedy implementation?

Common Pitfalls or Variants

Common pitfalls

Failing to correctly implement binary search over the valid answer space can lead to inefficient solutions.
Not handling edge cases where the group sizes or the usage limits are very small or large.
Ignoring the importance of sorting the usageLimits array to facilitate the greedy approach.

Follow-up variants

Handling different ranges of usageLimits values to see how the solution scales.
Optimizing the solution further using advanced techniques, such as dynamic programming.
Exploring alternative greedy approaches for edge cases where the binary search may not be optimal.

FAQ

What is the optimal solution for the Maximum Number of Groups With Increasing Length problem?

The optimal solution involves using binary search over the number of possible groups and validating each mid-point number of groups using a greedy approach.

Can we use a greedy approach alone to solve the problem?

A greedy approach alone is not enough; binary search is necessary to efficiently explore the valid number of groups and optimize the solution.

Why is sorting necessary for this problem?

Sorting the usageLimits array helps manage the group formation process, as it ensures that smaller limits are handled first, preventing overuse of any integer.

What is the time complexity of the solution?

The time complexity is O(n log n) due to binary search and the greedy validation process, where n is the length of the usageLimits array.

How does binary search apply to this problem?

Binary search is applied to find the maximum number of groups by checking if it's possible to form that number of groups with the given constraints.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def maxIncreasingGroups(self, usageLimits: List[int]) -> int:
        usageLimits.sort()
        k, n = 0, len(usageLimits)
        for i in range(n):
            if usageLimits[i] > k:
                k += 1
                usageLimits[i] -= k
            if i + 1 < n:
                usageLimits[i + 1] += usageLimits[i]
        return k

Solution 2

#### Python3

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class Solution:
    def maxIncreasingGroups(self, usageLimits: List[int]) -> int:
        usageLimits.sort()
        k, n = 0, len(usageLimits)
        for i in range(n):
            if usageLimits[i] > k:
                k += 1
                usageLimits[i] -= k
            if i + 1 < n:
                usageLimits[i + 1] += usageLimits[i]
        return k

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