How do you solve Maximum Number of Groups With Increasing Length on LeetCode?

Maximum Number of Groups With Increasing Length (LeetCode #2790) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Sorting the usage limits allows for a more efficient allocation of numbers to groups.

What is the time complexity of Maximum Number of Groups With Increasing Length?

The optimal solution for Maximum Number of Groups With Increasing Length runs in O(n log n) time and uses O(1) space. The time complexity is O(n log n) due to the sorting step, while the space complexity is O(1) as we only use a few additional variables.

What is the brute force solution for Maximum Number of Groups With Increasing Length?

The brute force approach for Maximum Number of Groups With Increasing Length is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying to create groups of increasing lengths while adhering to the usage limits. We will keep track of how many times each number can be used and attempt to form groups one by one until we can no longer satisfy the conditions. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Number of Groups With Increasing Length?

Maximum Number of Groups With Increasing Length uses the following concepts: Array, Math, Binary Search, Greedy, Sorting. The recommended patterns to study are: Greedy, Sorting.

What are the interview tips for Maximum Number of Groups With Increasing Length?

Always consider sorting as a first step when dealing with constraints on values. Think about how you can use greedy algorithms to maximize or minimize values based on certain conditions. Be ready to explain your thought process clearly, especially when transitioning from brute force to optimal solutions.

What are common mistakes when solving Maximum Number of Groups With Increasing Length?

Failing to sort the usageLimits before attempting to form groups. Not correctly tracking the current length of the groups, leading to incorrect group formation.

#2790

Maximum Number of Groups With Increasing Length

Hard

ArrayMathBinary SearchGreedySortingGreedySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

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Intuition

Time O(n log n)Space O(1)

The optimal approach involves sorting the usage limits and using a greedy strategy to maximize the number of groups. By sorting, we can efficiently allocate the numbers to groups while ensuring the increasing length condition is satisfied.

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Algorithm

4 steps

1Step 1: Sort the usageLimits array in non-decreasing order.
2Step 2: Initialize a counter for groups and a variable to track the current group length.
3Step 3: Iterate through the sorted usageLimits, trying to form groups of increasing lengths.
4Step 4: For each group, check if we can use the numbers without exceeding their limits, and update the usage count accordingly.

solution.py15 lines

1# Full working Python code
2usageLimits = [1, 2, 5]
3
4def maxGroups(usageLimits):
5    usageLimits.sort()
6    groups = 0
7    currentLength = 1
8
9    for usage in usageLimits:
10        if usage >= currentLength:
11            groups += 1
12            currentLength += 1
13    return groups
14
15print(maxGroups(usageLimits))

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the space complexity is O(1) as we only use a few additional variables.

1Sorting the usage limits allows for a more efficient allocation of numbers to groups.
2The greedy approach of forming groups based on the current length ensures that we maximize the number of groups.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.