Leetcode #2790: Maximum Number of Groups With Increasing Length
In this guide, we solve Leetcode #2790 Maximum Number of Groups With Increasing Length in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Greedy, Array, Math, Binary Search, Sorting
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
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.
Python Solution
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
Complexity
The time complexity is O(log n) or O(n log n). The space complexity is O(1).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.