Scale Engineer

Maximum Number of Groups With Increasing Length

Hard
#2502Time: O(2^N * N!) or worse. The number of subsets of numbers is `2^N`, and for each subset, we can arrange them in groups. The branching factor is huge. This is computationally infeasible.Space: O(N * k_max) for the recursion stack, where `k_max` is the maximum number of groups.
Patterns
MathGreedy
Algorithms
Binary SearchSorting
Data structures
Array

Prompt

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.

 

Example 1:

usageLimits

Example 2:

usageLimits

Example 3:

usageLimits

 

Constraints:

  • 1 <= usageLimits.length <= 105
  • 1 <= usageLimits[i] <= 109

Approaches

3 approaches with complexity analysis and trade-offs.

This approach explores the problem space by trying to build groups one by one recursively. It tries all possible valid group compositions at each step. A recursive function would try to form the k-th group, given the state of usage of numbers and the size of the (k-1)-th group.

Algorithm

  • Define a recursive function solve(k, prev_size, current_usages) where k is the group number to form, prev_size is the size of group k-1, and current_usages tracks usage of each number.
  • The function aims to maximize the number of groups starting from k.
  • Initialize max_groups = 0.
  • Iterate through all possible sizes s for group k, from prev_size + 1 to n.
  • For each size s, generate all combinations of s distinct numbers from 0 to n-1.
  • For each combination:
    • Check if this group is valid by comparing required usages with usageLimits - current_usages.
    • If valid, create next_usages by updating current_usages.
    • Recursively call res = 1 + solve(k + 1, s, next_usages).
    • Update max_groups = max(max_groups, res).
  • Return max_groups.
  • The initial call would be solve(1, 0, initial_usages).

Walkthrough

The core idea is to define a recursive function, say findMaxGroups(groupNum, prevGroupSize, usage), which tries to find the maximum number of additional groups that can be formed starting from group groupNum.

  • groupNum: The index of the current group to form (e.g., 1, 2, 3...).
  • prevGroupSize: The size of the previously formed group. The current group must be larger than this.
  • usage: An array or map to track the current usage of each number i, ensuring it doesn't exceed usageLimits[i].

The function would work as follows:

  1. Iterate through all possible sizes s for the current group, where s > prevGroupSize.
  2. For each size s, iterate through all combinations of s distinct numbers.
  3. For each combination, check if using these numbers is valid (i.e., their usage doesn't exceed their limits).
  4. If a valid group can be formed, make a recursive call: 1 + findMaxGroups(groupNum + 1, s, updatedUsage).
  5. The function returns the maximum value found among all valid choices. The base case is when no valid group can be formed, in which case it returns 0.

This approach is highly inefficient because it explores a vast number of possibilities (combinations of sizes and numbers) and would lead to a Time Limit Exceeded verdict on any reasonably sized input.

Complexity

Time

O(2^N * N!) or worse. The number of subsets of numbers is `2^N`, and for each subset, we can arrange them in groups. The branching factor is huge. This is computationally infeasible.

Space

O(N * k_max) for the recursion stack, where `k_max` is the maximum number of groups.

Trade-offs

Pros

  • Conceptually simple in terms of exploring all possibilities.

Cons

  • Extremely inefficient and will time out for all but the smallest inputs.

  • Complex to implement correctly due to managing combinations and state.

Solutions

class Solution {public  int maxIncreasingGroups(List<Integer> usageLimits) {    Collections.sort(usageLimits);    int k = 0;    long s = 0;    for (int x : usageLimits) {      s += x;      if (s > k) {        ++k;        s -= k;      }    }    return k;  }}

Video walkthrough

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