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Maximum Number of Groups With Increasing Length

Number: 2919

Difficulty: Hard

Paid? No

Companies: Amazon

Problem Description

You are given a 0-indexed array usageLimits of length n where each element usageLimits[i] represents the maximum number of times you can use number i (i ranges from 0 to n - 1) across all groups. You need to form groups such that:

  • Each group consists of distinct numbers.
  • Except for the first group, the length of each group is strictly greater than the previous group.

Return the maximum number of groups you can create while satisfying these conditions.

Key Insights

  • Even though each group must have distinct numbers, the same number can appear multiple times across different groups as long as its total usage does not exceed usageLimits[i].
  • For k groups with strictly increasing lengths, the groups must have sizes 1, 2, 3, …, k. Hence, the total number of numbers used is k*(k+1)/2.
  • A number with a usage limit x can contribute at most min(x, k) times (it can appear in at most one number per group).
  • To determine if it is feasible to form k groups, ensure that the sum of min(usageLimits[i], k) over all numbers is at least k*(k+1)/2.
  • With n numbers, note that even if every number were available in every group, the maximum contribution sum would be nk. Since k(k+1)/2 must be <= nk, k is bounded by roughly 2n. This forms a natural upper bound for binary search.

Space and Time Complexity

Time Complexity: O(n log n) — For each candidate value k (searched via binary search), you iterate over usageLimits (n elements). Space Complexity: O(1) (ignoring the space required for the input) — Only constant extra space is used.

Solution

We use binary search over the possible number of groups k. For a candidate k, we compute the total available usages by summing min(usageLimits[i], k) for each number. If this sum is at least k*(k+1)/2, then it is possible to form k groups. We adjust our binary search based on this condition. The key idea is to reduce the search space by leveraging the bound on k and checking the feasibility for each candidate via a greedy summation strategy.

Code Solutions

Below are code solutions in Python, JavaScript, C++, and Java with inline comments for clarity.

def maxGroups(usageLimits):
    n = len(usageLimits)
    # Lower bound of groups is 0, upper bound is set to 2*n based on our observation.
    low, high = 0, 2 * n  
    ans = 0
    while low <= high:
        mid = (low + high) // 2  # Candidate number of groups
        total = 0
        # Compute the total contribution from each number limited by mid
        for limit in usageLimits:
            total += min(limit, mid)
            # Early exit if total already meets the required sum
            if total >= mid * (mid + 1) // 2:
                break
        # Check if it is possible to form mid groups
        if total >= mid * (mid + 1) // 2:
            ans = mid  # mid is feasible; try a higher number of groups
            low = mid + 1
        else:
            high = mid - 1  # mid is not feasible; try lower
    return ans

# Example usage:
print(maxGroups([1, 2, 5]))  # Expected output: 3
print(maxGroups([2, 1, 2]))  # Expected output: 2
print(maxGroups([1, 1]))     # Expected output: 1
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