2790. Maximum Number of Groups With Increasing Length


Problem Description

In this problem, you are provided with an array called usageLimits which is 0-indexed and has a length of n. This array represents the maximum times each number, from 0 to n-1, can be used to create groups. The goal is to form the maximum number of groups under certain conditions:

  • Each group is made up of distinct numbers, meaning a particular number can’t be included more than once within the same group.
  • Each group must have more members than the previous one, except for the very first group.

You are required to determine the maximum number of groups that can be created while adhering to the above constraints.

Intuition

To tackle this problem, we need a strategy that systematically utilizes the availability constraints specified by usageLimits. Since we need each subsequent group to have more elements than the previous, we could benefit from starting with the least usage limits.

Hence, the first step in our approach is to sort the usageLimits array. This way, we start forming groups with numbers which have the lowest usage limits, thereby preserving the more usable numbers for later, potentially larger, groups.

Now, we have to keep track of how many groups we can form. To do this, we iterate through the sorted usageLimits array and aggregate a sum (s) of the elements, which helps us to understand the total available "slots" for forming groups at each step. The number of groups (k) is initially set to 0.

As we iterate, if the total available slots (sum s) are greater than the number of groups formed so far (k), this indicates that there is enough capacity to form a new group with one more member than the last. So we increment our group count (k) and then adjust our available slots by subtracting the size of the new group (k) from our sum (s).

The iteration continues until we've processed all elements in usageLimits. The final value of k will be the maximum number of groups we can form.

This approach leverages greedy and sorting techniques to efficiently satisfy the conditions of distinct numbers and strictly increasing group sizes.

Learn more about Greedy, Math, Binary Search and Sorting patterns.

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Solution Approach

The implementation of the solution uses a sorting technique, iteration, and simple arithmetic calculations:

  1. Sorting: We start by sorting the usageLimits array. Sorting is crucial because it ensures that we use the numbers with smaller limits first, allowing us to potentially create more groups.

  2. Iteration: After sorting, we iterate over each element in the usageLimits array. This step helps us to accumulate the total number of times we can use the numbers to form groups.

  3. Group Formation and Incrementation: We use two variables, k for the number of groups formed, and s for the aggregated sum of usageLimits. With each iteration, we check if the current sum s is greater than the number of groups k. If so, we can form a new group and we increment k. This check is essentially asking, "Do we have enough available numbers to form a group larger than the current largest group?"

  4. Sum Adjustment: Once we increment k to add a new group, we must also adjust the sum s to reflect the creation of this group. This is achieved by subtracting k from s. The subtraction of k from s signifies that we've allocated k distinct numbers for the new largest group, and these numbers cannot be re-used for another group.

The strategy employs a greedy approach, where we "greedily" form new groups as soon as we have enough capacity to do so, and always ensure that the new group complies with the conditions (distinct numbers and strictly greater size than the previous group). The final k value when we exit the loop is the maximum number of groups we can form.

Here, we don’t need any complex data structures. Array manipulation, following the sorting, and the use of a couple of integer variables to keep track of the sum and group count suffice in implementing the solution.

In conclusion, the combination of sorting, greedy algorithms with a simple condition check and variable adjustment within a single scan of the array yield the solution for the maximum number of groups that can be formed under the given constraints.

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Example Walkthrough

Let's illustrate the solution approach with a small example. Suppose we have the following usageLimits array:

1usageLimits = [3, 1, 2, 0]

Here’s how we might walk through the solution:

  1. Sorting: First, we sort the usageLimits array in non-decreasing order:

    1usageLimits = [0, 1, 2, 3]
  2. Iteration: Next, we begin to iterate over the sorted usageLimits array while keeping track of the sum s of the elements, and the number of formed groups k. Initially, s = 0 and k = 0.

  3. Group Formation and Incrementation:

    • At the first iteration, s = 0 + 0 = 0 and k = 0. We cannot form a group yet because there are not enough distinct numbers.
    • At the second iteration, s = 0 + 1 = 1 and k = 0. We now have enough numbers to form a group with 1 distinct number, so k is incremented to 1.
    • At the third iteration, s = 1 + 2 = 3 and k = 1. There are enough numbers to form another group with 2 distinct numbers (k + 1), so k is incremented to 2.
  4. Sum Adjustment: After each group formation, we need to adjust s for the new group size. After the third iteration, we subtract k from s:

    1s = 3 - 2 = 1
  5. Continuing Iteration:

    • At the fourth and final iteration, s = 1 + 3 = 4 and k = 2. Here, we can form one last group with 3 distinct numbers since s (4) is greater than k (2), so k is incremented to 3.
    • We adjust s one last time:
    1s = 4 - 3 = 1
  6. Result: Since there are no more elements in usageLimits, our iteration ends. The final value of k, which is 3, is the maximum number of groups that can be formed given the constraints. Each group having 1, 2, and 3 distinct members respectively.

The example demonstrates how by sorting the array and using a greedy approach, systematically checking and updating counts, we maximize the number of distinct groups that can be created following the rules specified.

Solution Implementation

1from typing import List
2
3class Solution:
4    def maxIncreasingGroups(self, usage_limits: List[int]) -> int:
5        # First, sort the usage limits in non-decreasing order.
6        usage_limits.sort()
7
8        # Initialize the number of groups and the current sum of the group limits.
9        groups_count = 0
10        current_sum = 0
11
12        # Iterate through the sorted usage limits.
13        for limit in usage_limits:
14            # Add the current limit to the current sum.
15            current_sum += limit
16          
17            # If the current sum is greater than the number of groups,
18            # it means we can form a new group.
19            if current_sum > groups_count:
20                # Increment the number of groups.
21                groups_count += 1
22                # Adjust the current sum by subtracting the new group's count.
23                current_sum -= groups_count
24      
25        # Return the total number of groups formed.
26        return groups_count
27
1import java.util.Collections;
2import java.util.List;
3
4class Solution {
5    public int maxIncreasingGroups(List<Integer> usageLimits) {
6        // Sort the usage limits list in ascending order
7        Collections.sort(usageLimits);
8      
9        // 'k' represents the number of increasing groups formed
10        int k = 0;
11      
12        // 'sum' represents the cumulative sum of the usage limits
13        long sum = 0;
14      
15        // Iterate over the sorted usage limits
16        for (int usageLimit : usageLimits) {
17            // Add the current usage limit to 'sum'
18            sum += usageLimit;
19          
20            // If the sum is greater than the number of groups formed,
21            // we can form a new group by increasing 'k'
22            // and then adjust the sum by subtracting 'k'
23            if (sum > k) {
24                k++;         // Increment the number of groups
25                sum -= k;    // Decrement the sum by the new number of groups
26            }
27        }
28        // Return the maximum number of increasing groups formed
29        return k;
30    }
31}
32
1class Solution {
2public:
3    // Method to calculate the maximum number of increasing groups
4    int maxIncreasingGroups(vector<int>& usageLimits) {
5        // Sort the usage limits in non-decreasing order
6        sort(usageLimits.begin(), usageLimits.end());
7
8        // Initialize the number of groups that can be formed
9        int numberOfGroups = 0;
10
11        // Initialize a sum variable to calculate the cumulative sum of elements
12        long long sum = 0;
13
14        // Iterate over each usage limit
15        for (int limit : usageLimits) {
16            // Add the current limit to the cumulative sum
17            sum += limit;
18
19            // Check if after adding the current element, the sum becomes
20            // greater than the number of groups we have so far.
21            // If yes, it means we can form a new group with higher limit.
22            if (sum > numberOfGroups) {
23                // Increment the number of groups that can be formed
24                numberOfGroups++;
25              
26                // Decrease the sum by the number of groups since
27                // we allocate each element of a group with a distinct size
28                sum -= numberOfGroups;
29            }
30        }
31
32        // Return the maximum number of increasing groups that can be formed
33        return numberOfGroups;
34    }
35};
36
1function maxIncreasingGroups(usageLimits: number[]): number {
2    // Sort the input array of usage limits in non-decreasing order
3    usageLimits.sort((a, b) => a - b);
4  
5    // Initialize the variable 'groups' to count the maximum number of increasing groups
6    let groups = 0;
7
8    // Initialize the variable 'sumOfGroupLimits' to keep the sum of limits in the current group
9    let sumOfGroupLimits = 0;
10
11    // Iterate through each usage limit in the sorted array
12    for (const limit of usageLimits) {
13        // Add the current limit to the sum of group limits
14        sumOfGroupLimits += limit;
15
16        // If the sum of group limits exceeds the current number of groups,
17        // this indicates the formation of a new increasing group
18        if (sumOfGroupLimits > groups) {
19            // Increment the number of groups
20            groups++;
21            // Subtract the number of groups from the sum of group limits
22            // to prepare for the next potential group
23            sumOfGroupLimits -= groups;
24        }
25    }
26
27    // Return the total number of increasing groups formed
28    return groups;
29}
30
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Time and Space Complexity

The given Python code is a function that aims to determine the maximum number of increasing groups from a list of usage limits. Here's the analysis of its time and space complexity:

Time Complexity

The most expensive operation in the algorithm is the sort operation, which has a time complexity of O(n log n), where n is the number of elements in usageLimits. After sorting, the function then iterates over the sorted list once, producing an additional O(n) complexity. Therefore, the total time complexity of the algorithm is dominated by the sorting step, resulting in O(n log n) overall.

Space Complexity

As for space complexity, the sorting operation can be O(n) in the worst case or O(log n) if an in-place sort like Timsort (used by Python's .sort() method) is applied efficiently. The rest of the algorithm uses a fixed amount of additional variables (k and s) and does not rely on any additional data structures that depend on the input size. Thus, the space complexity is O(1) for the additional variables used, but due to the sort's potential additional space, it could be O(n) or O(log n). Since Python's sort is typically implemented in a way that aims to minimize space overhead, the expected space complexity in practical use cases would likely be closer to O(log n).

Therefore, the final space complexity of the algorithm is O(log n) under typical conditions.

Note: The space complexity stated above assumes typical conditions based on Python's sorting implementation. The actual space complexity could vary depending on the specific details of the sort implementation in the environment where the code is executed.

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