1655. Distribute Repeating Integers

Problem Description

You have been given two arrays: nums and quantity. The array nums consists of n integers which may have at most 50 unique values. The quantity array has m different customer order quantities, where quantity[i] signifies the number of integers the ith customer needs to order. The challenge is to determine whether it is possible to distribute the integers from the nums array to fulfill the following conditions:

1. Each ith customer should receive exactly quantity[i] number of integers.
2. All the integers a customer receives must be identical.
3. All customers should have their orders satisfied as per the first two conditions.

The goal is to return true if it is indeed possible to disseminate the integers contained in nums to satisfy all customers as described by the conditions.

Intuition

To find the solution to this problem, we can think in terms of subsets and dynamic programming. Here are the key steps on how we can approach the problem:

1. Generate all possible subsets of order quantities that can be fulfilled.
2. Use a counter or map to keep track of the number of times each unique value appears in the nums array since each customer needs to receive the same number of identical integers.
3. Utilize dynamic programming where each state represents whether a given subset of customer orders can be satisfied with a selection of integers from the nums array.
4. For every unique integer in the nums array, iterate over all possible customer order subsets to determine if the current integer can fulfill some or all of the orders in the subset.
5. Since the dynamic programming dimension is 2^m (where m is the number of customers/orders), we need to iterate over all states and try to partition the state into two parts: the one which could be satisfied by previous numbers (if not the first) and the one which needs to be satisfied by the current number.

By iterating over the nums array and updating our dynamic programming table to keep track if we can satisfy customers' orders with the quantities we have, we will finally reach a conclusion. If the final state—representing all customers' orders being fulfilled—can be reached, we return true, otherwise, false.

The provided code implements this approach with an efficient bit manipulation strategy to manage subsets and dynamic programming states.

Learn more about Dynamic Programming, Backtracking and Bitmask patterns.

Solution Approach

The solution uses bit manipulation and dynamic programming to approach this problem.

Firstly, the solution calculates all possible sums of quantity using bit manipulation. The s array is used to store sums for different combinations of customers’ order quantities (quantity) based on the subset of customers chosen, which can be represented by a bit mask. This is done using the first for loop.

Then, a counter for nums is used to determine how many times each unique value appears. This is important as it dictates the maximum number of orders that can be fulfilled by that integer. The values from the counter are then stored in the arr list.

The dynamic programming table f is initialized as a 2D array filled with False, with dimensions [n][1 << m], where n is the number of unique integers in nums and m is the number of customer orders. Here, f[i][j] means whether it is possible to fulfill the subset of orders represented by j using the first i integers.

The first column of the DP table is set to True to represent that the subset of orders with no orders (empty subset) can always be fulfilled regardless of the available integers.

By looping through each integer x available (arr) and for each subset of orders, the code updates the DP table f. The nested loops go over each combination j of orders and attempt to reduce it to a smaller subset k by fulfilling some orders using the current integer x.

• If j is equal to k (or if f[i-1][j^k] is True for non-first integers), it means previous integers (up to i-1) can already fulfill the other part of orders (j^k).
• If s[k] is less than or equal to x, it means x can be used to fulfill the orders in subset k.

If both conditions are satisfied, f[i][j] becomes True, signaling that it's possible to satisfy the subset of orders j using integers up to i. The while loop uses the bit manipulation trick k = (k - 1) & j to iterate through all non-empty subsets of j.

The last step is to check f[-1][-1], which tells us whether all the orders can be satisfied using all available unique numbers in nums. If f[-1][-1] is True, the solution returns true, indicating that it is indeed possible to distribute nums according to the order quantities in quantity.

Example Walkthrough

Let's take a small example to illustrate the solution approach.

Assume nums = [1, 2, 3, 4] and quantity = [2, 2]. The array nums consists of integers which have at most 2 different quantities: 2 of 1 and 2 of 2. The quantity array contains two customer order quantities: the first customer needs 2 integers and the second customer also needs 2 integers.

Firstly, we calculate all possible sums of quantity using bit manipulation. Here the possible sums are 0 (no customers), 2 (either customer), and 4 (both customers):

• Subset bitmask 00 represents no customers, so sum s[0] is 0.
• Subset bitmask 01 represents the first customer, so sum s[1] is 2.
• Subset bitmask 10 represents the second customer, so sum s[2] is 2.
• Subset bitmask 11 represents both customers, so sum s[3] is 4.

Then, we count how many times each number exists in nums:

• Number 1 appears 1 time.
• Number 2 appears 1 time.
• Number 3 appears 1 time.
• Number 4 appears 1 time.

The counter for nums shows each unique integer occurs just once, so our arr list contains [1,1,1,1].

The dynamic programming (DP) table f will have dimensions [4][1 << 2] (since we have 4 unique integers in nums and 2 customers), which becomes [4][4]. All values in f are initialized to False except f[i][0] for each row i, which is set to True.

Now, we loop through each x in arr (each being 1 in our example) and update the DP table f. We check if the current integer x can fulfill some combination of orders represented by j. We iterate over all non-empty subsets k of j. For our example, there aren't any integers in arr that can fulfill a sum of 2 or 4. However, here's how the DP table updating logic would look like if we had an integer x that could:

• For x able to satisfy a sum of 2, f[i][01 or 10] would be set to True.
• For x able to satisfy a sum of 4 as well, f[i][11] would be set to True.

In our hypothetical situation, if we found a large enough x in arr, and after iterating through the nested loops, we could possibly have the DP table f showing f[i][11] = True, indicating we can satisfy both customers. But since in our actual case x is 1 and cannot satisfy either customer's quantity demand of 2, the final state f[-1][-1] stays False.

Thus, the return value for our example will be False, indicating that it's not possible to distribute the numbers in nums to satisfy the quantities in quantity.

In summary, our example input nums = [1, 2, 3, 4] and quantity = [2, 2] does not satisfy the conditions for the customers’ orders since no integer in nums is available in sufficient quantity to fulfill a quantity of 2.

Solution Implementation

Time and Space Complexity

The given code is an implementation of a solution to the problem of checking whether it is possible to distribute nums into quantity. Let's analyze the time and space complexity of the provided code:

Time Complexity

The time complexity can be analyzed as follows:

• There are two nested loops at the beginning which build an array s of size 2^m where m is the length of quantity. This takes O(m * 2^m) time since for every subset, it calculates the sum of quantities in that subset.
• The cnt = Counter(nums) operation has a time complexity of O(n) where n is the number of elements in nums.
• The double nested loops that build the f table iterate over every element in arr (which is at most n) and over every possible combination of quantity (2^m possibilities), making it O(n * 2^m).
• Inside the innermost loop, there is another while-loop that can iterate up to m times for each bit-set in k.

Putting this all together, the dominant part of the time complexity is the double nested loops, leading to an overall time complexity of O(n * 2^m + m * 2^m), which simplifies to O((n + m) * 2^m).

Space Complexity

The space complexity is determined by:

• The extra array s, which has a space complexity of O(2^m).
• The cnt object, which in the worst case, contains as many keys as there are unique elements in nums. In the worst case, this is O(n).
• The f table, which is a 2D table of n rows and 2^m columns, thus having a space complexity of O(n * 2^m).

Therefore, the total space complexity is dominated by the f table, giving us a space complexity of O(n * 2^m).

Learn more about how to find time and space complexity quickly using problem constraints.