2964. Number of Divisible Triplet Sums

Problem Description

The challenge is to find out the number of unique triplets within an array nums, where each triplet consists of different indices (i, j, k) that follow a specific criterion. This criterion requires that the indices form a strictly increasing sequence (i < j < k) and the sum of the elements at these indices is divisible by a given integer d. In other words, nums[i] + nums[j] + nums[k] should be an integer multiple of d.

To tackle this problem, we need to identify combinations of three different array elements whose indices are in ascending order and check for the divisibility of their sum by d. It's a computational challenge that requires efficient enumeration of the possible triplets to avoid a brute force approach that would take too long.

Intuition

To optimize the process of finding these triplets, the solution leverages a hash table strategy to avoid redundant calculations. The main idea behind the solution is to use a hash table, denoted as cnt, to keep track of how many times each remainder (when elements of nums are divided by d) occurs up to the current index being considered. As we iterate through the array, we calculate what remainder we would need from the nums[i] (where i < j) to ensure that the sum of nums[i], nums[j], and nums[k] is divisible by d.

Here's the thinking process:

1. As we move through the array, with each nums[j], we look ahead to all future nums[k] where k > j. For each of these pairs, we calculate the remainder that would be needed by a preceding nums[i] to make the sum of nums[i] + nums[j] + nums[k] divisible by d. We use the aforementioned hash table to quickly check the count of such potential nums[i] elements.

2. We then add the count from our hash table to our answer (accumulating the number of triplets that meet our criteria up to the current j).

3. After considering pairs of nums[j] and nums[k], we increment the count of the remainder of nums[j] itself in the hash table before continuing to the next j.

By doing this, we avoid calculating the same remainder over and over for each potential triplet, which drastically reduces the number of operations from what would be required in a brute force solution.

Solution Approach

The solution to this LeetCode problem is centered around a clever use of a hash table, specifically a defaultdict from Python's collections module, which allows us to automatically initialize missing keys with an integer (initialized to 0 in this case). This helps us to track the frequency of remained parts of numbers modulo d.

The algorithm can be broken down into the following steps:

1. First, we iterate over the elements of the nums array while calculating the remainder when each element nums[j] is divided by d (i.e., nums[j] % d). We use this to determine what the corresponding nums[i]'s remainder should be in order to have (nums[i] + nums[j] + nums[k]) % d == 0.

2. For any given index j, we look ahead to the elements nums[k] for all k such that k > j and calculate x, which is equal to (d - (nums[j] + nums[k]) % d) % d. This represents the remainder we need from some nums[i] (where i < j) so that the sum of the three elements is divisible by d.

3. The calculated x is then used to check in our cnt hash table how many times we've seen such a remainder before index j. We sum up these occurrences in a variable ans, which ultimately holds the total number of triplets that satisfy the problem's condition.

4. Before we move on to the next j, we increase the count of nums[j]'s remainder in the hash table by 1, i.e., cnt[nums[j] % d] += 1, representing that we have seen another occurrence of this particular remainder.

5. Once we exhaust all possibilities for j and its corresponding k, the variable ans will hold the correct answer, which is then returned.

One of the clever patterns employed in this solution is the recognition that for triplets (i, j, k) to satisfy our condition, it is not necessary to track each i explicitly. By using remainders and counting their occurrences, we implicitly handle all possible i candidates while iterating over j and k. This avoids the need for a full, expensive three-level loop and thus significantly improves the time complexity.

The use of mathematics to track the needed remainder part instead of the raw sum also reduces the memory complexity as we only need to store counts for each possible remainder range from 0 to d-1. This is much smaller than the potential range of the sums if d is small relative to the values in nums.

This algorithm runs in O(n^2) time complexity, with n being the size of the array nums, which is much more efficient than the brute force O(n^3). The space complexity is O(d) due to the hash table storing at most d different remainders.

Example Walkthrough

Let's illustrate the solution approach using an example:

Consider nums = [2, 3, 5, 7, 11] and d = 5.

We initiate a defaultdict(int) which will serve as the cnt hash table to store the counts of seen remainders.

As there are no values in cnt yet, all remainders start with a count of 0.

We start by iterating through the array:

1. For nums[0] = 2, the remainder when divided by d is 2.

   • There are no previous elements, so we just move to the next index.
   • Update cnt[2] to 1 (since 2 % 5 = 2).

2. At nums[1] = 3, the remainder is 3.

   • We are looking for a triplet that includes nums[i], nums[j]=3, and some future nums[k].
   • Update cnt[3] to 1 (3 % 5 = 3).

3. Moving on to nums[2] = 5, the remainder is 0.

   • Now, let's look ahead for future k values:
     • For a potential k with nums[k] = 7 (next element), we need a remainder x (needed from some previous nums[i]), which is (5 - (3 + 7) % 5) % 5 = 0. We look into cnt and see that cnt[0] = 0 (0 hasn't been seen yet as a remainder before index 2), so no triplets can be formed here.
   • Update cnt[0] to 1 (5 % 5 = 0).

4. At nums[3] = 7, the remainder when divided by d is 2.

   • Let's look ahead:
     • For a future k with nums[k] = 11 (next element), we need a remainder x from some previous nums[i] which is (5 - (3 + 11) % 5) % 5 = 1. We don't have any elements that left a remainder of 1 until now (cnt[1] is 0), so no triplet can be formed.
   • Update cnt[2] to 2.

5. Finally, at nums[4] = 11, the remainder is 1.

   • No need to look ahead because 11 is the last element.
   • Update cnt[1] to 1 (11 % 5 = 1).

Since there are no triplets that satisfy (nums[i] + nums[j] + nums[k]) % 5 == 0, our answer thus far is 0.

In this example, we failed to find any valid triplets, but the process demonstrates how we would systematically check for them using the remainders and the cnt hash table to avoid redundancy. In cases where nums contains the right combinations, the cnt table would help us tally up the valid triplet counts quickly and efficiently.

Solution Implementation

Time and Space Complexity

The given Python code defines a method divisibleTripletCount within a Solution class that counts triplets in an array, where the sum of each triplet is divisible by a given integer d. The code primarily involves two nested loops: the outer loop iterates over each element j of the array, and the inner loop iterates over the elements following j (k). Here's a breakdown of the complexities:

Time Complexity:

The time complexity is determined by the number of nested iterations over the input list nums.

• The outer loop runs for n iterations, with n being the length of nums.
• For each iteration of the outer loop, the inner loop executes n - j - 1 times, which results in an average case of n/2 times per iteration of the outer loop.

This creates a total of around n * (n/2) comparisons, simplifying to (n^2)/2 which is in the order of O(n^2). This is because constants are ignored in Big O notation.

Therefore, the time complexity of the code is O(n^2).

Space Complexity:

The space complexity refers to the amount of additional memory used by the program in relation to the input size.

• A defaultdict(int) is created to store counts of nums[j] % d. In the worst case, we would have to store a number for every unique value of nums[j] % d. However, since % d creates d possible remainders, the defaultdict will at most contain d entries.

• We also have a few integer variables (ans, n, x), but these do not scale with input size n, and thus contribute a constant amount to the space complexity.

The space complexity of the code is therefore dictated by the defaultdict size, along with a small constant for the variables used, so it is O(d). However, given that d is a single integer value and not related to the size of the input array nums, the d in the space complexity could be considered a constant factor.

Thus, the space complexity of the code is O(1) if d is considered constant with respect to n. However, since the reference answer suggests that the space complexity is O(n), it seems there might be a presumption that the array might contain up to n distinct values modulo d, binding the space complexity to the length of the array nums. Under this assumption, the space complexity would indeed be O(n).

In conclusion, the time complexity of the code is O(n^2), and the space complexity, depending on the context, is either O(1) or O(n) based on the aforementioned reasoning.

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