Solution Explanation: Number of Single Divisor Triplets

This problem asks us to find the number of triplets (i, j, k) in a given array nums such that nums[i] + nums[j] + nums[k] is divisible by exactly one of nums[i], nums[j], or nums[k].

Approach: Counting and Enumeration

The most straightforward approach is to iterate through all possible triplets and check the divisibility condition. Since the problem constraints limit the values in nums to the range [1, 100], we can optimize by pre-counting the occurrences of each number using a Counter (Python) or a frequency array (other languages).

The algorithm proceeds as follows:

1. Count Frequencies: Count the occurrences of each number in nums. This step takes O(n) time where n is the length of nums.

2. Iterate through Combinations: Iterate through all possible combinations of three numbers (a, b, c) from the range [1, 100]. This nested loop takes O(M^3) time where M=100 is the upper bound of the numbers. For each combination, check if its sum (s = a + b + c) satisfies the single divisor condition.

3. Check Divisibility Condition: For each combination (a, b, c) and their sum s, count how many of a, b, and c divide s. If the count is exactly 1, it's a single divisor triplet.

4. Calculate Count of Triplets: For each valid triplet (a,b,c), we need to calculate how many such triplets exist in the original array nums. This depends on whether a, b, and c are distinct or not. If they are all distinct, the number of triplets is count(a) * count(b) * count(c). If two are the same, we adjust the calculation accordingly (e.g., if a=b, it becomes count(a) * (count(a)-1) * count(c) to avoid counting the same triplet multiple times).

5. Sum Up Triplets: Finally, sum up the counts of all valid triplets.

Time and Space Complexity Analysis

• Time Complexity: The dominant factor in the time complexity is the three nested loops iterating through possible combinations of numbers from 1 to 100. This leads to a time complexity of O(M^3), where M = 100. The initial frequency counting step is O(n), which is less significant than O(M^3).

• Space Complexity: The space complexity is determined primarily by the frequency count array or Counter. This array has a fixed size of 101 (to accommodate numbers from 1 to 100), making the space complexity O(M).

Code Examples (Python, Java, C++, Go, TypeScript)

The code examples provided in the original response are efficient implementations of this approach. They effectively handle the counting, combination generation, divisibility check, and triplet counting steps. The use of a Counter or a frequency array makes the code cleaner and more efficient than directly iterating through nums for each triplet check.

The code in each language follows the same basic algorithm and shares the same time and space complexity. They differ only in syntax and data structure usage. The comments in the code clarify the steps.