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The maximum value of nums.length is very large, but the maximum value of nums[i] is not.
Count the number of times each value appears in nums. Brute force through every possible combination of values and count how many single divisor triplets can be made with that combination of values.
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A brute-force solution checks all possible triplets of indices, which results in O(n^3) time complexity. This becomes slow as the array grows, whereas counting value combinations significantly reduces the number of checks.
Problems with similar mathematical counting and frequency optimization techniques are common in FAANG interviews. While this exact problem may vary, the underlying concepts of combinatorics and modular checks are frequently tested.
A frequency array or hash map works best for this problem. It allows efficient counting of how many times each value appears and helps compute the number of valid triplet permutations without checking every index combination.
The optimal approach uses frequency counting to avoid iterating over every index triplet. Instead, iterate over distinct values and evaluate the divisibility conditions while using their frequencies to compute the number of valid permutations.