Create number-of-single-divisor-triplets.cpp

Author: kamyu104

C++/number-of-single-divisor-triplets.cpp

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+// Time:  O(d^3), d is the number of distinct nums
+// Space: O(d)
+
+// combinatorics
+class Solution {
+public:
+    long long singleDivisorTriplet(vector<int>& nums) {
+        const auto& check = [](int a, int b, int c) {
+            const int total = a + b + c;
+            return (total % a == 0) + (total % b == 0) + (total % c == 0) == 1;
+        };
+        unordered_map<int, int64_t> cnt;
+        for (const auto& x : nums) {
+            ++cnt[x];
+        }
+        vector<int> unique_nums;
+        for (const auto& [k, _] : cnt) {
+            unique_nums.emplace_back(k);
+        }
+        int64_t result = 0;
+        for (int i = 0; i < size(unique_nums); ++i) {
+            for (int j = i + 1; j < size(unique_nums); ++j) {
+                for (int k = j + 1; k < size(unique_nums); ++k) {
+                    if (check(unique_nums[i], unique_nums[j], unique_nums[k])) {
+                        result += cnt[unique_nums[i]] * cnt[unique_nums[j]] * cnt[unique_nums[k]];
+                    }
+                }
+            }
+        }
+        for (int i = 0; i < size(unique_nums); ++i) {
+            for (int j = 0; j < size(unique_nums); ++j) {
+                if (check(unique_nums[i], unique_nums[i], unique_nums[j])) {
+                    result += cnt[unique_nums[i]] * (cnt[unique_nums[i]] - 1) / 2 * cnt[unique_nums[j]];
+                }
+            }
+        }
+        return result * 6;
+    }
+};