Given an array of n integers arr, and a sum k, you need to return count of triplets with the sum k(Duplicate triplets are allowed).\n```\narr[] : {1,3,2,4,1,3,4,2}, k = 5\nOutput: 4 //{1,2,2}, {2,1,2}, {1,3,1}, {1,1,3}\nabove 4 Outputs does not include permutations, its 4 times because every time we are considering different index of elements. (1,2 and 3 appear twice, so considering different indices everytime, we get output 4)\n```\n\nI tried 2Sum Approach using HashTable but I am a bit confused when should I insert in HashTable, because the 2Sum Approach I follow is:\n\n```\nint twoSums(int nums[], int n, int k){\n\tint count = 0;\n\tunordered_set<int> hashSet;\n\tfor(int i = 0; i < n;i++){\n\t\t//if Sum - currElement is present in hashset\n\t\tif(hashSet.find(k - nums[i]) != hashSet.end())\n\t\t\tcount++;\n\t\t//Once this element is checked, now add to hashSet\n\t\thashSet.insert(nums[i]);\n\t}\n\treturn  count;\n}\n```\n\n\n

Given an array of n integers arr, and a sum k, you need to return count of triplets with the sum k(Duplicate triplets are allowed).\n\narr[] : {1,3,2,4,1,3,4,2}