Unique Solution using Set Nlog(N)!!!
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    int findPairs(vector<int>& nums, int k) {\n        set<pair<int, int>> st;\n        set<int> seen;\n        for (int i = 0; i < nums.size(); i++) {\n            if (seen.count(nums[i] - k)) {\n                st.insert(make_pair(nums[i], nums[i]-k));\n            }\n            if (seen.count(nums[i] + k)) {\n                st.insert(make_pair(nums[i]+k, nums[i] ));\n            }\n            seen.insert(nums[i]);\n        }\n        return st.size();\n    }\n};\n\n```
['C++']