Easy Solution using 2 pointer approach C++
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    O(nlogn), O(N^2)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n    O(1)\n\n# Code\n```\nclass Solution {\npublic:\n    int countPairs(vector<int>& nums, int target) {\n        sort(nums.begin(), nums.end());\n        int left = 0, right = nums.size()-1, count=0;\n        while(left < right){\n            if(nums[left]+nums[right] < target){\n                count+=right - left;\n                left++;\n            }\n            else right--;\n        }\n        return count;\n    }\n};\n```\n```\nclass Solution {\npublic:\n    int countPairs(vector<int>& nums, int target) {\n        int count = 0;\n        for(int i=0; i < nums.size()-1; i++){\n            for(int j=i+1; j < nums.size(); j++){\n                if(nums[i]+nums[j] < target) count++;\n            }\n        }\n        return count;\n    }\n};\n```
['C++']