Easy cpp;
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 mod=1e9+7;\n    \n    int countWays(vector<vector<int>>& ranges) {\n        // Check if the input vector is empty.\n          int ways = 1;\n        if (ranges.empty()) {\n            return 0;\n        }\n\n        // Sort the ranges by their start points.\n        sort(ranges.begin(), ranges.end());\n\n        // Initialize the variables.\n        \n        int start = ranges[0][0];\n        int end = ranges[0][1];\n     \n        // Iterate over the ranges.\n        for (int i = 1; i < ranges.size(); i++) {\n            // If the current range overlaps with the previous range,\n            // update the end point.\n            if (end >= ranges[i][0]) {\n                end = max(end, ranges[i][1]);\n            } else {\n                // Otherwise, we need to start a new range.\n                ways++;\n                end = ranges[i][1];\n            }\n        }\n\n        // Calculate the total number of ways to cover all intervals.\n     long long res = 1 ;\n\n        for( int  i = 1 ;i <= ways ; i++ ){\n\n            res = ( res * 2 ) % mod ;\n\n        }\n        \n        \n        return res % mod ;\n\n    }\n};\n// Now, we have to distribute every unoverlapped interval in 2 groups i.e. either in group 1 or in group 2 and \n// we can do that by\n//  multiplying the answer with 2 every single we get an unoverlapped interval.\n\n/*\nclass Solution {\npublic:\n    \n    int countWays(vector<vector<int>>&v) {\n        long long i=0,n=v.size(),mod=1e9+7;\n        sort(v.begin(),v.end());\n        int mergesize=1; // size=1 because we started to iterate from i=1 i.e. second interval.\n        long long ans=2; // distributing the first interval in 2 ways\n        //int first = v[0][0];\n        int second = v[0][1];\n        for(i=1;i<n;i++){\n            if(v[i][0]<=second)second = max(second,v[i][1]);\n            else{\n                mergesize++;\n                //first=v[i][0]; (incase,if u want to store the merged intervals)\n                second=v[i][1];\n                ans=(ans*2)%mod;  // distributing the current unoverlapped interval we got either in group 1 or in group 2.\n            }\n        }\n        return ans;\n    }\n};\n*/\n```
['C++']