C++ || Line sweep || O(N) || Very easy
Approach : Line sweep\n Describe your approach to solving the problem. \n\n# Complexity\n- Time complexity: O(n)\n Add your time complexity here, e.g. O(n) \n\n
# Approach : Line sweep\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    int maximumPopulation(vector<vector<int>>& logs) {\n        int p[101] = {0};\n        \n        for(int i = 0; i<logs.size(); i++){\n            int l = logs[i][0] % 1950;\n            int r = logs[i][1] % 1950;\n\n            p[l]+=1;\n            p[r]-=1;\n        }\n        \n        int indx = 0;\n        for(int i = 1; i<101; i++){\n            p[i] = p[i-1] + p[i];\n            if(p[i] > p[indx]){\n                indx = i;\n            }\n        }\n        return 1950+indx;\n    }\n};\n```
['C++']