Pigeon Hole Principle
Intuition\n Suppose we have m pigeons and n holes where m>n and we want to put m pigeons into n holes, then atleast one hole will have more than one pigeon. \n 
# Intuition\n* Suppose we have m pigeons and n holes where m>n and we want to put m pigeons into n holes, then atleast one hole will have more than one pigeon. \n* Similarly, if we have a number with frequency > 2 and we want to put this number into one of the 2 arrays then atleast one array will have the duplicate number.\n\n\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    bool isPossibleToSplit(vector<int>& nums) {\n        int n = nums.size();\n        unordered_map<int,int> mp;\n        \n        for(auto i:nums){\n            mp[i]++;\n            if(mp[i]>2) return false;\n        }\n        \n        return true;\n        \n    }\n};\n```
['C++']