C++ || Easy || Simple
Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\no(k)\n\n# Code\n\nclass Solution {\npublic:\n    bool digitCount(string nums) {\n        unordered_
# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\no(k)\n\n# Code\n```\nclass Solution {\npublic:\n    bool digitCount(string nums) {\n        unordered_map<int,int>mp;\n        int n=nums.size();\n        for(int i=0;i<n;i++)\n        {\n            mp[nums[i]-\'0\']++;\n        }\n        for(int i=0;i<n;i++)\n        {\n            if(mp[i]!=nums[i]-\'0\')\n                return 0;\n        }\n        return 1;\n    }\n};\n```
['C++']