Simple C++ solution
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    bool po(vector<int>&prefsum,int j,int i){\n        j++,i++;\n        if(prefsum[j]-prefsum[i]==0) return 1;\n        else return 0;\n    }\n    bool po1(vector<vector<int>>&charsum,int j,int i,int k){\n        j++,i++;\n        for(int l=0;l<26;l++){\n            int freq=charsum[j][l]-charsum[i-1][l];\n            if(freq!=0 && freq!=k) return 0;\n        }\n        return 1;\n    }\n    int countCompleteSubstrings(string word, int k) {\n        int n=word.size();\n        int ans=0;\n        vector<int>prefsum(n+1,0);\n        vector<vector<int>>charsum(n+1,vector<int>(26,0));\n        for(int i=1;i<=n;i++){\n            for(int j=0;j<26;j++) charsum[i][j]+=charsum[i-1][j];\n            charsum[i][word[i-1]-\'a\']++;\n        }\n        for(int i=2;i<=n;i++){\n            prefsum[i]+=prefsum[i-1];\n            if(abs(word[i-1]-word[i-2])>2) prefsum[i]+=1;\n        }\n\n        for(int i=0;i<n;i++){\n            int j=i+k-1;\n            for(int l=0;j<n && l<26;l++,j+=k){\n              if(po(prefsum,j,i)){\n                  if(po1(charsum,j,i,k)) ans++;\n              }\n                   \n            }\n        }\n        return ans;\n\n    }\n};\n```
['C++']