Simple Solution Credits: CodeStoryWithMik on youtube
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: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n        int countCompleteSubstrings(string word, int k) {\n        int n = word.length();\n\n        int result = 0;\n        int last = 0;\n\n        for(int i = 1; i <= n; i++) {\n            if(i == n || abs(word[i] - word[i-1]) > 2) { \n                result += solve(last, i-1, word, k); \n                last = i;\n            }\n        }\n\n        return result;\n    }\n\n    int solve(int start,int end,string& word,int k){\n        int subAns=0;\n        for(int uniqueCh = 1; uniqueCh<=26 && uniqueCh*k<=end-start+1;uniqueCh++){\n            int ws = uniqueCh*k;\n            vector<int> count(26);\n            int i=start;\n            int good=0;\n            for(int j=start;j<=end;j++){\n                char ch = word[j];\n                count[ch-\'a\']++;\n                if(count[ch-\'a\']==k)\n                    good++;\n                else if(count[ch-\'a\']==k+1)\n                    good--;\n\n                if(j-i+1>ws){\n                    if(count[word[i]-\'a\']==k+1)\n                        good++;\n                    else if(count[word[i]-\'a\']==k)\n                        good--;\n                    count[word[i]-\'a\']--;\n                    i++;\n                }\n                if(good==uniqueCh)\n                    subAns++;\n            }\n        }\n        return subAns;\n    }\n};\n\n\n\n/*\n\ni   g   i   g   e   e\n                i   j\n\nans = 3\n\nmap\ne -> 2\n\n*/\n```
['C++']