try it--beat 90%(upvote if useful)
Intuition\nThe problem is to find the count of all valid substrings of word1 that contain at least the characters in word2. We can solve this efficiently using 
Intuition\nThe problem is to find the count of all valid substrings of word1 that contain at least the characters in word2. We can solve this efficiently using a sliding window approach, where we track the characters in the current window of word1 and compare it with the character frequencies of word2. The idea is to keep expanding the window until it contains all the characters from word2, and then count all valid substrings starting from the current window.\n\nApproach\nPreprocessing word2:\nWe create an array word2Count to store the frequency of each character in word2. This helps us know how many times each character needs to appear in a valid substring.\nSliding window on word1:\nWe maintain a sliding window over word1 using two pointers, start and end.\nWe also maintain an array windowCount to track the frequency of characters in the current window.\nWe expand the window by moving the end pointer, adding characters to windowCount. For each character, if the count matches its required frequency in word2, we increase the matches count.\nCheck if the window is valid:\nA valid window is one where all characters in word2 have been matched (i.e., matches == required).\nOnce the window is valid, all substrings starting from start to end are valid because any longer substring will still contain the characters of word2.\nShrink the window:\nAfter counting the valid substrings for the current window, we move the start pointer to shrink the window and try to maintain validity by updating windowCount and reducing the matches count if necessary.\nComplexity\nTime complexity:\nThe time complexity is O(n), where n is the length of word1. The sliding window approach ensures that each character is processed at most twice (once when expanding the window and once when shrinking it).\n\nSpace complexity:\nThe space complexity is O(1) since the space used for word2Count and windowCount is constant (both arrays of size 26).\n\n# Code\n```cpp []\nclass Solution {\npublic:\n    long long validSubstringCount(string word1, string word2) {\n        vector<int> word2Count(26, 0);\n        for (char c : word2) {\n            word2Count[c - \'a\']++;\n        }\n\n        int word2Length = word2.size();\n        long long result = 0;\n        int required = 0;\n        vector<int> windowCount(26, 0);\n\n        // Count how many unique characters in word2 need to be satisfied\n        for (int count : word2Count) {\n            if (count > 0) required++;\n        }\n\n        int start = 0, matches = 0;\n\n        // Sliding window over word1\n        for (int end = 0; end < word1.size(); end++) {\n            // Add current character to the window\n            char c = word1[end];\n            windowCount[c - \'a\']++;\n\n            // Check if the current character satisfies the requirement from word2\n            if (windowCount[c - \'a\'] == word2Count[c - \'a\']) {\n                matches++;\n            }\n\n            // Shrink the window until it is valid\n            while (matches == required) {\n                // Once the window size is large enough, all substrings starting from `start` to `end` are valid\n                result += word1.size() - end;\n\n                // Shrink the window from the left\n                char leftChar = word1[start];\n                windowCount[leftChar - \'a\']--;\n\n                // If the character no longer satisfies the word2 requirement, decrease match count\n                if (windowCount[leftChar - \'a\'] < word2Count[leftChar - \'a\']) {\n                    matches--;\n                }\n                start++;\n            }\n        }\n\n        return result;\n    }\n};\n\n```\n
['C++']