EASY IMPLEMENTATION
*Bold# Intuition***\nThe goal is to match the characters of two strings, word1 and word2, while being allowed to replace exactly one character in word1. My init
# *****Bold**# Intuition***\nThe goal is to match the characters of two strings, word1 and word2, while being allowed to replace exactly one character in word1. My initial thought is to use a combination of greedy matching and suffix matching:\n\nGreedy Matching: We try to match characters from left to right between the two strings, keeping track of any mismatches.\nSuffix Matching: We precompute a "suffix match count" array for word1 to help us decide if the remaining part of word2 can match a suffix of word1. This helps in identifying whether a replacement in word1 would make the remaining characters match with word2.\n\n# Approach\nSuffix Array: We create a suffixcnt array where each index i stores how many characters of word2 can match a suffix of word1 starting from index i. We compute this by iterating from the end of both strings and counting matching characters.\n\nGreedy Iteration: Starting from the beginning of both strings, we try to match the characters of word1 with word2. When a mismatch occurs:\n\nWe check if replacing the current character in word1 will make the rest of the string match word2 using the suffixcnt array.\nWe ensure only one replacement happens by maintaining a flag.\nResult Construction: We store the indices in word1 where matches or replacements happen. If the entire word2 can be matched, we return these indices; otherwise, we return an empty result.\n\n# Complexity\n- Time complexity:\nThe time complexity of this approach is o(n)\n\nO(n), where n is the length of word1.\nComputing the suffixcnt array takes linear time, i.e.,\nO(n).\nThe greedy iteration through word1 to find the matching or replacement indices also takes\nO(n) time.\n\n- Space complexity:\nThe space complexity is\nO(n).\nWe use a suffixcnt array of size n and a result vector that at most contains m elements (but typically not more than n).\n\n# Code\n```cpp []\nclass Solution {\npublic:\n    vector<int> validSequence(string word1, string word2) {\n\n          int n = word1.length();\n         int m = word2.length();\n        vector<int>suffixcnt(n,0);\n        int i =n-1;\n        int j =m-1;\n        while(i >=0 && j >=0){\n            if(word1[i] == word2[j]){\n               suffixcnt[i] = (m-j); // here m = 3 its size BUT its index would be j =2 as its is zero based indexing \n                j--;\n                i--;\n            }else{\n                i--;\n            }\n        }\n        for(int i = n-2; i>=0; i--){\n            suffixcnt[i] = max(suffixcnt[i], suffixcnt[i+1]);\n        }\n// now iterating from the begining to check the possibility of word replacement if its next \n// word of the exist if not dont replace simplt move forward but if it does store it index\n// in the result vector as its final answer;\n\n        vector<int> res;\n        i =0;\n        j=0;\n        int flag =0;\n        while(i<n && j<m){\n            if(word1[i] == word2[j]){\n                res.push_back(i);\n                i++;\n                j++;\n            }\n            else{\n              // here check if next to the current string does the left word of the word2 exist\n              // if it does so in question it is given that we can replace exactly one \n             // character in word1 to make it identical\n             int remainingLength = (m-j-1); // check the remaining length of the word2 left\n             // does the left word will exist upcoming if yes proceed and change the char\n             if ( i+1 < n && suffixcnt[i + 1] >= remainingLength && flag == 0) {\n                res.push_back(i); // Mark the index where replacement happens\n                i++;\n                j++;\n                flag = 1; // Set flag to indicate a replacement has been made\n            } else {\n                i++; // Move forward in word1 if replacement isn\'t possible\n            }\n        }\n        }\n        if(res.size() == m)\n        return res;\n        else\n        return{};\n\n    }\n};\n```
['C++']