✌️C++ Easiest Approach With Best Explanation Beats 100% (Using DP)
Intuition\nThe problem requires finding the minimum number of operations required to convert one string to another. We can perform three types of operations: in
# Intuition\nThe problem requires finding the minimum number of operations required to convert one string to another. We can perform three types of operations: insert a character, delete a character, or replace a character.\n\nTo solve this problem, we can use dynamic programming. We can create a matrix where the rows represent the characters in the first string and the columns represent the characters in the second string. We can fill the matrix using a bottom-up approach.\n\nThe base case is when one of the strings is empty. In this case, the minimum number of operations required is equal to the length of the non-empty string.\n\nFor the general case, we can use the following recurrence relation: if the current characters in the two strings match, the minimum number of operations required is the same as the minimum number of operations required to convert the first i-1 characters in the first string to the first j-1 characters in the second string. Otherwise, we can either insert a character, delete a character, or replace a character. We can take the minimum of these three options and add one to get the minimum number of operations required to convert the first i characters in the first string to the first j characters in the second string.\n\nThe final answer is stored in the bottom-right corner of the matrix.<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nHere\'s the step-by-step approach:\n\n1. Initialize a matrix dp of size (m+1) x (n+1), where m and n are the lengths of word1 and word2, respectively.\n2. Fill in the base cases:\ndp[i][0] = i for i from 0 to m.\ndp[0][j] = j for j from 0 to n.\n3. For each i from 1 to m and each j from 1 to n, compute the value of dp[i][j] as follows:\n4. If word1[i-1] is equal to word2[j-1], then dp[i][j] = dp[i-1][j-1]. Otherwise, dp[i][j] is the minimum of the following three values plus 1:\na) dp[i][j-1], which represents inserting a character in word1.\nb) dp[i-1][j], which represents deleting a character in word1.\nc) dp[i-1][j-1], which represents replacing a character in word1.\n5. Return dp[m][n], which is the minimum number of operations required to convert word1 to word2.\n\n\n\n\n\n# Complexity\n- Time complexity:\nThe time complexity of the solution is O(mn), where m and n are the lengths of the input strings. This is because we need to fill in an (m+1) x (n+1) matrix, and each cell takes constant time to fill.\n\n- Space complexity: O(mn)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    int minDistance(string word1, string word2) {\n        int m = word1.length();\n        int n = word2.length();\n        \n        vector<vector<int>> dp(m+1, vector<int>(n+1));\n        \n        for (int i = 0; i <= m; i++) {\n            dp[i][0] = i;\n        }\n        \n        for (int j = 0; j <= n; j++) {\n            dp[0][j] = j;\n        }\n        \n        for (int i = 1; i <= m; i++) {\n            for (int j = 1; j <= n; j++) {\n                if (word1[i-1] == word2[j-1]) {\n                    dp[i][j] = dp[i-1][j-1];\n                } else {\n                    dp[i][j] = 1 + min({dp[i][j-1], dp[i-1][j], dp[i-1][j-1]});\n                }\n            }\n        }\n        \n        return dp[m][n];\n    }\n};\n//\uD83D\uDC47please upvote if you liked my approach and explanation\n\n```
['C++']