Clean Codes🔥🔥|| Full Explanation✅|| Dynamic Programming✅|| C++|| Java|| Python3
Intuition :\n- Here we have to find the minimum edit distance problem between two strings word1 and word2. \n- The minimum edit distance is defined as the minim
# Intuition :\n- Here we have to find the minimum edit distance problem between two strings word1 and word2. \n- The minimum edit distance is defined as the minimum number of operations required to transform one string into another.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach :\n- The approach here that I am using is dynamic programming. The idea is to build a 2D matrix dp where `dp[i][j] `represents the minimum number of operations required to transform the substring `word1[0...i-1]` into the substring `word2[0...j-1].`\n# How is Matrix built :\n- The matrix is built iteratively using the following recurrence relation:\n1. If `word1[i-1] == word2[j-1]`, then `dp[i][j] = dp[i-1][j-1]`. That is, no operation is required because the characters at positions `i-1` and `j-1` are already the same.\n2. Otherwise, `dp[i][j]` is the minimum of the following three values:\n- `dp[i-1][j-1] + 1`: replace the character at position `i-1` in `word1` with the character at position `j-1` in` word2`.\n- `dp[i-1][j] + 1`: delete the character at position `i-1` in `word1.`\n- `dp[i][j-1] + 1`: insert the character at position `j-1` in `word2` into `word1` at position `i`.\n# The base cases are:\n- `dp[i][0] = i`: transforming `word1[0...i-1]` into an empty string requires `i` deletions.\n- `dp[0][j] = j`: transforming an empty string into `word2[0...j-1] `requires `j` insertions.\n<!-- Describe your approach to solving the problem. -->\n# Final Step :\n- Finally, return `dp[m][n]`, which represents the minimum number of operations required to transform `word1 `into `word2`, where `m` is the length of `word1` and `n` is the length of `word2`.\n\n# Complexity\n- Time complexity : O(mn)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity : O(mn)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Please Upvote\uD83D\uDC4D\uD83D\uDC4D\n```\nThanks for visiting my solution.\uD83D\uDE0A\n```\n# Codes [C++ |Java |Python3] : With Comments\n```C++ []\nclass Solution {\n public:\n  int minDistance(string word1, string word2) {\n    const int m = word1.length();//first word length\n    const int n = word2.length();//second word length\n    // dp[i][j] := min # of operations to convert word1[0..i) to word2[0..j)\n    vector<vector<int>> dp(m + 1, vector<int>(n + 1));\n\n    for (int i = 1; i <= m; ++i)\n      dp[i][0] = i;\n\n    for (int j = 1; j <= n; ++j)\n      dp[0][j] = j;\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])//same characters\n          dp[i][j] = dp[i - 1][j - 1];//no operation\n        else\n          dp[i][j] = min({dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]}) + 1;\n                             //replace       //delete        //insert\n    return dp[m][n];\n  }\n};\n```\n```Java []\nclass Solution {\n  public int minDistance(String word1, String word2) {\n    final int m = word1.length();//first word length\n    final int n = word2.length();///second word length\n    // dp[i][j] := min # of operations to convert word1[0..i) to word2[0..j)\n    int[][] dp = new int[m + 1][n + 1];\n\n    for (int i = 1; i <= m; ++i)\n      dp[i][0] = i;\n\n    for (int j = 1; j <= n; ++j)\n      dp[0][j] = j;\n\n    for (int i = 1; i <= m; ++i)\n      for (int j = 1; j <= n; ++j)\n        if (word1.charAt(i - 1) == word2.charAt(j - 1))//same characters\n          dp[i][j] = dp[i - 1][j - 1];//no operation\n        else\n          dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1])) + 1;                      //replace               //delete        //insert\n\n    return dp[m][n];\n  }\n}\n\n```\n```Python []\nclass Solution:\n  def minDistance(self, word1: str, word2: str) -> int:\n    m = len(word1)\n    n = len(word2)\n    # dp[i][j] := min # Of operations to convert word1[0..i) to word2[0..j)\n    dp = [[0] * (n + 1) for _ in range(m + 1)]\n\n    for i in range(1, m + 1):\n      dp[i][0] = i\n\n    for j in range(1, n + 1):\n      dp[0][j] = j\n\n    for i in range(1, m + 1):\n      for j in range(1, n + 1):\n        if word1[i - 1] == word2[j - 1]:\n          dp[i][j] = dp[i - 1][j - 1]\n        else:\n          dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1\n\n    return dp[m][n]\n```\n# Please Upvote\uD83D\uDC4D\uD83D\uDC4D\n![ezgif-3-22a360561c.gif](https://assets.leetcode.com/users/images/399bb888-6ae6-4614-a338-fe33088c118d_1677378017.7506878.gif)\n\n
['Dynamic Programming', 'Python', 'C++', 'Java', 'Python3']