Easy Solution DP Based C++ 95% fast solution
\n  int minDistance(string word1, string word2) {\n        \n        int m=word1.length(),n=word2.length();\n        \n        int dp[m+1][n+1];\n        \n    
```\n  int minDistance(string word1, string word2) {\n        \n        int m=word1.length(),n=word2.length();\n        \n        int dp[m+1][n+1];\n        \n        for(int i=0;i<=m;i++)\n            dp[i][0]=i;\n        \n        for(int i=0;i<=n;i++)\n            dp[0][i]=i;\n        \n        for(int i=1;i<=m;i++)\n        {\n            for(int j=1;j<=n;j++)\n            {\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], min(dp[i - 1][j], dp[i - 1][j - 1]));\n                }\n            }\n        }\n        return dp[m][n];\n    }\n\n```
['Dynamic Programming', 'C']