Concise JAVA DP solution with comments
public int minDistance(String word1, String word2) {\n    \t// dp[i][j] : minimum steps to convert i long word1 and j long word2\n    \tint dp[][] = new int[wor
    public int minDistance(String word1, String word2) {\n    \t// dp[i][j] : minimum steps to convert i long word1 and j long word2\n    \tint dp[][] = new int[word1.length() + 1][word2.length() + 1];\n    \n    \tfor (int i = 0; i <= word1.length(); i++) dp[i][0] = i;    \t\n    \tfor (int j = 0; j <= word2.length(); j++) dp[0][j] = j; \n    \t \n    \tfor (int i = 1;i <= word1.length(); i++) {\n    \t\tfor (int j = 1; j<= word2.length(); j++) {\n    \t\t\tif (word1.charAt(i-1) == word2.charAt(j-1))// <--\n    \t\t\t\tdp[i][j] = dp[i-1][j-1];\n    \t\t\telse \n                    // dp[i-1][j-1] : replace word1(i) with word2(j), because word1(0, i-1) == word2(0, j-1);\n                    // dp[i  ][j-1] : delete word(j)\n                    // dp[i-1][j  ] : delete word(i), because word1(0, i-1) == word2(0, j)\n    \t\t\t\tdp[i][j] = Math.min(dp[i-1][j-1], Math.min(dp[i][j-1], dp[i-1][j])) + 1; \n    \t\t}\n    \t}\n    \treturn dp[word1.length()][word2.length()];\n    }
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