# Concise JAVA DP solution with comments

• ``````public int minDistance(String word1, String word2) {
// dp[i][j] : minimum steps to convert i long word1 and j long word2
int dp[][] = new int[word1.length() + 1][word2.length() + 1];

for (int i = 0; i <= word1.length(); i++) dp[i][0] = i;
for (int j = 0; j <= word2.length(); j++) dp[0][j] = j;

for (int i = 1;i <= word1.length(); i++) {
for (int j = 1; j<= word2.length(); j++) {
if (word1.charAt(i-1) == word2.charAt(j-1))// <--
dp[i][j] = dp[i-1][j-1];
else
// dp[i-1][j-1] : replace word1(i) with word2(j), because word1(0, i-1) == word2(0, j-1);
// dp[i  ][j-1] : delete word(j)
// dp[i-1][j  ] : delete word(i), because word1(0, i-1) == word2(0, j)
dp[i][j] = Math.min(dp[i-1][j-1], Math.min(dp[i][j-1], dp[i-1][j])) + 1;
}
}
return dp[word1.length()][word2.length()];
}``````

• Clear code and good comments! Nice work!

• Your comments make your solution much easier to understand. 1000 times better than those pure coding without any explanation. Thanks !

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