@shawngao said in Java DP Solution, same as Edit Distance:

Math.min(Math.min(dp[i - 1][j - 1] + 2, dp[i - 1][j] + 1), dp[i][j - 1] + 1);

Nice solution. Since we're permitted to operate only one character at a time, checking to delete both characters is not required, although it produces same result since you add 2.

dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + 1;