471. Encode String with Shortest Length
Problem:
Given a non-empty string, encode the string such that its encoded length is the shortest.
The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note:
- k will be a positive integer and encoded string will not be empty or have extra space.
- You may assume that the input string contains only lowercase English letters. The string's length is at most 160.
- If an encoding process does not make the string shorter, then do not encode it. If there are several solutions, return any of them is fine.
Example 1:
Input: "aaa"
Output: "aaa"
Explanation: There is no way to encode it such that it is shorter than the input string, so we do not encode it.
Example 2:
Input: "aaaaa"
Output: "5[a]"
Explanation: "5[a]" is shorter than "aaaaa" by 1 character.
Example 3:
Input: "aaaaaaaaaa"
Output: "10[a]"
Explanation: "a9[a]" or "9[a]a" are also valid solutions, both of them have the same length = 5, which is the same as "10[a]".
Example 4:
Input: "aabcaabcd"
Output: "2[aabc]d"
Explanation: "aabc" occurs twice, so one answer can be "2[aabc]d".
Example 5:
Input: "abbbabbbcabbbabbbc"
Output: "2[2[abbb]c]"
Explanation: "abbbabbbc" occurs twice, but "abbbabbbc" can also be encoded to "2[abbb]c", so one answer can be "2[2[abbb]c]".
Solutions:
public class Solution {
public String encode(String s) {
String[][] dp = new String[s.length()][s.length()];
for (int l = 1; l <= s.length(); l ++) {
for (int i = 0; i + l <= s.length(); i ++) {
int j = i + l - 1;
dp[i][j] = s.substring(i, j + 1);
for (int k = i; k < j ; k ++) {
String cand = dp[i][k] + dp[k + 1][j];
if (cand.length() < dp[i][j].length()) {
dp[i][j] = cand;
}
}
String range = s.substring(i, j + 1);
String drange = range + range;
int cut = drange.indexOf(range, 1);
if (cut != -1 && cut < range.length()) {
String cand = range.length() / cut + "[" + dp[i][i + cut - 1] +"]";
if (cand.length() < dp[i][j].length()) {
dp[i][j] = cand;
}
}
}
}
String res = dp[0][dp.length - 1];
return res;
}
}