Formatted question description: https://leetcode.ca/all/471.html

# 471. Encode String with Shortest Length (Hard)

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:

1. k will be a positive integer and encoded string will not be empty or have extra space.
2. You may assume that the input string contains only lowercase English letters. The string's length is at most 160.
3. 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]".


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Related Topics:
Dynamic Programming

Similar Questions:

## TLE Solution

The TLE case "slkjdfbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb".

Why not simply traverse the string and merge substring of the same letters? Because for this case "aaabaaaaaabaaa", the result should be "2[aaabaaa]". If we merge it to "aaab6[a]baaa", we can’t get the right answer.

// OJ: https://leetcode.com/problems/encode-string-with-shortest-length/

// Time: O(N^3)
// Space: O(N)
class Solution {
private:
bool compare(string &s, int s1, int s2, int len) {
if (s1 + len > s.size() || s2 + len > s.size()) return false;
int i = 0;
for (; i < len && s[s1 + i] == s[s2 + i]; ++i);
return i == len;
}
public:
string encode(string s) {
if (s.size() <= 4) return s;
string ans = s;
for (int len = s.size() / 2; len >= 1; --len) {
for (int i = 0; i + len <= s.size(); ++i) {
int cnt = 1;
while (compare(s, i, i + cnt * len, len)) ++cnt;
if (cnt == 1) continue;
string prefix = i > 0 ? encode(s.substr(0, i)) : "";
string suffix = i + cnt * len < s.size() ? encode(s.substr(i + cnt * len)) : "";
string middle = to_string(cnt) + "[" + encode(s.substr(i, len)) + "]";
string encoded = prefix + middle + suffix;
if (ans == "" || encoded.size() < ans.size()) ans = encoded;
}
}
return ans;
}
};


Java

class Solution {
public String encode(String s) {
int length = s.length();
String[][] dp = new String[length][length];
for (int i = length - 1; i >= 0; i--) {
for (int j = i; j < length; j++) {
int curLength = j - i + 1;
String substring = s.substring(i, j + 1);
if (curLength <= 4)
dp[i][j] = substring;
else {
int subLength = (substring + substring).indexOf(substring, 1);
if (subLength < curLength)
dp[i][j] = String.valueOf(curLength / subLength) + "[" + dp[i][i + subLength - 1] + "]";
else {
dp[i][j] = substring;
for (int k = i; k < j; k++) {
if ((dp[i][k] + dp[k + 1][j]).length() < dp[i][j].length())
dp[i][j] = dp[i][k] + dp[k + 1][j];
}
}
}
}
}
return dp[length - 1];
}
}