Formatted question description: https://leetcode.ca/all/1531.html

1531. String Compression II (Hard)

Run-length encoding is a string compression method that works by replacing consecutive identical characters (repeated 2 or more times) with the concatenation of the character and the number marking the count of the characters (length of the run). For example, to compress the string "aabccc" we replace "aa" by "a2" and replace "ccc" by "c3". Thus the compressed string becomes "a2bc3".

Notice that in this problem, we are not adding '1' after single characters.

Given a string s and an integer k. You need to delete at most k characters from s such that the run-length encoded version of s has minimum length.

Find the minimum length of the run-length encoded version of s after deleting at most k characters.

 

Example 1:

Input: s = "aaabcccd", k = 2
Output: 4
Explanation: Compressing s without deleting anything will give us "a3bc3d" of length 6. Deleting any of the characters 'a' or 'c' would at most decrease the length of the compressed string to 5, for instance delete 2 'a' then we will have s = "abcccd" which compressed is abc3d. Therefore, the optimal way is to delete 'b' and 'd', then the compressed version of s will be "a3c3" of length 4.

Example 2:

Input: s = "aabbaa", k = 2
Output: 2
Explanation: If we delete both 'b' characters, the resulting compressed string would be "a4" of length 2.

Example 3:

Input: s = "aaaaaaaaaaa", k = 0
Output: 3
Explanation: Since k is zero, we cannot delete anything. The compressed string is "a11" of length 3.

 

Constraints:

  • 1 <= s.length <= 100
  • 0 <= k <= s.length
  • s contains only lowercase English letters.

Related Topics:
String, Dynamic Programming

Solution 1. DP

Let dp[i][j] be the answer to the sub-problem with s[0..(i-1)] and j deletions.

For dp[i][j], we have two options:

  1. delete s[i - 1]. Then the result is dp[i - 1][j - 1].
  2. keep s[i - 1]. Then we can try to contruct the optimal solution by scanning backward and delete those letters are different from s[i - 1]. Those letters that are same as s[i - 1] will be merged using run-length encoding
    1. We scan t from i to 1.
    2. If s[t - 1] == s[i - 1] then we merge them together; otherwise we delete s[t - 1]. We store the merge count using cnt and the deleted count using del.
    3. We can update dp[i][j] using dp[t - 1][j - del] and 1 + (cnt >= 100 ? 3 : cnt >= 10 ? 2 : cnt >= 2 ? 1 : 0)) which is the number of characters required to encode the merged s[i - 1] section.
// OJ: https://leetcode.com/problems/string-compression-ii/

// Time: O(N^2 * K)
// Space: O(N^2)
int dp[101][101];
class Solution {
public:
    int getLengthOfOptimalCompression(string s, int k) {
        int N = s.size();
        memset(dp, 0x3f, sizeof(dp));
        dp[0][0] = 0;
        for (int i = 1; i <= N; ++i) {
            for (int j = 0; j <= k; ++j) {
                if (j) dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);
                int cnt = 0, del = 0;
                for (int t = i; t > 0; --t) {
                    if (s[t - 1] == s[i - 1]) ++cnt;
                    else ++del;
                    if (j - del < 0) break;
                    dp[i][j] = min(dp[i][j], dp[t - 1][j - del] + 1 + (cnt >= 100 ? 3 : cnt >= 10 ? 2 : cnt >= 2 ? 1 : 0));
                }
            }
        }
        return dp[N][k];
    }
};

Java

class Solution {
    public int getLengthOfOptimalCompression(String s, int k) {
        int length = s.length();
        int[][] dp = new int[length + 1][k + 1];
        for (int i = 0; i <= length; i++)
            Arrays.fill(dp[i], Integer.MAX_VALUE >> 1);
        dp[0][0] = 0;
        for (int i = 1; i <= length; i++) {
            int max = Math.min(i, k);
            for (int j = 0; j <= max; j++) {
                if (j > 0)
                    dp[i][j] = dp[i - 1][j - 1];
                int same = 0, different = 0;
                for (int m = i; m >= 1 && different <= j; m--) {
                    if (s.charAt(m - 1) == s.charAt(i - 1)) {
                        same++;
                        dp[i][j] = Math.min(dp[i][j], dp[m - 1][j - different] + calculateLength(same));
                    } else
                        different++;
                }
            }
        }
        return dp[length][k];
    }

    public int calculateLength(int x) {
        if (x == 1)
            return 1;
        else if (x < 10)
            return 2;
        else if (x < 100)
            return 3;
        else
            return 4;
    }
}

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