This approach uses dynamic programming to minimize the length of the compressed string after removing up to k characters. We'll iteratively decide for each position in the string whether to remove the character or keep it, and if kept, determine the optimal compression length.
The idea is to use a DP table where dp[i][j] stores the minimal compression length achievable for the substring s[0...i] with j deletions.
Time Complexity: O(n^3). For each starting point, we may need to process every subsequent character and make decisions up to k deletions.
Space Complexity: O(n*k). Space used by the DP table.
1function compressLength(c) {
2 if (c === 0) return 0;
3 if (c === 1) return 1;
4 if (c < 10) return 2;
5 if (c < 100) return 3;
6 return 4;
7}
8
9function minLength(s, k) {
10 let n = s.length;
11 let dp = Array.from({ length: n + 1 }, () => Array(k + 1).fill(Infinity));
12 dp[0][0] = 0;
13 for (let i = 0; i < n; i++) {
14 for (let j = 0; j <= k; j++) {
15 if (dp[i][j] === Infinity) continue;
16 let count = 0;
17 for (let l = i; l < n; l++) {
18 if (s[l] === s[i]) count++;
19 let needDelete = l - i + 1 - count;
20 if (j + needDelete <= k) {
21 dp[l + 1][j + needDelete] = Math.min(dp[l + 1][j + needDelete], dp[i][j] + compressLength(count));
22 }
23 }
24 }
25 }
26 return Math.min(...dp[n]);
27}
28
29// Example usage
30const s = "aaabcccd";
31const k = 2;
32console.log(minLength(s, k));JavaScript employs a similar strategy to the other languages, with utility functions and nested loops to fill the DP table, ultimately computing the minimal encoded length after deleting up to k characters.
In this greedy approach, we identify segments of identical characters and evaluate them individually, considering their entire group in the adjustments. By counting the length of these segments, we can make informed decisions about deletions that do not depend on the widely traversed DP table, reducing some overhead complexity.
Time Complexity: O(n + m log m) where n is the length of the string and m = 26 (constant time, simplified in this case).
Space Complexity: O(m) for frequency arrays.
1import java.util.Arrays;
2
3public class StringCompressionGreedy {
4 public static int minLength(String s, int k) {
5 int[] freq = new int[26];
6 for (char c : s.toCharArray()) freq[c - 'a']++;
7
8 Arrays.sort(freq);
9 for (int i = 25; i >= 0 && k > 0; i--) {
10 if (freq[i] > 0) {
11 int deleteCount = freq[i] - (freq[i] == 1 ? 0 : 1);
12 deleteCount = Math.min(deleteCount, k);
13 freq[i] -= deleteCount;
14 k -= deleteCount;
15 }
16 }
17
18 int length = 0;
19 for (int f : freq) {
20 if (f > 0) {
21 length += f == 1 ? 1 : 1 + (f < 10 ? 2 : f < 100 ? 3 : 4);
22 }
23 }
24
25 return length;
26 }
27
28 public static void main(String[] args) {
29 String s = "aaabcccd";
30 int k = 2;
31 System.out.println(minLength(s, k));
32 }
33}Java implements the greedy method through an array to track character frequencies and sorts them to prioritize reductions on the largest group combinations. This reduces the compressed sequence's length effectively.