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.
1import math
2
3def compress_length(c):
4 if c == 0:
5 return 0
6 if c == 1:
7 return 1
8 if c < 10:
9 return 2
10 if c < 100:
11 return 3
12 return 4
13
14def min_length(s, k):
15 n = len(s)
16 dp = [[math.inf] * (k + 1) for _ in range(n + 1)]
17 dp[0][0] = 0
18 for i in range(n):
19 for j in range(k + 1):
20 if dp[i][j] == math.inf:
21 continue
22 count = 0
23 for l in range(i, n):
24 if s[l] == s[i]:
25 count += 1
26 need_delete = l - i + 1 - count
27 if j + need_delete <= k:
28 dp[l + 1][j + need_delete] = min(dp[l + 1][j + need_delete], dp[i][j] + compress_length(count))
29 return min(dp[n])
30
31if __name__ == "__main__":
32 s = "aaabcccd"
33 k = 2
34 print(min_length(s, k))The Python solution breaks down the problem using a helper function to determine the compression length. It uses a list of lists to maintain the DP table and employs nested loops to fill it according to the defined rules.
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.
1#include <iostream>
2#include <vector>
3#include <algorithm>
4using namespace std;
5
6int minLength(const string& s, int k) {
7 vector<int> freq(26, 0);
8 for (char c : s) freq[c - 'a']++;
9
10 sort(freq.rbegin(), freq.rend());
11 for (int i = 0; i < 26 && k > 0; i++) {
12 if (freq[i] > 0) {
13 int deleteCount = freq[i] - (freq[i] == 1 ? 0 : 1);
14 deleteCount = min(deleteCount, k);
15 freq[i] -= deleteCount;
16 k -= deleteCount;
17 }
18 }
19
20 int length = 0;
21 for (int f : freq) {
22 if (f > 0) {
23 length += f == 1 ? 1 : 1 + (f < 10 ? 2 : f < 100 ? 3 : 4);
24 }
25 }
26
27 return length;
28}
29
30int main() {
31 string s = "aaabcccd";
32 int k = 2;
33 cout << minLength(s, k) << endl;
34 return 0;
35}The C++ program computes minimum length by using a greedy approach on frequency counts. By sorting and then reducing the highest frequencies, we simplify the calculation of the final compressed length. We make use of the STL to sort and calculate efficiently.