How do you solve String Compression II on LeetCode?

String Compression II (LeetCode #1531) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n² * k) time complexity and O(n * k) space complexity. Key insight: Understanding run-length encoding is crucial for solving this problem.

What is the brute force solution for String Compression II?

The brute force approach for String Compression II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible strings by deleting up to k characters and then calculating the run-length encoding for each string. This method is straightforward but inefficient due to the exponential number of combinations. The optimal Optimal Solution approach improves this to O(n² * k).

What algorithm or data structure is used to solve String Compression II?

String Compression II uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for String Compression II?

Clearly explain your thought process and how you arrived at each step. Be prepared to discuss the trade-offs between brute-force and optimal solutions. Practice similar problems to become comfortable with dynamic programming concepts.

What are common mistakes when solving String Compression II?

Failing to consider the impact of deletions on the run-length encoding. Not properly initializing the DP table or misunderstanding its dimensions.

#1531

String Compression II

Hard

StringDynamic ProgrammingDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n² * k)
Space	O(1)	O(n * k)

💡

Intuition

Time O(n² * k)Space O(n * k)

The optimal solution uses dynamic programming to efficiently compute the minimum length of the run-length encoded string after deleting up to k characters. It builds on the idea of keeping track of the current index and the remaining deletions, leveraging previously computed results to avoid redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP table where dp[i][j] represents the minimum compressed length of the substring s[0:i] with j deletions allowed.
2Step 2: For each character in the string, calculate the compressed length considering the current character and the previous runs.
3Step 3: Update the DP table based on whether to delete characters or keep them, ensuring to consider the run-length encoding.

solution.py28 lines

1def get_compressed_length(s, start, end):
2    length = 0
3    count = 0
4    for i in range(start, end + 1):
5        count += 1
6        if i == end or s[i] != s[i + 1]:
7            length += 1 + (len(str(count)) if count > 1 else 0)
8            count = 0
9    return length
10
11def compress_string(s, k):
12    n = len(s)
13    dp = [[float('inf')] * (k + 1) for _ in range(n + 1)]
14    dp[0][0] = 0
15    for i in range(1, n + 1):
16        for j in range(k + 1):
17            dp[i][j] = dp[i - 1][j] + 1
18            count = 0
19            for m in range(i - 1, -1, -1):
20                count += 1
21                if m < i - 1 and s[m] != s[i - 1]:
22                    break
23                if j - (count - 1) >= 0:
24                    dp[i][j] = min(dp[i][j], dp[m][j - (count - 1)] + get_compressed_length(s, m, i - 1))
25    return dp[n][k]
26
27# Example usage:
28print(compress_string('aaabcccd', 2))  # Output: 4

ℹ

Complexity note: The time complexity is O(n² * k) because we iterate through the string and for each character, we may consider all previous characters, while also considering up to k deletions.

1Understanding run-length encoding is crucial for solving this problem.
2Dynamic programming can significantly reduce the complexity by avoiding redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.