What is the time complexity of String Compression II?

The optimal solution has a time complexity of O(n^2*k).

What is the space complexity of String Compression II?

The space complexity is O(n*k).

What data structures are used to solve String Compression II?

This problem uses String, Dynamic Programming. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving String Compression II?

Insight 1: Dynamic programming is crucial because we need to explore different combinations of character deletions and keep track of the minimum compressed length. Insight 2: The core idea is to consider two options for each character: either delete it or keep it and form a run.

Which companies ask LeetCode 1531. String Compression II in interviews?

This problem is frequently asked at Toptal.

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String Compression II - LeetCode 1531 Solution

String Compression II - Complete Solution Guide

String Compression II is LeetCode problem 1531, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 .

Detailed Explanation

The problem asks us to compress a given string 's' using run-length encoding, but with the added ability to delete at most 'k' characters from the string before compression. The goal is to find the minimum length of the run-length encoded string after performing the optimal deletions. Run-length encoding replaces consecutive identical characters with the character and the count of repetitions (if the count is greater than 1). For example, 'aaabcccd' becomes 'a3bc3d'. We need to determine which characters to delete to minimize the final encoded string's length.

Solution Approach

The solution employs a top-down dynamic programming approach with memoization. A recursive function `solve(start, k_rem)` is defined, where `start` is the starting index of the substring being considered, and `k_rem` is the number of characters that can still be deleted. The function returns the minimum length of the run-length encoded string for the substring starting at `start` after deleting at most `k_rem` characters. The function explores two options for each character: delete it or keep it. If the character is kept, it iterates through the remaining string, forming a run and calculating the cost of encoding that run. The minimum cost among all possible runs is then returned.

Step-by-Step Algorithm

Step 1: Define a recursive function `solve(start, k_rem)` to calculate the minimum length of the compressed string starting from index `start` with `k_rem` deletions remaining.
Step 2: Base Cases: If `k_rem` becomes negative, return infinity (or a large value) as it's an invalid state. If `start + k_rem` is greater than or equal to the length of the string 'n', it means all remaining characters can be deleted, so return 0.
Step 3: Memoization: Check if the result for the current state (start, k_rem) is already cached. If so, return the cached value.
Step 4: Option 1: Delete the character at index `start`. Recursively call `solve(start + 1, k_rem - 1)` and store the result.
Step 5: Option 2: Keep the character at index `start` and form a run. Iterate from `start` to the end of the string, counting the number of identical characters (count) and the number of different characters (deletions).
Step 6: If the number of deletions exceeds `k_rem`, break the inner loop.
Step 7: Calculate the cost of encoding the current run. The cost is 1 if count is 1, 2 if count is between 2 and 9, 3 if count is between 10 and 99, and 4 if count is 100.
Step 8: Recursively call `solve(j + 1, k_rem - deletions)` and add the cost of the current run to the result. Update the minimum result.
Step 9: Store the minimum result in the cache and return it.

Key Insights

Insight 1: Dynamic programming is crucial because we need to explore different combinations of character deletions and keep track of the minimum compressed length.
Insight 2: The core idea is to consider two options for each character: either delete it or keep it and form a run.
Insight 3: Memoization or caching is necessary to avoid redundant calculations, as the same subproblems will be encountered multiple times due to overlapping subproblems.
Insight 4: Carefully track the number of deletions available and the length of the run that is being formed to make the decision on whether to delete or keep the current character.

Complexity Analysis

Time Complexity: O(n^2*k)

Space Complexity: O(n*k)

Topics

This problem involves: String, Dynamic Programming.

Companies

Asked at: Toptal.

String Compression II | LeetCode 1531

Hard

String Dynamic Programming

Time: O(n^2*k)Space: O(n*k)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    private String s;
    private int n;
    private int[][] memo;

    public int getLengthOfOptimalCompression(String s, int k) {
        this.s = s;
        this.n = s.length();
        this.memo = new int[n][k + 1];
        for (int i = 0; i < n; i++) {
            Arrays.fill(memo[i], -1);
        }
        return solve(0, k);
    }

    private int solve(int start, int k_rem) {
        if (k_rem < 0) {
            return Integer.MAX_VALUE / 2; // Avoid overflow
        }

        if (start + k_rem >= n) {
            return 0;
        }

        if (memo[start][k_rem] != -1) {
            return memo[start][k_rem];
        }

        // Option 1: Delete s[start]
        int res = solve(start + 1, k_rem - 1);

        // Option 2: Keep s[start] and form a run
        int count = 0;
        int deletions = 0;
        for (int j = start; j < n; j++) {
            if (s.charAt(j) == s.charAt(start)) {
                count++;
            } else {
                deletions++;
            }

            if (deletions > k_rem) {
                break;
            }

            // Calculate the length of this run's encoding
            int cost = 0;
            if (count == 1) {
                cost = 1;
            } else if (count < 10) {
                cost = 2;
            } else if (count < 100) {
                cost = 3;
            } else { // count == 100
                cost = 4;
            }

            // Update result with the cost of this run plus the optimal
            // solution for the rest of the string
            res = Math.min(res, cost + solve(j + 1, k_rem - deletions));
        }

        memo[start][k_rem] = res;
        return res;
    }
}

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Problem Info

Description

Problem 1531: String Compression II - Detailed explanation and solution approaches.

Topics

String Dynamic Programming

Companies

Toptal

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1531

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Define a recursive function `solve(start, k_rem)` to calculate the minimum length of the compressed string starting from index `start` with `k_rem` deletions remaining.
2
Step 2: Base Cases: If `k_rem` becomes negative, return infinity (or a large value) as it's an invalid state. If `start + k_rem` is greater than or equal to the length of the string 'n', it means all remaining characters can be deleted, so return 0.
3
Step 3: Memoization: Check if the result for the current state (start, k_rem) is already cached. If so, return the cached value.
4
Step 4: Option 1: Delete the character at index `start`. Recursively call `solve(start + 1, k_rem - 1)` and store the result.
5
Step 5: Option 2: Keep the character at index `start` and form a run. Iterate from `start` to the end of the string, counting the number of identical characters (count) and the number of different characters (deletions).
6
Step 6: If the number of deletions exceeds `k_rem`, break the inner loop.
7
Step 7: Calculate the cost of encoding the current run. The cost is 1 if count is 1, 2 if count is between 2 and 9, 3 if count is between 10 and 99, and 4 if count is 100.
8
Step 8: Recursively call `solve(j + 1, k_rem - deletions)` and add the cost of the current run to the result. Update the minimum result.
9
Step 9: Store the minimum result in the cache and return it.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2*k)

💾

Space Complexity

O(n*k)

📝

Why?

The time complexity comes from the nested loops within the recursive function. The outer loop (recursion) can go up to `n * k` times due to the `start` index and `k_rem`. Inside the recursive function, the inner loop runs in `O(n)` time in the worst case for computing the run length. The space complexity is primarily due to the `dp` table that caches the results of the recursive calls.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Bottom-Up Dynamic Programming

Instead of using recursion with memoization, a bottom-up dynamic programming approach can be used. A 2D table `dp[i][j]` would store the minimum compressed length for the substring `s[0...i]` with `j` deletions. The table would be filled iteratively, starting with base cases and building up to the final result.

Time: O(n^2*k)Space: O(n*k)

Trade-offs: Bottom-up DP can sometimes be slightly more efficient in practice due to the absence of recursion overhead. However, the code might be a bit more verbose and harder to understand initially.

✨

Key Insights

1Insight 1: Dynamic programming is crucial because we need to explore different combinations of character deletions and keep track of the minimum compressed length.
2Insight 2: The core idea is to consider two options for each character: either delete it or keep it and form a run.
3Insight 3: Memoization or caching is necessary to avoid redundant calculations, as the same subproblems will be encountered multiple times due to overlapping subproblems.
4Insight 4: Carefully track the number of deletions available and the length of the run that is being formed to make the decision on whether to delete or keep the current character.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: k = 0 (no deletions allowed). The solution correctly handles this by exploring only the 'keep' option, effectively performing run-length encoding on the original string.
Edge case 2: k >= n (all characters can be deleted). The solution handles this by having a base case that returns 0 when the number of characters that can be deleted is greater or equal to the number of remaining characters in the string.
Edge case 3: String with only one character. The solution correctly compresses this to a length of 1, if no deletions are made. If all characters can be deleted, length becomes 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using memoization/caching. This leads to exponential time complexity and TLE (Time Limit Exceeded) errors.

✗Common mistake 2: Incorrectly calculating the cost of encoding a run. Forgetting to consider the number of digits required to represent the count is a common mistake (1 digit for counts 2-9, 2 digits for 10-99, and 3 digits for 100).

✗Common mistake 3: Not handling the edge cases where all characters can be deleted correctly.

✗Common mistake 4: Overflow in calculation when assigning a large value as infinity, especially in languages like Java and C/C++. It is advisable to use Integer.MAX_VALUE / 2 (or similar) to avoid integer overflow issues during minimization.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data compression in image and video formats. Run-length encoding (RLE) is a simple compression technique that can be used as a building block in more complex compression algorithms.

🏦Real-world scenario 2: Text compression and indexing. RLE can be used to compress repetitive sequences of characters in text data, which can improve storage efficiency and indexing performance.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 31, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 29

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "String Compression II"?

For this hard-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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