What is the time complexity of Maximize the Number of Partitions After Operations?

The optimal solution has a time complexity of O(n*26*2^26).

What data structures are used to solve Maximize the Number of Partitions After Operations?

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

What are the key insights for solving Maximize the Number of Partitions After Operations?

Insight 1: Dynamic Programming: Since we want to find the optimal solution by making a decision at each position of the string, DP is a natural fit. Insight 2: Bitmasking: Since `k` is at most 26 (number of lowercase English letters), we can efficiently represent the set of distinct characters encountered so far using a bitmask.

Which companies ask LeetCode 3003. Maximize the Number of Partitions After Operations in interviews?

This problem is frequently asked at HiLabs, ThoughtWorks.

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Maximize the Number of Partitions After Operations - LeetCode 3003 Solution

Q: What is the space complexity of Maximize the Number of Partitions After Operations?

The space complexity is O(n*2^26).

Maximize the Number of Partitions After Operations - Complete Solution Guide

Maximize the Number of Partitions After Operations is LeetCode problem 3003, 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

You are given a string s and an integer k . First, you are allowed to change at most one index in s to another lowercase English letter. After that, do the following partitioning operation until s is empty : Choose the longest prefix of s containing at most k distinct characters. Delete the prefix from s and increase the number of partitions by one. The remaining characters (if any) in s maintain their initial order. Return an integer denoting the maximum number of resulting partitions after the

Detailed Explanation

The problem asks us to find the maximum number of partitions we can create from a string `s` after making at most one change to a character in `s`. A partition is formed by repeatedly taking the longest prefix of the string that contains at most `k` distinct characters and removing it. We need to determine the optimal index to potentially change (or not change any) to maximize the number of such partitions.

Solution Approach

The provided code uses a top-down dynamic programming approach with memoization to solve this problem. It iterates through the string, keeping track of the distinct characters encountered so far using a bitmask. At each index, it explores two possibilities: either keep the current character or change it to another lowercase English letter (if allowed). The DP function `dp(i, current_mask, can_change)` calculates the maximum number of partitions possible from index `i` onwards, given the current bitmask `current_mask` of distinct characters and a boolean flag `can_change` indicating whether a change is still allowed. The base case is when the index `i` reaches the end of the string, in which case, it returns 1 (representing the last partition). The result of each subproblem is stored in the memoization table `memo` to avoid redundant calculations.

Step-by-Step Algorithm

Step 1: Define a recursive DP function `dp(i, current_mask, can_change)` where `i` is the current index in the string, `current_mask` is a bitmask representing the distinct characters seen so far, and `can_change` indicates if we can still change a character.
Step 2: Base Case: If `i` reaches the end of the string (`i == n`), return 1, indicating one partition.
Step 3: Memoization Check: Check if the result for the current state `(i, current_mask, can_change)` is already stored in the `memo` table. If so, return the stored value.
Step 4: Option 1 (Don't Change): Calculate the new bitmask `new_mask` by adding the current character `s[i]` to `current_mask`.
Step 5: If the number of distinct characters in `new_mask` exceeds `k`, increment the number of partitions by 1 and recursively call `dp` with the next index `i + 1` and a bitmask containing only the current character's bit set to 1. Pass on the can_change parameter. Otherwise, recursively call `dp` with the next index `i + 1`, `new_mask`, and `can_change`.
Step 6: Option 2 (Change, if allowed): If `can_change` is true, iterate through all 26 possible lowercase English letters. For each letter, calculate the `changed_mask` by adding that letter to the `current_mask`.
Step 7: If the number of distinct characters in the `changed_mask` exceeds `k`, increment the number of partitions by 1 and recursively call `dp` with the next index `i + 1`, a bitmask containing only the new character's bit set to 1, and `can_change` set to false. Otherwise, recursively call `dp` with the next index `i + 1`, `changed_mask`, and `can_change` set to false.
Step 8: Store the maximum number of partitions obtained from either option in the `memo` table for the current state `(i, current_mask, can_change)`.
Step 9: Return the stored result.

Key Insights

Insight 1: Dynamic Programming: Since we want to find the optimal solution by making a decision at each position of the string, DP is a natural fit.
Insight 2: Bitmasking: Since `k` is at most 26 (number of lowercase English letters), we can efficiently represent the set of distinct characters encountered so far using a bitmask.
Insight 3: Memoization: Overlapping subproblems exist, so memoization is crucial for optimizing the DP solution's runtime. Without memoization, the repeated calculations would lead to exponential time complexity.
Insight 4: Only one change matters. We either make it or we don't. This single change is at the very beginning and the boolean `can_change` will record whether we have performed the change or not.

Complexity Analysis

Time Complexity: O(n*26*2^26)

Space Complexity: O(n*2^26)

Topics

This problem involves: String, Dynamic Programming, Bit Manipulation, Bitmask.

Companies

Asked at: HiLabs, ThoughtWorks.

Maximize the Number of Partitions After Operations | LeetCode 3003

Hard

String Dynamic Programming Bit Manipulation Bitmask

Time: O(n*26*2^26)Space: O(n*2^26)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int maxPartitionsAfterOperations(String s, int k) {
        int n = s.length();
        Map<String, Integer> memo = new HashMap<>();

        return dp(0, 0, true, s, k, n, memo);
    }

    private int dp(int i, int currentMask, boolean canChange, String s, int k, int n, Map<String, Integer> memo) {
        String state = i + "," + currentMask + "," + canChange;
        if (i == n) {
            return 1;
        }
        if (memo.containsKey(state)) {
            return memo.get(state);
        }

        // Option 1: Don't change s[i]
        int charOrd = s.charAt(i) - 'a';
        int newMask = currentMask | (1 << charOrd);

        int res = 0;
        if (Integer.bitCount(newMask) > k) {
            res = 1 + dp(i + 1, (1 << charOrd), canChange, s, k, n, memo);
        } else {
            res = dp(i + 1, newMask, canChange, s, k, n, memo);
        }

        // Option 2: Change s[i], if allowed
        if (canChange) {
            for (int newCharOrd = 0; newCharOrd < 26; newCharOrd++) {
                int changedMask = currentMask | (1 << newCharOrd);
                if (Integer.bitCount(changedMask) > k) {
                    res = Math.max(res, 1 + dp(i + 1, (1 << newCharOrd), false, s, k, n, memo));
                } else {
                    res = Math.max(res, dp(i + 1, changedMask, false, s, k, n, memo));
                }
            }
        }

        memo.put(state, res);
        return res;
    }
}

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Problem Info

Description

Problem 3003: Maximize the Number of Partitions After Operations - Detailed explanation and solution approaches.

Topics

String Dynamic Programming Bit Manipulation Bitmask

Companies

HiLabsThoughtWorks

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:3003

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 DP function `dp(i, current_mask, can_change)` where `i` is the current index in the string, `current_mask` is a bitmask representing the distinct characters seen so far, and `can_change` indicates if we can still change a character.
2
Step 2: Base Case: If `i` reaches the end of the string (`i == n`), return 1, indicating one partition.
3
Step 3: Memoization Check: Check if the result for the current state `(i, current_mask, can_change)` is already stored in the `memo` table. If so, return the stored value.
4
Step 4: Option 1 (Don't Change): Calculate the new bitmask `new_mask` by adding the current character `s[i]` to `current_mask`.
5
Step 5: If the number of distinct characters in `new_mask` exceeds `k`, increment the number of partitions by 1 and recursively call `dp` with the next index `i + 1` and a bitmask containing only the current character's bit set to 1. Pass on the can_change parameter. Otherwise, recursively call `dp` with the next index `i + 1`, `new_mask`, and `can_change`.
6
Step 6: Option 2 (Change, if allowed): If `can_change` is true, iterate through all 26 possible lowercase English letters. For each letter, calculate the `changed_mask` by adding that letter to the `current_mask`.
7
Step 7: If the number of distinct characters in the `changed_mask` exceeds `k`, increment the number of partitions by 1 and recursively call `dp` with the next index `i + 1`, a bitmask containing only the new character's bit set to 1, and `can_change` set to false. Otherwise, recursively call `dp` with the next index `i + 1`, `changed_mask`, and `can_change` set to false.
8
Step 8: Store the maximum number of partitions obtained from either option in the `memo` table for the current state `(i, current_mask, can_change)`.
9
Step 9: Return the stored result.

💡

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*26*2^26)

💾

Space Complexity

O(n*2^26)

📝

Why?

The DP function `dp(i, current_mask, can_change)` has three parameters: `i` (index of the string, up to `n`), `current_mask` (26 bits, representing 2^26 possible combinations), and `can_change` (boolean). In the worst case, the `dp` function will be called for all possible combinations of `i`, `current_mask`, and `can_change`. Inside `dp`, we iterate through 26 possible characters if `can_change` is true. Hence, the time complexity is O(n * 26 * 2^26). The space complexity is dominated by the `memo` table, which stores the results for all possible states. Therefore, the space complexity is O(n * 2^26).

📖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

Greedy Approach

A greedy approach might involve trying all possible character changes and calculating the number of partitions for each changed string and the original. However, it is difficult to prove optimality with a simple greedy strategy.

Time: O(26*n)Space: O(1)

Trade-offs: A greedy approach is likely to be much faster, but it's hard to ensure optimality. It might be suitable as an initial attempt but would likely fail on more complex test cases. Dynamic programming provides the correct solution at the cost of greater complexity.

✨

Key Insights

1Insight 1: Dynamic Programming: Since we want to find the optimal solution by making a decision at each position of the string, DP is a natural fit.
2Insight 2: Bitmasking: Since `k` is at most 26 (number of lowercase English letters), we can efficiently represent the set of distinct characters encountered so far using a bitmask.
3Insight 3: Memoization: Overlapping subproblems exist, so memoization is crucial for optimizing the DP solution's runtime. Without memoization, the repeated calculations would lead to exponential time complexity.
4Insight 4: Only one change matters. We either make it or we don't. This single change is at the very beginning and the boolean `can_change` will record whether we have performed the change or not.

📚

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: Empty string or `k` is greater than or equal to 26. If `s` is empty, the question constraints indicate it won't be. If `k` is greater than or equal to 26, any change we make won't result in more partitions, so effectively `s` becomes a single partition.
Edge case 2: `k = 1`. If k=1 and s contains more than one distinct character, then we need to change the string so that consecutive characters are different. We can potentially obtain more partitions.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting Memoization: Without memoization, the time complexity explodes, leading to TLE (Time Limit Exceeded) errors.

✗Common mistake 2: Incorrect Bitmasking: Incorrect handling of the bitmask, especially when adding or checking the number of distinct characters, can lead to incorrect results.

✗Common mistake 3: Not handling the change parameter `can_change` correctly. We should only change once throughout the execution.

✅

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 algorithms might use similar partitioning techniques to divide data into segments with limited variability before applying compression.

🏦Real-world scenario 2: Network traffic analysis could use this kind of logic to segment traffic based on packet types up to a certain limit, before switching to a new segment.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: October 2, 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

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

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 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 "Maximize the Number of Partitions After Operations"?

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.

🤖 Need help with this problem?