What is the time complexity of Minimum Length of String After Operations?

The optimal solution has a time complexity of O(N).

What data structures are used to solve Minimum Length of String After Operations?

This problem uses Hash Table, String, Counting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Length of String After Operations?

Insight 1: The order in which we perform the operations doesn't matter. If a valid 'i' exists, we can always perform the operation. This is because deleting elements can only enable new operations. Insight 2: The crucial observation is that we can treat each character independently. The problem's constraint ensures that if a character appears an even number of times, we can reduce it to two. If a character appears an odd number of times, we can reduce it to one. This allows simplifying the string.

Which companies ask LeetCode 3223. Minimum Length of String After Operations in interviews?

This problem is frequently asked at IBM.

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Minimum Length of String After Operations - LeetCode 3223 Solution

Minimum Length of String After Operations - Complete Solution Guide

Minimum Length of String After Operations is LeetCode problem 3223, a Medium 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 . You can perform the following process on s any number of times: Choose an index i in the string such that there is at least one character to the left of index i that is equal to s[i] , and at least one character to the right that is also equal to s[i] . Delete the closest occurrence of s[i] located to the left of i . Delete the closest occurrence of s[i] located to the right of i . Return the minimum length of the final string s that you can achieve. Example 1: Input:

Detailed Explanation

The problem asks us to find the minimum possible length of a string 's' after performing a series of deletion operations. The operation involves selecting an index 'i' such that 's[i]' has at least one identical character to its left and one to its right. When such an index is found, we remove the closest character to the left of 'i' that equals 's[i]' and the closest character to the right of 'i' that also equals 's[i]'. We can repeat this operation any number of times until no such index 'i' exists. The final length of the string 's' after these operations is the output.

Solution Approach

The provided code utilizes a counting approach. It iterates through the input string 's', counting the occurrences of each character. Then, it iterates through the counts of each character. If the count of a character is odd, it adds 1 to the minimum length. If the count is even, it adds 2 to the minimum length. This is based on the realization explained in Key Insights, that operations can always simplify a string to 1 or 2 characters of each type.

Step-by-Step Algorithm

Step 1: Initialize a data structure (e.g., a hash map or array) to store character counts.
Step 2: Iterate through the input string 's', updating the counts for each character in the data structure.
Step 3: Initialize a variable 'min_len' to 0.
Step 4: Iterate through the character counts.
Step 5: For each character count, check if it is even or odd.
Step 6: If the count is odd, add 1 to 'min_len'.
Step 7: If the count is even, add 2 to 'min_len'.
Step 8: Return 'min_len'.

Key Insights

Insight 1: The order in which we perform the operations doesn't matter. If a valid 'i' exists, we can always perform the operation. This is because deleting elements can only enable new operations.
Insight 2: The crucial observation is that we can treat each character independently. The problem's constraint ensures that if a character appears an even number of times, we can reduce it to two. If a character appears an odd number of times, we can reduce it to one. This allows simplifying the string.
Insight 3: The core idea is to count the occurrences of each character. If a character's count is even, it contributes 2 to the minimum length. If odd, it contributes 1.

Complexity Analysis

Time Complexity: O(N)

Space Complexity: O(N)

Topics

This problem involves: Hash Table, String, Counting.

Companies

Asked at: IBM.

Minimum Length of String After Operations | LeetCode 3223

Medium

Hash Table String Counting

Time: O(N)Space: O(N)

Solution Code

Available in:

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

class Solution {
    public int minimumLength(String s) {
        Map<Character, Integer> counts = new HashMap<>();
        for (char c : s.toCharArray()) {
            counts.put(c, counts.getOrDefault(c, 0) + 1);
        }
        
        int min_len = 0;
        for (int count : counts.values()) {
            if (count % 2 == 1) {
                min_len += 1;
            } else {
                min_len += 2;
            }
        }
        
        return min_len;
    }
}

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Problem Info

Description

Problem 3223: Minimum Length of String After Operations - Detailed explanation and solution approaches.

Topics

Hash Table String Counting

Companies

IBM

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3223

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

8 steps

1
Step 1: Initialize a data structure (e.g., a hash map or array) to store character counts.
2
Step 2: Iterate through the input string 's', updating the counts for each character in the data structure.
3
Step 3: Initialize a variable 'min_len' to 0.
4
Step 4: Iterate through the character counts.
5
Step 5: For each character count, check if it is even or odd.
6
Step 6: If the count is odd, add 1 to 'min_len'.
7
Step 7: If the count is even, add 2 to 'min_len'.
8
Step 8: Return 'min_len'.

💡

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)

💾

Space Complexity

O(N)

📝

Why?

The time complexity is dominated by the initial iteration to count characters. The space complexity depends on the number of unique characters which is bounded by the alphabet size and thus is O(1).

📖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

Using Array for Counting (C)

Instead of a hash map, we can use a fixed-size array to count character frequencies, assuming the character set is limited (e.g., lowercase English letters). This approach is demonstrated in the C code.

Time: O(N)Space: O(1)

Trade-offs: Slightly faster than using a hash map, especially in C, as array access is generally more efficient. However, it requires knowing the character set beforehand and has a fixed size.

✨

Key Insights

1Insight 1: The order in which we perform the operations doesn't matter. If a valid 'i' exists, we can always perform the operation. This is because deleting elements can only enable new operations.
2Insight 2: The crucial observation is that we can treat each character independently. The problem's constraint ensures that if a character appears an even number of times, we can reduce it to two. If a character appears an odd number of times, we can reduce it to one. This allows simplifying the string.
3Insight 3: The core idea is to count the occurrences of each character. If a character's count is even, it contributes 2 to the minimum length. If odd, it contributes 1.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

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

Edge Cases

Edge case 1: Empty string. The code implicitly handles this case, since the loop will not execute, and the initial value of `min_len` (0) is returned.
Edge case 2: String with only one distinct character (e.g., "aaaa"). The code will correctly calculate min_len as 2 if the count is even and 1 if the count is odd.
Edge case 3: String with all unique characters (e.g., "abcdefg"). In this case, all character counts will be 1, and the result will be the number of unique characters.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Trying to simulate the deletion process directly. This leads to a much more complex and inefficient solution with high time complexity.

✗Common mistake 2: Forgetting to account for edge cases such as an empty string or a string with only one unique character.

✅

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. This algorithm can be seen as a simplified form of run-length encoding or other compression techniques where repeating characters are grouped together.

🏦Real-world scenario 2: Analyzing data streams. Similar counting techniques are used to identify frequent elements in a stream of data, which has applications in network monitoring and web analytics.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

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: November 11, 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

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

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.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Minimum Length of String After Operations"?

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

🤖 Need help with this problem?