What is the time complexity of Minimum Number of Steps to Make Two Strings Anagram II?

The optimal solution has a time complexity of O(n+m).

What data structures are used to solve Minimum Number of Steps to Make Two Strings Anagram II?

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 Number of Steps to Make Two Strings Anagram II?

Insight 1: The core idea is to count the frequency of each character in both strings. The difference in frequencies determines how many characters need to be added to either string to make them anagrams. Insight 2: We only need to track the *difference* in character counts between the two strings. Characters with the same counts in both strings do not contribute to the number of required steps.

Which companies ask LeetCode 2186. Minimum Number of Steps to Make Two Strings Anagram II in interviews?

This problem is frequently asked at Wealthfront.

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Minimum Number of Steps to Make Two Strings Anagram II - LeetCode 2186 Solution

Minimum Number of Steps to Make Two Strings Anagram II - Complete Solution Guide

Minimum Number of Steps to Make Two Strings Anagram II is LeetCode problem 2186, 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 two strings s and t . In one step, you can append any character to either s or t . Return the minimum number of steps to make s and t anagrams of each other. An anagram of a string is a string that contains the same characters with a different (or the same) ordering. Example 1: Input: s = " lee tco de ", t = "co a t s " Output: 7 Explanation: - In 2 steps, we can append the letters in "as" onto s = "leetcode", forming s = "leetcode as ". - In 5 steps, we can append the letters in "

Detailed Explanation

The problem asks us to find the minimum number of single-character append operations needed to make two strings, `s` and `t`, anagrams of each other. Anagrams are strings containing the same characters with the same frequencies but possibly in different orders. The constraint is that we can only append characters to either string, not remove or replace them.

Solution Approach

The solution uses a frequency counting approach. An array (or a hash map, though array is more efficient here) of size 26 is used to store the difference in frequencies of each lowercase English alphabet character between the two strings. The code iterates through the first string, incrementing the corresponding count in the frequency array. Then, it iterates through the second string, decrementing the corresponding count. Finally, it calculates the sum of the absolute values of the counts in the array. This sum represents the minimum number of append operations needed.

Step-by-Step Algorithm

Step 1: Initialize a frequency array `counts` of size 26, with all elements set to 0. This array will store the difference in character counts between the two strings.
Step 2: Iterate through the string `s`. For each character in `s`, calculate its index in the `counts` array by subtracting the ASCII value of 'a' from the character's ASCII value. Increment the value at this index in the `counts` array.
Step 3: Iterate through the string `t`. For each character in `t`, calculate its index in the `counts` array similarly. Decrement the value at this index in the `counts` array.
Step 4: Initialize a variable `totalSteps` to 0. Iterate through the `counts` array. For each element in the array, add its absolute value to `totalSteps`.
Step 5: Return the value of `totalSteps`. This represents the minimum number of steps required to make the two strings anagrams.

Key Insights

Insight 1: The core idea is to count the frequency of each character in both strings. The difference in frequencies determines how many characters need to be added to either string to make them anagrams.
Insight 2: We only need to track the *difference* in character counts between the two strings. Characters with the same counts in both strings do not contribute to the number of required steps.
Insight 3: Minimizing steps translates to minimizing the changes needed to equalize character frequencies. This is achieved by simply summing the absolute differences in the counts for each character.

Complexity Analysis

Time Complexity: O(n+m)

Space Complexity: O(1)

Topics

This problem involves: Hash Table, String, Counting.

Companies

Asked at: Wealthfront.

Minimum Number of Steps to Make Two Strings Anagram II | LeetCode 2186

Medium

Hash Table String Counting

Time: O(n+m)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minSteps(String s, String t) {
        /**
         * Calculates the minimum number of steps to make two strings anagrams.
         * An anagram must have the same characters with the same frequencies.
         * A step consists of appending a character to either string.
         *
         * The final state of both strings, after apppending characters, must have
         * an identical character frequency map. Let's call this final frequency map F.
         *
         * For any character 'c', its count in the final strings, F['c'], must be at least
         * its count in the original strings s and t.
         * So, F['c'] >= count(s, 'c') and F['c'] >= count(t, 'c').
         * To minimize the number of appended characters (steps), we should choose the smallest
         * possible F['c'], which is max(count(s, 'c'), count(t, 'c')).
         *
         * The number of 'c's to add to s is: max(count(s, 'c'), count(t, 'c')) - count(s, 'c').
         * The number of 'c's to add to t is: max(count(s, 'c'), count(t, 'c')) - count(t, 'c').
         *
         * The total steps for character 'c' is the sum of these two, which simplifies to:
         * abs(count(s, 'c') - count(t, 'c')).
         *
         * The total minimum steps is the sum of these absolute differences over all
         * 26 lowercase English letters.
         *
         * We can implement this by:
         * 1. Creating a frequency array/map of size 26.
         * 2. Iterate through s and increment the count for each character.
         * 3. Iterate through t and decrement the count for each character.
         * 4. The final array will hold the difference in frequencies (count_s - count_t).
         * 5. The sum of the absolute values of these differences is the total number of steps.
         */

        // Array to store the frequency difference for each character 'a' through 'z'.
        int[] counts = new int[26];

        // Populate the frequency map for string s (incrementing counts).
        for (char char_s : s.toCharArray()) {
            counts[char_s - 'a']++;
        }

        // Update the frequency map for string t (decrementing counts).
        // The result is an array of frequency differences.
        for (char char_t : t.toCharArray()) {
            counts[char_t - 'a']--;
        }

        // The total number of steps is the sum of absolute differences.
        // A positive value counts[i] means character i is in excess in s and needs
        // to be added to t. The number of additions is counts[i].
        // A negative value counts[i] means character i is in excess in t and needs
        // to be added to s. The number of additions is -counts[i].
        // In both cases, it's the absolute value of the difference.
        int steps = 0;
        for (int count : counts) {
            steps += Math.abs(count);
        }
        return steps;
    }
}

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Problem Info

Description

Problem 2186: Minimum Number of Steps to Make Two Strings Anagram II - Detailed explanation and solution approaches.

Topics

Hash Table String Counting

Companies

Wealthfront

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2186

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

5 steps

1
Step 1: Initialize a frequency array `counts` of size 26, with all elements set to 0. This array will store the difference in character counts between the two strings.
2
Step 2: Iterate through the string `s`. For each character in `s`, calculate its index in the `counts` array by subtracting the ASCII value of 'a' from the character's ASCII value. Increment the value at this index in the `counts` array.
3
Step 3: Iterate through the string `t`. For each character in `t`, calculate its index in the `counts` array similarly. Decrement the value at this index in the `counts` array.
4
Step 4: Initialize a variable `totalSteps` to 0. Iterate through the `counts` array. For each element in the array, add its absolute value to `totalSteps`.
5
Step 5: Return the value of `totalSteps`. This represents the minimum number of steps required to make the two strings anagrams.

💡

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+m)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the linear traversal of the two input strings. The space complexity is constant because the size of the frequency array `counts` is fixed, depending only on the number of possible lowercase English characters (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

Hash Map Approach

Instead of using an array, a hash map (dictionary) can be used to store character frequencies. This approach is more flexible if the character set is not limited to lowercase English letters but would be slightly less efficient in this particular scenario due to overhead of hash map operations.

Time: O(n + m)Space: O(k), where k is the number of distinct characters in both strings.

Trade-offs: Using a hash map provides more flexibility if the character set is larger or unknown. However, it has slightly higher overhead compared to using an array when the character set is fixed (as in this problem). For this specific problem, the array approach is preferable due to its simplicity and efficiency.

✨

Key Insights

1Insight 1: The core idea is to count the frequency of each character in both strings. The difference in frequencies determines how many characters need to be added to either string to make them anagrams.
2Insight 2: We only need to track the *difference* in character counts between the two strings. Characters with the same counts in both strings do not contribute to the number of required steps.
3Insight 3: Minimizing steps translates to minimizing the changes needed to equalize character frequencies. This is achieved by simply summing the absolute differences in the counts for each character.

📚

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 strings. If both strings are empty, the code will correctly return 0, as the frequency array will remain all zeros, and the sum of absolute values will be 0.
Edge case 2: Strings are already anagrams. If the strings are anagrams, the frequency array will become all zeros after processing both strings, and the code will correctly return 0.
Edge case 3: One string is empty and the other is not. The algorithm will correctly count the characters in the non-empty string and return the number of characters as the number of required steps.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using separate frequency maps for each string and then comparing them, instead of directly calculating the difference. This requires more memory and can be less efficient.

✗Common mistake 2: Forgetting to take the absolute value of the frequency differences. This will result in an incorrect answer, as negative counts represent characters that are more frequent in one string than the other.

✅

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: Spell checking. Determining the minimum number of edits (insertions or deletions) to correct a misspelled word to match a word in a dictionary can be modeled using a similar approach.

🏦Real-world scenario 2: Bioinformatics. Analyzing DNA sequences and finding the minimum number of mutations needed to transform one sequence into another can utilize frequency analysis and differences to find relatedness.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: June 27, 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 14

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

@InterviewProJul 7

❓

Frequently Asked Questions

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Number of Steps to Make Two Strings Anagram II"?

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.

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