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

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

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

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?

Insight 1: We need to determine the frequency differences between the characters in `s` and `t`. Characters that appear more frequently in `t` than in `s` need to be replaced to match the frequency of `s`. Insight 2: We can efficiently use a frequency array (or hash table) to store the character counts. Since the input is lowercase English letters, an array of size 26 is sufficient and optimal.

Which companies ask LeetCode 1347. Minimum Number of Steps to Make Two Strings Anagram in interviews?

This problem is frequently asked at DoorDash, IXL, SoFi, X.

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Minimum Number of Steps to Make Two Strings Anagram - LeetCode 1347 Solution

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

Minimum Number of Steps to Make Two Strings Anagram is LeetCode problem 1347, 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 of the same length s and t . In one step you can choose any character of t and replace it with another character . Return the minimum number of steps to make t an anagram of s . An Anagram of a string is a string that contains the same characters with a different (or the same) ordering. Example 1: Input: s = "bab", t = "aba" Output: 1 Explanation: Replace the first 'a' in t with b, t = "bba" which is anagram of s. Example 2: Input: s = "leetcode", t = "practice" Output:

Detailed Explanation

The problem asks us to find the minimum number of character replacement steps required to transform string `t` into an anagram of string `s`. Both strings have the same length and consist of lowercase English letters. An anagram is a rearrangement of characters, so `t` must have the same character frequencies as `s` after the replacements.

Solution Approach

The solution uses a frequency array `counts` of size 26 to keep track of the character frequency differences between `s` and `t`. We iterate through both strings simultaneously. For each character in `s`, we increment the corresponding count in the `counts` array. For each character in `t`, we decrement the count. After iterating through both strings, the `counts` array holds the net difference in the frequency of each character. Finally, we iterate through the `counts` array and sum up all the positive counts. This sum represents the minimum number of replacements needed in `t` to make it an anagram of `s`.

Step-by-Step Algorithm

Step 1: Initialize an integer array `counts` of size 26 with all elements set to 0. This array will store the frequency difference for each lowercase letter.
Step 2: Iterate through strings `s` and `t` simultaneously using a single loop (since they have the same length).
Step 3: For each index `i`, increment `counts[s[i] - 'a']` and decrement `counts[t[i] - 'a']`. This effectively calculates the difference in frequency for each character.
Step 4: Initialize an integer variable `steps` to 0. This variable will store the minimum number of steps needed.
Step 5: Iterate through the `counts` array. For each element `count` in `counts`, if `count` is greater than 0, add `count` to `steps`.
Step 6: Return `steps`.

Key Insights

Insight 1: We need to determine the frequency differences between the characters in `s` and `t`. Characters that appear more frequently in `t` than in `s` need to be replaced to match the frequency of `s`.
Insight 2: We can efficiently use a frequency array (or hash table) to store the character counts. Since the input is lowercase English letters, an array of size 26 is sufficient and optimal.
Insight 3: Only characters present more often in `s` than in `t` contribute to the number of required steps. We need to only sum up the positive differences in character counts.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Hash Table, String, Counting.

Companies

Asked at: DoorDash, IXL, SoFi, X.

Minimum Number of Steps to Make Two Strings Anagram | LeetCode 1347

Medium

Hash Table String Counting

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

Solution Code

Available in:

class Solution {
    public int minSteps(String s, String t) {
        int[] counts = new int[26];
        for (int i = 0; i < s.length(); i++) {
            counts[s.charAt(i) - 'a'] += 1;
            counts[t.charAt(i) - 'a'] -= 1;
        }
        
        int steps = 0;
        for (int count : counts) {
            if (count > 0) {
                steps += count;
            }
        }
        
        return steps;
    }
}

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Problem Info

Description

Problem 1347: Minimum Number of Steps to Make Two Strings Anagram - Detailed explanation and solution approaches.

Topics

Hash Table String Counting

Companies

DoorDashIXLSoFiX

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:1347

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

6 steps

1
Step 1: Initialize an integer array `counts` of size 26 with all elements set to 0. This array will store the frequency difference for each lowercase letter.
2
Step 2: Iterate through strings `s` and `t` simultaneously using a single loop (since they have the same length).
3
Step 3: For each index `i`, increment `counts[s[i] - 'a']` and decrement `counts[t[i] - 'a']`. This effectively calculates the difference in frequency for each character.
4
Step 4: Initialize an integer variable `steps` to 0. This variable will store the minimum number of steps needed.
5
Step 5: Iterate through the `counts` array. For each element `count` in `counts`, if `count` is greater than 0, add `count` to `steps`.
6
Step 6: Return `steps`.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant operation in terms of time complexity is the iteration through the strings, which takes O(n) time. The space complexity is determined by the `counts` array, which is of constant size O(26), thus 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 Hash Maps (Dictionaries)

Instead of using an array, we could use hash maps (dictionaries) to store the character frequencies. This approach is more flexible if the character set is not limited to lowercase English letters.

Time: O(n) on average, where n is the length of the strings. Worst case would be O(n^2) if there were hash collisions.Space: O(1) as the number of possible characters (lowercase English letters) is fixed to 26. In a very specific corner case of significantly larger unicode character range, it could become closer to O(n), but for this problem, it remains O(1).

Trade-offs: Hash maps offer more flexibility for larger character sets, but they might have slightly higher overhead compared to a fixed-size array for lowercase English letters.

✨

Key Insights

1Insight 1: We need to determine the frequency differences between the characters in `s` and `t`. Characters that appear more frequently in `t` than in `s` need to be replaced to match the frequency of `s`.
2Insight 2: We can efficiently use a frequency array (or hash table) to store the character counts. Since the input is lowercase English letters, an array of size 26 is sufficient and optimal.
3Insight 3: Only characters present more often in `s` than in `t` contribute to the number of required steps. We need to only sum up the positive differences in character counts.

📚

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: When strings `s` and `t` are already anagrams. In this case, all elements in the `counts` array will be 0, and the algorithm will correctly return 0.
Edge case 2: Empty strings (although the problem states length is > 0, it's good to think about). If both strings were empty the code would still function correctly, returning 0 after initializing count array.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the strings are already anagrams correctly. The algorithm should return 0 in this case.

✗Common mistake 2: Incorrectly calculating the frequency differences. It's important to increment the count for characters in `s` and decrement the count for characters in `t`.

✅

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: Text processing and data analysis, where it's necessary to compare the character frequencies of different strings or documents.

🏦Real-world scenario 2: Cryptography, where anagrams or character frequency analysis might be used to break codes or analyze ciphertext patterns.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: August 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

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

@InterviewProJul 13

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

@CodeNinjaJul 6

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

@TechLeadJun 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 Number of Steps to Make Two Strings Anagram"?

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?