What is the time complexity of Permutation Difference between Two Strings?

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

What data structures are used to solve Permutation Difference between Two Strings?

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

What are the key insights for solving Permutation Difference between Two Strings?

Insight 1: Efficiently finding the index of each character in string `t` is crucial for performance. A naive linear search for each character in `s` within `t` leads to O(n^2) time complexity. Insight 2: Using a hash map (or dictionary in Python) to store the indices of characters in `t` can improve the search time to O(1) on average, resulting in an overall linear time solution.

Which companies ask LeetCode 3146. Permutation Difference between Two Strings in interviews?

This problem is frequently asked at Accenture.

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Permutation Difference between Two Strings - LeetCode 3146 Solution

Permutation Difference between Two Strings - Complete Solution Guide

Permutation Difference between Two Strings is LeetCode problem 3146, a Easy 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 such that every character occurs at most once in s and t is a permutation of s . The permutation difference between s and t is defined as the sum of the absolute difference between the index of the occurrence of each character in s and the index of the occurrence of the same character in t . Return the permutation difference between s and t . Example 1: Input: s = "abc", t = "bac" Output: 2 Explanation: For s = "abc" and t = "bac" , the permutation difference of

Detailed Explanation

The problem asks you to calculate the 'permutation difference' between two strings, `s` and `t`. Both strings contain unique characters, and `t` is a permutation of `s` (meaning it contains the same characters but in a different order). The permutation difference is computed by summing the absolute differences between the indices of each character in `s` and its corresponding index in `t`. For example, if `s = "abc"` and `t = "bca"`, the permutation difference is |0-1| + |1-2| + |2-0| = 1 + 1 + 2 = 4. The problem emphasizes that each character appears only once in each string.

Solution Approach

The provided solutions all iterate through string `s`. For each character in `s`, they find its index in `t` and calculate the absolute difference between the indices. The C++ and Java solutions use the `find()` and `indexOf()` methods respectively, which are essentially optimized linear searches. The Python solution uses nested loops, resulting in a less efficient O(n^2) time complexity. The sum of these absolute differences is the final permutation difference.

Step-by-Step Algorithm

Step 1: Iterate through each character in string `s` using a loop (index `i`).
Step 2: For each character in `s`, find its index (`j`) in string `t` (this can be done using an optimized linear search or a hash map for better performance).
Step 3: Calculate the absolute difference between the indices `i` and `j` using `abs(i - j)`.
Step 4: Add the absolute difference to a running total representing the permutation difference.
Step 5: After iterating through all characters in `s`, return the final permutation difference.

Key Insights

Insight 1: Efficiently finding the index of each character in string `t` is crucial for performance. A naive linear search for each character in `s` within `t` leads to O(n^2) time complexity.
Insight 2: Using a hash map (or dictionary in Python) to store the indices of characters in `t` can improve the search time to O(1) on average, resulting in an overall linear time solution.
Insight 3: The problem constraints (characters only appear once) simplify the solution since we don't have to handle duplicate characters and their indices.

Complexity Analysis

Time Complexity: O(mn)

Space Complexity: O(1)

Topics

This problem involves: Hash Table, String.

Companies

Asked at: Accenture.

Permutation Difference between Two Strings | LeetCode 3146

Easy

Hash Table String

Time: O(mn)Space: O(1)

Solution Code

Available in:

class Solution {
    public int findPermutationDifference(String s, String t) {
        int permutationDifference = 0;
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            int indexInS = i;
            int indexInT = t.indexOf(c);
            permutationDifference += Math.abs(indexInS - indexInT);
        }
        return permutationDifference;
    }
}

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Problem Info

Description

Problem 3146: Permutation Difference between Two Strings - Detailed explanation and solution approaches.

Topics

Hash Table String

Companies

Accenture

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3146

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: Iterate through each character in string `s` using a loop (index `i`).
2
Step 2: For each character in `s`, find its index (`j`) in string `t` (this can be done using an optimized linear search or a hash map for better performance).
3
Step 3: Calculate the absolute difference between the indices `i` and `j` using `abs(i - j)`.
4
Step 4: Add the absolute difference to a running total representing the permutation difference.
5
Step 5: After iterating through all characters in `s`, return the final permutation difference.

💡

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(mn)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The Python solution has nested loops resulting in O(n*m) time complexity where n and m are lengths of s and t. However, the other provided solutions (Java and C++) use optimized string methods (`indexOf` and `find`) which have an average-case time complexity of O(n) where n is the length of the string. The space complexity is O(1) because the algorithms use a constant amount of extra space regardless of the input size. The space used is only for a few integer variables to keep track of indices and the difference.

📖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

Create a hash map to store the index of each character in `t`. This allows finding the index of each character in `t` in O(1) on average. This will provide the optimal time complexity of O(n).

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

Trade-offs: The hash map approach uses extra space (O(n)) but significantly improves the time complexity from O(n^2) to O(n), making it more efficient for larger inputs. The added space cost is generally worth it for performance gains.

✨

Key Insights

1Insight 1: Efficiently finding the index of each character in string `t` is crucial for performance. A naive linear search for each character in `s` within `t` leads to O(n^2) time complexity.
2Insight 2: Using a hash map (or dictionary in Python) to store the indices of characters in `t` can improve the search time to O(1) on average, resulting in an overall linear time solution.
3Insight 3: The problem constraints (characters only appear once) simplify the solution since we don't have to handle duplicate characters and their indices.

📚

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. The code should handle empty strings gracefully; the permutation difference would be 0.
Edge case 2: Strings with only one character. The permutation difference will always be 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling edge cases such as empty strings or strings with only one character properly.

✗Common mistake 2: Incorrectly calculating the absolute difference between indices or accumulating the total difference.

✗Common mistake 3: Using nested loops in Python without understanding the O(n^2) time complexity implication.

✅

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: Comparing the similarity between two DNA sequences by considering the positional differences of nucleotides.

🏦Real-world scenario 2: Analyzing the difference in character arrangement between two encrypted messages, potentially helping in cryptanalysis.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ 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 26, 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

❓

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.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Permutation Difference between Two Strings"?

For this easy-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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