What is the time complexity of Odd String Difference?

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

What is the space complexity of Odd String Difference?

The space complexity is O(n*m).

What data structures are used to solve Odd String Difference?

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

What are the key insights for solving Odd String Difference?

The problem can be efficiently solved using a hash table (dictionary in Python) to store the difference arrays and their corresponding string indices. This allows for quick lookup of existing difference arrays. Converting the difference array into a tuple (in Python) or other hashable data structure is crucial for using it as a key in the hash table. Arrays themselves are often not hashable.

Which companies ask LeetCode 2451. Odd String Difference in interviews?

This problem is frequently asked at Datadog, IBM, Visa.

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Odd String Difference - LeetCode 2451 Solution

Q: How to solve Odd String Difference?

The Python solution efficiently solves this problem by first computing the difference arrays for all strings. It then uses a dictionary to store each unique difference array as a key, and a list of the indices of strings with that difference array as the value. Finally, it iterates through the dictionary to find the key (difference array) that appears only once, indicating the odd string out. The index of that string is then used to retrieve the corresponding string from the input array.

Odd String Difference - Complete Solution Guide

Odd String Difference is LeetCode problem 2451, 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 an array of equal-length strings words . Assume that the length of each string is n . Each string words[i] can be converted into a difference integer array difference[i] of length n - 1 where difference[i][j] = words[i][j+1] - words[i][j] where 0 <= j <= n - 2 . Note that the difference between two letters is the difference between their positions in the alphabet i.e. the position of 'a' is 0 , 'b' is 1 , and 'z' is 25 . For example, for the string "acb" , the difference integer ar

Detailed Explanation

The problem asks you to find the string in a given array of strings that has a different difference integer array compared to the others. Each string is converted into a difference integer array by calculating the difference between consecutive characters' ASCII values. The differences are calculated as `words[i][j+1] - words[i][j]`. All strings, except for one, will have the same difference integer array. The task is to identify and return the 'odd string out'.

Solution Approach

The Python solution efficiently solves this problem by first computing the difference arrays for all strings. It then uses a dictionary to store each unique difference array as a key, and a list of the indices of strings with that difference array as the value. Finally, it iterates through the dictionary to find the key (difference array) that appears only once, indicating the odd string out. The index of that string is then used to retrieve the corresponding string from the input array.

Step-by-Step Algorithm

Step 1: Create an empty list called `diffs` to store the difference arrays for each string.
Step 2: Iterate through each `word` in the input `words` array.
Step 3: For each `word`, create a new `diff` list and compute the difference between consecutive characters (using ASCII values).
Step 4: Append the computed `diff` list to the `diffs` list.
Step 5: Create an empty dictionary called `counts` to store difference arrays as keys and a list of corresponding string indices as values.
Step 6: Iterate through the `diffs` list.
Step 7: For each `diff`, convert it to a tuple (to make it hashable) and store it as a key in `counts`. If the key already exists, append the current index to its value list; otherwise, create a new entry in `counts`.
Step 8: Iterate through the `counts` dictionary.
Step 9: If a key (difference array) has only one index associated with it (length of value list is 1), return the corresponding string from the `words` array using that index.

Key Insights

The problem can be efficiently solved using a hash table (dictionary in Python) to store the difference arrays and their corresponding string indices. This allows for quick lookup of existing difference arrays.
Converting the difference array into a tuple (in Python) or other hashable data structure is crucial for using it as a key in the hash table. Arrays themselves are often not hashable.
The solution leverages the fact that only one string will have a unique difference array. This simplifies the search process once the difference arrays are stored in the hash table.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(n*m)

Topics

This problem involves: Array, Hash Table, String.

Companies

Asked at: Datadog, IBM, Visa.

Odd String Difference | LeetCode 2451

Easy

Array Hash Table String

Time: O(n*m)Space: O(n*m)

Solution Code

Available in:

class Solution {
    public String oddString(String[] words) {
        int n = words.length;
        int len = words[0].length();

        int[] diff0 = new int[len - 1];
        for (int i = 0; i < len - 1; i++) {
            diff0[i] = words[0].charAt(i + 1) - words[0].charAt(i);
        }

        int[] diff1 = new int[len - 1];
        for (int i = 0; i < len - 1; i++) {
            diff1[i] = words[1].charAt(i + 1) - words[1].charAt(i);
        }

        int[] diff2 = new int[len - 1];
        for (int i = 0; i < len - 1; i++) {
            diff2[i] = words[2].charAt(i + 1) - words[2].charAt(i);
        }

        boolean firstEqualsSecond = true;
        for (int i = 0; i < len - 1; i++) {
            if (diff0[i] != diff1[i]) {
                firstEqualsSecond = false;
                break;
            }
        }

        boolean firstEqualsThird = true;
        for (int i = 0; i < len - 1; i++) {
            if (diff0[i] != diff2[i]) {
                firstEqualsThird = false;
                break;
            }
        }

        if (firstEqualsSecond && !firstEqualsThird) {
            return words[2];
        }

        if (!firstEqualsSecond && firstEqualsThird) {
            return words[1];
        }

        for (int i = 3; i < n; i++) {
            int[] diff = new int[len - 1];
            for (int j = 0; j < len - 1; j++) {
                diff[j] = words[i].charAt(j + 1) - words[i].charAt(j);
            }

            boolean equalsFirst = true;
            for (int j = 0; j < len - 1; j++) {
                if (diff0[j] != diff[j]) {
                    equalsFirst = false;
                    break;
                }
            }

            if (!equalsFirst) {
                return words[i];
            }
        }

        return words[0];
    }
}

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Problem Info

Description

Problem 2451: Odd String Difference - Detailed explanation and solution approaches.

Topics

Array Hash Table String

Companies

DatadogIBMVisa

Stats

Difficulty:Easy

Languages:4

Companies:3

Problem #:2451

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: Create an empty list called `diffs` to store the difference arrays for each string.
2
Step 2: Iterate through each `word` in the input `words` array.
3
Step 3: For each `word`, create a new `diff` list and compute the difference between consecutive characters (using ASCII values).
4
Step 4: Append the computed `diff` list to the `diffs` list.
5
Step 5: Create an empty dictionary called `counts` to store difference arrays as keys and a list of corresponding string indices as values.
6
Step 6: Iterate through the `diffs` list.
7
Step 7: For each `diff`, convert it to a tuple (to make it hashable) and store it as a key in `counts`. If the key already exists, append the current index to its value list; otherwise, create a new entry in `counts`.
8
Step 8: Iterate through the `counts` dictionary.
9
Step 9: If a key (difference array) has only one index associated with it (length of value list is 1), return the corresponding string from the `words` array using that index.

💡

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(n*m)

📝

Why?

The time complexity arises from iterating through each string (m) and calculating the differences within each string (n). The space complexity is due to storing the difference arrays in the `counts` dictionary. Each difference array has a size of n-1. In the worst-case scenario where all difference arrays are unique, it will be O(m*(n-1)) which simplifies to O(m*n).

📖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

Direct Comparison

Compare the difference array of the first string with all other strings. The first string with a different difference array is the odd one out.

Time: O(m*n)Space: O(n)

Trade-offs: This approach is simpler to understand but might be slightly less efficient than the hash table approach for a very large number of strings with many unique difference arrays.

✨

Key Insights

1The problem can be efficiently solved using a hash table (dictionary in Python) to store the difference arrays and their corresponding string indices. This allows for quick lookup of existing difference arrays.
2Converting the difference array into a tuple (in Python) or other hashable data structure is crucial for using it as a key in the hash table. Arrays themselves are often not hashable.
3The solution leverages the fact that only one string will have a unique difference array. This simplifies the search process once the difference arrays are stored in the hash table.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

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: All strings have the same difference array. In this case, the algorithm will return the first string as the code currently handles it. The last statement `return words[0]` should be handled better. A check before the loop should be implemented to handle this edge case more efficiently.
Edge case 2: The input list `words` has only one element or fewer than 3 elements. The code needs to handle this case by returning an error or by checking before calculations to prevent errors.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle edge cases like all strings having the same difference array or an insufficient number of strings.

✗Common mistake 2: Incorrectly calculating the difference array, for example, using string indices instead of ASCII values or making off-by-one errors in the loop bounds.

✅

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: Anomaly detection in data streams. If the strings represent sensor readings or other time-series data, the algorithm can identify unusual patterns.

🏦Real-world scenario 2: Identifying corrupted data in a dataset where data entries should follow a consistent pattern. The algorithm could be used to detect rogue entries.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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: September 20, 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

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

@CodeNinjaJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

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 should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Odd String Difference"?

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

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