What is the time complexity of Count Prefix and Suffix Pairs I?

The optimal solution has a time complexity of O(n^2 * m).

What data structures are used to solve Count Prefix and Suffix Pairs I?

This problem uses Array, String, Trie, Rolling Hash, String Matching, Hash Function. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Count Prefix and Suffix Pairs I?

Insight 1: The problem requires checking both prefix and suffix conditions simultaneously for each pair of strings. Insight 2: A nested loop is a straightforward way to iterate through all possible pairs of strings (i, j) where i < j.

Which companies ask LeetCode 3042. Count Prefix and Suffix Pairs I in interviews?

This problem is frequently asked at Autodesk, Capital One.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Count Prefix and Suffix Pairs I - LeetCode 3042 Solution

Q: How to solve Count Prefix and Suffix Pairs I?

The provided solutions use a brute-force approach. They iterate through all possible pairs of strings using nested loops. For each pair (words[i], words[j]), a helper function `isPrefixAndSuffix` checks if words[i] is both a prefix and a suffix of words[j]. If it is, the counter is incremented. Finally, the total count is returned.

Q: Which companies ask LeetCode 3042. Count Prefix and Suffix Pairs I in interviews?

This problem is frequently asked at Autodesk, Capital One.

Count Prefix and Suffix Pairs I - Complete Solution Guide

Count Prefix and Suffix Pairs I is LeetCode problem 3042, 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 a 0-indexed string array words . Let's define a boolean function isPrefixAndSuffix that takes two strings, str1 and str2 : isPrefixAndSuffix(str1, str2) returns true if str1 is both a prefix and a suffix of str2 , and false otherwise. For example, isPrefixAndSuffix("aba", "ababa") is true because "aba" is a prefix of "ababa" and also a suffix, but isPrefixAndSuffix("abc", "abcd") is false . Return an integer denoting the number of index pairs (i, j) such that i < j , and isPrefixAn

Detailed Explanation

The problem asks you to count the number of pairs of strings (words[i], words[j]) where i < j, and words[i] is both a prefix and a suffix of words[j]. The input is a list of strings, and the output is a single integer representing the count of such pairs. The problem emphasizes that words[i] must be a prefix AND a suffix of words[j], not just one or the other. Constraints limit the length of the input list and the length of individual strings.

Solution Approach

The provided solutions use a brute-force approach. They iterate through all possible pairs of strings using nested loops. For each pair (words[i], words[j]), a helper function `isPrefixAndSuffix` checks if words[i] is both a prefix and a suffix of words[j]. If it is, the counter is incremented. Finally, the total count is returned.

Step-by-Step Algorithm

Step 1: Initialize a counter `count` to 0.
Step 2: Iterate through the `words` array using a nested loop. The outer loop iterates from i = 0 to n-1, and the inner loop iterates from j = i+1 to n-1 (to ensure i < j).
Step 3: For each pair (words[i], words[j]), call the `isPrefixAndSuffix` function to check the prefix and suffix condition.
Step 4: If `isPrefixAndSuffix` returns true, increment the `count`.
Step 5: After iterating through all pairs, return the final value of `count`.

Key Insights

Insight 1: The problem requires checking both prefix and suffix conditions simultaneously for each pair of strings.
Insight 2: A nested loop is a straightforward way to iterate through all possible pairs of strings (i, j) where i < j.
Insight 3: Efficient string manipulation functions (like startsWith and endsWith in Python/Java) can significantly reduce the time taken to check the prefix and suffix conditions.

Complexity Analysis

Time Complexity: O(n^2 * m)

Space Complexity: O(1)

Topics

This problem involves: Array, String, Trie, Rolling Hash, String Matching, Hash Function.

Companies

Asked at: Autodesk, Capital One.

Count Prefix and Suffix Pairs I | LeetCode 3042

Easy

Array String Trie Rolling Hash String Matching Hash Function

Time: O(n^2 * m)Space: O(1)

Solution Code

Available in:

class Solution {
    public int countPrefixSuffixPairs(String[] words) {
        int count = 0;
        for (int i = 0; i < words.length; i++) {
            for (int j = i + 1; j < words.length; j++) {
                if (isPrefixAndSuffix(words[i], words[j])) {
                    count++;
                }
            }
        }
        return count;
    }

    private boolean isPrefixAndSuffix(String str1, String str2) {
        if (str2.length() < str1.length()) {
            return false;
        }
        if (str2.startsWith(str1) && str2.endsWith(str1)) {
            return true;
        }
        return false;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 3042: Count Prefix and Suffix Pairs I - Detailed explanation and solution approaches.

Topics

Array String Trie Rolling Hash String Matching Hash Function

Companies

AutodeskCapital One

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:3042

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 counter `count` to 0.
2
Step 2: Iterate through the `words` array using a nested loop. The outer loop iterates from i = 0 to n-1, and the inner loop iterates from j = i+1 to n-1 (to ensure i < j).
3
Step 3: For each pair (words[i], words[j]), call the `isPrefixAndSuffix` function to check the prefix and suffix condition.
4
Step 4: If `isPrefixAndSuffix` returns true, increment the `count`.
5
Step 5: After iterating through all pairs, return the final value of `count`.

💡

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^2 * m)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The nested loops iterate through approximately n * (n-1)/2 pairs of strings, where n is the length of the `words` array. For each pair, the `isPrefixAndSuffix` function takes O(m) time in the worst case, where m is the maximum length of a string. Therefore, the overall time complexity is O(n^2 * m). The space complexity is O(1) because the solution uses a constant amount of extra space regardless of the input size.

📖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

Trie-based approach

Construct a Trie from the reversed strings. For each string, check if its prefix exists in the Trie. This can improve the time complexity if the strings share many common prefixes and suffixes.

Time: O(n*m + n*m) - constructing trie is O(n*m) and searching O(n*m)Space: O(n*m) - to store the trie

Trade-offs: A Trie-based approach might be more efficient for a large number of strings with significant overlap. The space complexity is higher, though.

✨

Key Insights

1Insight 1: The problem requires checking both prefix and suffix conditions simultaneously for each pair of strings.
2Insight 2: A nested loop is a straightforward way to iterate through all possible pairs of strings (i, j) where i < j.
3Insight 3: Efficient string manipulation functions (like startsWith and endsWith in Python/Java) can significantly reduce the time taken to check the prefix and suffix conditions.

📚

Key Terms Explained

For Students

Array

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

String

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

Edge Cases

Edge case 1: Empty input array. The solution correctly handles this case by returning 0.
Edge case 2: Strings of different lengths. The `isPrefixAndSuffix` function correctly handles this by returning false if str1 is longer than str2.
Edge case 3: words[i] == words[j]. The solution correctly skips this as the inner loop starts at j=i+1. This avoids unnecessary comparisons

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check both prefix and suffix conditions separately. Only checking one condition would result in an incorrect count.

✗Common mistake 2: Inefficient string comparison. Using inefficient string comparison methods within the `isPrefixAndSuffix` function would significantly increase the time complexity.

✗Common mistake 3: Incorrect loop bounds in nested loops. Incorrect loop conditions could lead to index out of bound errors or incorrect pair counts.

✅

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: Analyzing code for potential bugs or code duplication. You can use this pattern to find code blocks that are both prefixes and suffixes of larger functions (potential for refactoring).

🏦Real-world scenario 2: Bioinformatics. Identifying DNA sequences that share common prefixes and suffixes might reveal clues to genetic relationships or functional similarities.

💡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: November 10, 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

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

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 "Count Prefix and Suffix Pairs I"?

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 string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?