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

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

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

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

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

Insight 1: Using brute-force string comparison for each pair of words would be inefficient, especially with the given constraints (up to 10^5 words, up to 10^5 length per word). Insight 2: Rolling hash is an efficient method for determining if two substrings are equal by comparing their hash values instead of directly comparing the strings. Using two different prime numbers for the hashing helps in avoiding collision issues.

Which companies ask LeetCode 3045. Count Prefix and Suffix Pairs II in interviews?

This problem is frequently asked at Autodesk, Capital One, Samsung.

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Count Prefix and Suffix Pairs II - LeetCode 3045 Solution

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

The provided solution uses a rolling hash technique with two different prime numbers to efficiently determine if one string is both a prefix and a suffix of another. For each word in the input array, the solution calculates prefix hashes and powers of the prime base. Then, for each possible length 'l' from 1 to the length of the current word, the solution calculates the hash of the prefix and the hash of the suffix of length 'l'. If these hashes match for both prime bases, it implies that the prefix and suffix are equal. A hash map is used to keep track of how many times each word has occurred, which helps to count the pairs efficiently. The solution avoids string comparisons and uses efficient hashing for comparing substrings.

Count Prefix and Suffix Pairs II - Complete Solution Guide

Count Prefix and Suffix Pairs II is LeetCode problem 3045, a Hard 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 isPrefixA

Detailed Explanation

The problem asks us to count the number of pairs of strings (words[i], words[j]) in a given array of strings 'words', where i < j, and 'words[i]' is both a prefix and a suffix of 'words[j]'. A prefix of a string is a substring that starts at the beginning of the string, and a suffix is a substring that ends at the end of the string. We need to find all pairs (i, j) satisfying i < j such that words[i] is both a prefix and a suffix of words[j], and return the total count of such pairs.

Solution Approach

The provided solution uses a rolling hash technique with two different prime numbers to efficiently determine if one string is both a prefix and a suffix of another. For each word in the input array, the solution calculates prefix hashes and powers of the prime base. Then, for each possible length 'l' from 1 to the length of the current word, the solution calculates the hash of the prefix and the hash of the suffix of length 'l'. If these hashes match for both prime bases, it implies that the prefix and suffix are equal. A hash map is used to keep track of how many times each word has occurred, which helps to count the pairs efficiently. The solution avoids string comparisons and uses efficient hashing for comparing substrings.

Step-by-Step Algorithm

Step 1: Initialize two prime numbers (P1, P2) and two modulo values (M1, M2) for hashing. Initialize a hash map 'counts' to store the frequency of each word (represented by its hash values) and a variable 'ans' to store the final count of pairs.
Step 2: Iterate through each 'word' in the input 'words' array.
Step 3: Calculate prefix hashes (h1_pref, h2_pref) and powers of prime bases (p1_powers, p2_powers) for the current 'word'.
Step 4: Iterate through all possible lengths 'l' from 1 to the length of the 'word'.
Step 5: Calculate the hash of the prefix of length 'l' (h1_p, h2_p) and the hash of the suffix of length 'l' (h1_s, h2_s).
Step 6: If h1_p == h1_s and h2_p == h2_s, it means that the prefix and suffix of length 'l' are equal. Increment 'ans' by the number of times the string represented by the hash values (h1_p, h2_p) has appeared previously (using the 'counts' hash map).
Step 7: After iterating through all possible lengths 'l', store the frequency of the current 'word' (represented by its hash values (h1_pref[L], h2_pref[L])) in the 'counts' hash map.
Step 8: After iterating through all 'words', return the final count 'ans'.

Key Insights

Insight 1: Using brute-force string comparison for each pair of words would be inefficient, especially with the given constraints (up to 10^5 words, up to 10^5 length per word).
Insight 2: Rolling hash is an efficient method for determining if two substrings are equal by comparing their hash values instead of directly comparing the strings. Using two different prime numbers for the hashing helps in avoiding collision issues.
Insight 3: Pre-computing the powers of the prime base and the prefix hashes allows for efficient calculation of suffix hashes.

Complexity Analysis

Time Complexity: O(n*m^2)

Space Complexity: O(n)

Topics

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

Companies

Asked at: Autodesk, Capital One, Samsung.

Count Prefix and Suffix Pairs II | LeetCode 3045

Hard

Array String Trie Rolling Hash String Matching Hash Function

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public long countPrefixSuffixPairs(String[] words) {
        int P1 = 31, M1 = (int)1e9 + 7;
        int P2 = 37, M2 = (int)1e9 + 9;
        
        Map<Pair<Integer, Integer>, Integer> counts = new HashMap<>();
        long ans = 0;
        
        for (String word : words) {
            int L = word.length();
            
            int[] h1_pref = new int[L + 1];
            int[] h2_pref = new int[L + 1];
            int[] p1_powers = new int[L + 1];
            int[] p2_powers = new int[L + 1];
            
            p1_powers[0] = 1;
            p2_powers[0] = 1;
            
            for (int i = 0; i < L; i++) {
                int char_val = word.charAt(i) - 'a' + 1;
                h1_pref[i+1] = (int)(((long)h1_pref[i] * P1 + char_val) % M1);
                h2_pref[i+1] = (int)(((long)h2_pref[i] * P2 + char_val) % M2);
                p1_powers[i+1] = (int)(((long)p1_powers[i] * P1) % M1);
                p2_powers[i+1] = (int)(((long)p2_powers[i] * P2) % M2);
            }
            
            for (int l = 1; l <= L; l++) {
                int h1_p = h1_pref[l];
                int h2_p = h2_pref[l];
                
                int h1_s = (int)(((long)h1_pref[L] - ((long)h1_pref[L-l] * p1_powers[l]) % M1 + M1) % M1);
                int h2_s = (int)(((long)h2_pref[L] - ((long)h2_pref[L-l] * p2_powers[l]) % M2 + M2) % M2);
                
                if (h1_p == h1_s && h2_p == h2_s) {
                    Pair<Integer, Integer> key = new Pair<>(h1_p, h2_p);
                    ans += counts.getOrDefault(key, 0);
                }
            }
            
            Pair<Integer, Integer> word_key = new Pair<>(h1_pref[L], h2_pref[L]);
            counts.put(word_key, counts.getOrDefault(word_key, 0) + 1);
        }
        
        return ans;
    }

    private static class Pair<U, V> {
        private final U first;
        private final V second;

        public Pair(U first, V second) {
            this.first = first;
            this.second = second;
        }

        public U getFirst() {
            return first;
        }

        public V getSecond() {
            return second;
        }

        @Override
        public boolean equals(Object obj) {
            if (this == obj) return true;
            if (obj == null || getClass() != obj.getClass()) return false;
            Pair<?, ?> pair = (Pair<?, ?>) obj;
            return java.util.Objects.equals(first, pair.first) && java.util.Objects.equals(second, pair.second);
        }

        @Override
        public int hashCode() {
            return java.util.Objects.hash(first, second);
        }
    }
}

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Problem Info

Description

Problem 3045: Count Prefix and Suffix Pairs II - Detailed explanation and solution approaches.

Topics

Array String Trie Rolling Hash String Matching Hash Function

Companies

AutodeskCapital OneSamsung

Stats

Difficulty:Hard

Languages:4

Companies:3

Problem #:3045

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

8 steps

1
Step 1: Initialize two prime numbers (P1, P2) and two modulo values (M1, M2) for hashing. Initialize a hash map 'counts' to store the frequency of each word (represented by its hash values) and a variable 'ans' to store the final count of pairs.
2
Step 2: Iterate through each 'word' in the input 'words' array.
3
Step 3: Calculate prefix hashes (h1_pref, h2_pref) and powers of prime bases (p1_powers, p2_powers) for the current 'word'.
4
Step 4: Iterate through all possible lengths 'l' from 1 to the length of the 'word'.
5
Step 5: Calculate the hash of the prefix of length 'l' (h1_p, h2_p) and the hash of the suffix of length 'l' (h1_s, h2_s).
6
Step 6: If h1_p == h1_s and h2_p == h2_s, it means that the prefix and suffix of length 'l' are equal. Increment 'ans' by the number of times the string represented by the hash values (h1_p, h2_p) has appeared previously (using the 'counts' hash map).
7
Step 7: After iterating through all possible lengths 'l', store the frequency of the current 'word' (represented by its hash values (h1_pref[L], h2_pref[L])) in the 'counts' hash map.
8
Step 8: After iterating through all 'words', return the final count 'ans'.

💡

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^2)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is derived from the nested loops: one for iterating through each word and another for checking all possible prefix/suffix lengths. The space complexity comes from the hash map storing word frequencies and the arrays used for calculating hash values.

📖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

Insert all reversed words into a trie. For each word, traverse the trie to see if the word is also a suffix.

Time: O(N*M), where N is the number of words and M is the length of the longest word.Space: O(N*M)

Trade-offs: Trie could be memory-intensive with larger datasets but might be slightly faster due to efficient prefix matching. This approach might perform better with large datasets where many words share common prefixes. But the given solution is efficient as well.

KMP String Matching Algorithm

For each pair of words (word1, word2) check if word1 is a prefix and suffix of word2 using the KMP algorithm.

Time: O(n*m), where n is the number of words, and m is the maximum length of words. Creating the lps array for KMP is O(m). String matching will take O(m) as well. Thus in total complexity becomes O(n*m)Space: O(m). KMP algorithm will require creating lps array which will be proportional to the longest word

Trade-offs: If there is high amount of repeated computation for matching patterns, KMP might be useful, but the current solution is efficient to solve the problem for the constraint. Rolling hash solution reduces number of string comparisions and efficient comparision

✨

Key Insights

1Insight 1: Using brute-force string comparison for each pair of words would be inefficient, especially with the given constraints (up to 10^5 words, up to 10^5 length per word).
2Insight 2: Rolling hash is an efficient method for determining if two substrings are equal by comparing their hash values instead of directly comparing the strings. Using two different prime numbers for the hashing helps in avoiding collision issues.
3Insight 3: Pre-computing the powers of the prime base and the prefix hashes allows for efficient calculation of suffix hashes.

📚

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 string - The code handles empty strings correctly because the loops are designed to work even when the length is zero or one.
Edge case 2: Single character strings - The code correctly identifies prefix/suffix pairs when words consist of single characters, such as 'a' being a prefix and suffix of 'aa'.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Hash collisions: Using a single hash function without considering collisions can lead to incorrect results. The solution mitigates this by using two different hash functions with different prime numbers and modulo values.

✗Common mistake 2: Integer overflow: The rolling hash calculation can potentially result in integer overflow. The solution prevents this by using modulo operations during the calculation.

✅

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: Detecting plagiarism: Identifying if one document contains sections that are both prefixes and suffixes of another document can be useful for plagiarism detection.

🏦Real-world scenario 2: Bioinformatics: Identifying repeating patterns or motifs within DNA sequences where certain sequences act as both prefixes and suffixes of larger sequences.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

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

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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 13, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

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

@InterviewProJun 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.

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.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Count Prefix and Suffix Pairs II"?

For this hard-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?