Chapter 3: FAANG-Style String Problems: Interview Prep for Top Tech Companies

Welcome to Chapter 3 of our DSA in Java journey! Today, we’re diving into strings—one of the most crucial topics for acing coding interviews. Mastering string problems is essential for success in technical interviews, as they test a wide range of skills from problem-solving to optimization.

Strings aren’t just simple sequences of characters—they serve as the foundation for some of the most challenging and creative algorithmic problems. From detecting patterns to handling large datasets efficiently, string problems help you sharpen your problem-solving abilities.

In this chapter, we’ll explore key string concepts, work through essential interview problems, and provide you with the tools to tackle any string-related challenge with confidence.

3.1:Introduction to Strings: The Backbone of Textual Data in Programming

In programming, a string is one of the most fundamental data types, representing a sequence of characters. Strings are everywhere—from storing user names and passwords to displaying error messages and processing large amounts of textual data.

3.2:What is a String?

A string is essentially a collection of characters. These characters can include letters, numbers, symbols, and even whitespace. Strings are enclosed in quotation marks, and depending on the programming language, these can be:

• Single quotes: 'Hello'

• Double quotes: "World"

• Triple quotes (in some languages like Python): '''Multiline String'''

3.3:Why Strings are Important.

Strings are used in countless ways, making them indispensable in programming. Whether you're designing a user interface, querying a database, or analyzing text, strings are at the core. Here are some common use cases:

1. User Input and Output: Displaying messages or accepting user responses.

2. Data Storage: Names, addresses, or any textual information.

3. Manipulation and Parsing: Processing and transforming text, such as splitting a sentence into words.

3.4:Key Features of Strings.

1. Immutable Nature:
   In many programming languages, strings are immutable. This means once you create a string, its contents cannot be altered directly. For example:

    javaCopy codeString str = "Hello";
    str = "World"; // Creates a new string; the original "Hello" remains unchanged.
   

2. Index-Based Access:
   Strings are indexed, allowing you to access individual characters based on their position. Most programming languages use zero-based indexing:

    javaCopy codeString name = "John";
    System.out.println(name.charAt(0)); // Output: J
   

3. Built-in Methods:
   Strings come with a suite of methods to make text processing easier, such as concatenation, case conversion, and searching:

    javaCopy codeString greeting = "Hello";
    greeting = greeting.concat(" World!");  
    System.out.println(greeting); // Output: Hello World!
   

Example: Working with Strings in Java.

Let’s consider a simple example to illustrate string operations in Java:

javaCopy codepublic class Main {
    public static void main(String[] args) {
        String message = "Hello, World!";
        System.out.println("Original Message: " + message);
        System.out.println("Length: " + message.length());
        System.out.println("Uppercase: " + message.toUpperCase());
        System.out.println("Substring: " + message.substring(7, 12));
    }
}

Output:

Original Message: Hello, World!  
Length: 13  
Uppercase: HELLO, WORLD!  
Substring: World

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3. Common String Problems in Interviews.

Problem 1:Reverse Words in a String.

Problem Statement:

Given a string, reverse the words in the string while preserving spaces between them. You need to remove leading and trailing spaces before performing the reversal.

Example:

• Input: " Java programming is fun "

• Output: "fun is programming Java"

Approach:

• Trim leading and trailing spaces from the string (if any).

• Split the string into words. This can be done using Java's split() method which splits the string based on spaces.

• Reverse the words. After splitting the string into words, reverse the order of words in the list or array.

• Join the words back together with a single space between each word.

Java Code Implementation:

public class ReverseWordsInString {
    public static String reverseWords(String s) {
        // Step 1: Trim leading and trailing spaces
        s = s.trim();

        // Step 2: Split the string by spaces into an array of words
        String[] words = s.split("\\s+");

        // Step 3: Reverse the array of words
        StringBuilder reversedString = new StringBuilder();
        for (int i = words.length - 1; i >= 0; i--) {
            reversedString.append(words[i]);
            if (i != 0) {
                reversedString.append(" ");
            }
        }

        // Step 4: Return the reversed string
        return reversedString.toString();
    }

    public static void main(String[] args) {
        String input = "  Java   programming   is fun  ";
        String output = reverseWords(input);
        System.out.println(output);  // Output: "fun is programming Java"
    }
}

Time Complexity:

• O(n), where n is the length of the string. We traverse the string to split it into words, reverse the words, and join them back together, all of which take linear time.

Space Complexity:

• O(n), due to the space required for storing the words in the array and the final reversed string.

Example Walkthrough:

1. Input string: " Java programming is fun "

2. Trimmed string: "Java programming is fun"

3. Words after splitting: ["Java", "programming", "is", "fun"]

4. Words after reversing: ["fun", "is", "programming", "Java"]

5. Final output after joining: "fun is programming Java"

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Problem 2:Check if a String is a Palindrome .

Problem Statement:

Given a string, your task is to determine whether it reads the same backward as forward. If the string is a palindrome, return true; otherwise, return false.

Example:

• Input: "racecar"

• Output: true

• Input: "hello"

• Output: false

Approach:

To solve this problem, we can use the following steps:

1. Check for Palindrome:

   • Compare characters from the beginning and end of the string.

   • If all corresponding characters match, return true.

   • If any character does not match, return false.

Java Code Implementation:

public class Palindrome {
    public static boolean isPalindrome(String str) {
        // Step 1: Compare characters from both ends
        int n = str.length();
        for (int i = 0; i < n / 2; i++) {
            if (str.charAt(i) != str.charAt(n - 1 - i)) {
                return false;
            }
        }

        // Step 2: If all characters match, it's a palindrome
        return true;
    }

    public static void main(String[] args) {
        String str = "Madam";
        System.out.println(isPalindrome(str));  // Output: True
    }
}

Output: true

Explanation Of Code:

1. Palindrome Check:

   • We use a for loop to compare characters from the beginning and end of the string.

   • If the characters at any position don't match, the string is not a palindrome, and we return false.

   • If we successfully loop through half of the string and find no mismatches, we return true.

2. Main Method:

   • In the main() method, we pass the string "A man a plan a canal Panama" to the isPalindrome method and print the result.

Time Complexity:

• O(n), where n is the length of the string. The loop iterates through half of the string (since we're comparing characters from both ends).

Space Complexity:

• O(1), as we are not using any extra space for modifications (other than the string itself).

Example Walkthrough:

Example 1:

• Input: "A man a plan a canal Panama"

  • Step 1 (Compare Characters):

    • A does not match a, so the string is not a palindrome.

  • Step 2 (Return Result): The string is not a palindrome, so the output is false.

Example 2:

• Input: "hello"

  • Step 1 (Compare Characters):

    • h does not match o, so the string is not a palindrome.

  • Step 2 (Return Result): The string is not a palindrome, so the output is false.

Example 3:

• Input: "" (empty string)

  • Step 1 (Compare Characters): No characters to compare.

  • Step 2 (Return Result): An empty string is considered a palindrome, so the output is true.

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Problem 3:Zigzag Conversion.

Problem Statement:

Given a string s and an integer numRows, convert the string s into a zigzag pattern on numRows rows. Then, return the string that is read from top to bottom, left to right, line by line.

Example:

Consider the input string "PAYPALISHIRING" and numRows = 3. The string would be arranged in a zigzag pattern across 3 rows as follows:

P   A   H   N
A P L S I I G
Y   I   R

Reading the string from top to bottom, left to right, row by row, we get the result: "PAHNAPLSIIGYIR".

Approach:

To solve this problem, we will:

1. Use an array of StringBuilder objects to represent each row of the zigzag pattern.

2. Traverse through the input string s character by character and simulate the zigzag pattern:

   • Move down when reaching the bottom row.

   • Move up when reaching the top row.

3. After placing all characters in the rows, concatenate the contents of each row and return the resulting string.

Java Code Implementation:

public class ZigzagConversion {

    public String convert(String s, int numRows) {
        // If numRows is 1 or numRows is greater than or equal to the string length, return the string as is.
        if (numRows == 1 || numRows >= s.length()) {
            return s;
        }

        // Create an array of StringBuilder to store characters for each row
        StringBuilder[] rows = new StringBuilder[numRows];
        for (int i = 0; i < numRows; i++) {
            rows[i] = new StringBuilder();
        }

        // Variables to keep track of the current row and the direction (down or up)
        int currentRow = 0;
        boolean goingDown = false;

        // Traverse through the string and place characters in the appropriate rows
        for (char c : s.toCharArray()) {
            rows[currentRow].append(c);

            // Change direction if we are at the first or last row
            if (currentRow == 0 || currentRow == numRows - 1) {
                goingDown = !goingDown;
            }

            // Move up or down based on the direction
            currentRow += goingDown ? 1 : -1;
        }

        // Combine all the rows to form the final string
        StringBuilder result = new StringBuilder();
        for (StringBuilder row : rows) {
            result.append(row);
        }

        return result.toString();
    }

    public static void main(String[] args) {
        ZigzagConversion zigzag = new ZigzagConversion();
        String input = "PAYPALISHIRING";
        int numRows = 3;
        System.out.println(zigzag.convert(input, numRows));  // Output: "PAHNAPLSIIGYIR"
    }
}

Output: PAHNAPLSIIGYIR

Explanation Of Code:

1. Edge Case Handling:

   • If numRows is 1 or greater than or equal to the length of the string, the string remains unchanged. There's no need for zigzag conversion in these cases.

2. StringBuilder Array:

   • We create an array of StringBuilder objects, where each element represents a row of the zigzag pattern.

   • The size of the array is equal to numRows, which is the number of rows in the zigzag pattern.

3. Traversing the String:

   • We iterate over each character in the string and append it to the appropriate row in the StringBuilder array.

   • We track the direction using a boolean variable goingDown. If we're at the top row (currentRow == 0) or the last row (currentRow == numRows - 1), we switch direction.

4. Combining Rows:

   • After all characters are placed in the rows, we concatenate the content of each row to form the final result.

Time Complexity:

• The time complexity is O(n), where n is the length of the input string. We iterate through the string once and perform constant-time operations for each character.

Space Complexity:

• The space complexity is O(n) because we are storing the result in an array of StringBuilder objects, each of which contains part of the final string.

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Problem 4: Valid Anagram.

Problem Overview:

The problem asks us to determine whether two given strings, s and t, are anagrams of each other.

What is an Anagram?

Two strings are considered anagrams if:

• They contain the same characters.

• Each character appears the same number of times in both strings.

• The order of the characters may differ.

For example:

• "anagram" and "nagaram" are anagrams because they contain the same characters in the same frequencies.

• "rat" and "car" are not anagrams because the characters differ in both strings.

Problem Statement:

Given two strings s and t, return true if t is an anagram of s, and false otherwise.

Example 1:

• Input: s = "anagram", t = "nagaram"

• Output: true

Example 2:

• Input: s = "rat", t = "car"

• Output: false

Approach:

A more efficient solution involves counting the frequency of each character in both strings using an array (for lowercase English letters, we can use a fixed-size array of 26 elements). If both strings have the same character frequencies, they are anagrams.

Java Code Implementation:

public class ValidAnagram {

    public boolean isAnagram(String s, String t) {
        // If lengths don't match, they can't be anagrams
        if (s.length() != t.length()) {
            return false;
        }

        // Frequency count array for characters a-z (26 letters)
        int[] count = new int[26];

        // Iterate through both strings and count the frequency of each character
        for (int i = 0; i < s.length(); i++) {
            count[s.charAt(i) - 'a']++;
            count[t.charAt(i) - 'a']--;
        }

        // If all counts are 0, then the strings are anagrams
        for (int c : count) {
            if (c != 0) {
                return false;
            }
        }

        return true;
    }

    public static void main(String[] args) {
        ValidAnagram anagram = new ValidAnagram();
        System.out.println(anagram.isAnagram("anagram", "nagaram"));  // Output: true
        System.out.println(anagram.isAnagram("rat", "car"));  // Output: false
    }
}

Explanation:

• Step 1: If the lengths of the strings are not equal, return false.

• Step 2: Create an integer array count of size 26 to count the frequency of each character in both strings.

• Step 3: Iterate over both strings simultaneously, incrementing the count for characters in string s and decrementing the count for characters in string t.

• Step 4: If all values in the count array are zero after processing both strings, then the strings are anagrams. Otherwise, they are not.

Time Complexity:

• The time complexity is O(n), where n is the length of the strings, because we iterate over the strings once to count characters.

Space Complexity:

• The space complexity is O(1), because the space required for the frequency array is constant (26 elements for lowercase English letters).

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Problem 5:Longest Substring Without Repeating Characters

Problem Statement:

Given a string s, find the length of the longest substring without repeating characters.

Example 1:

• Input: s = "abcabcbb"

• Output: 3

• Explanation: The answer is "abc", with a length of 3.

Example 2:

• Input: s = "bbbbb"

• Output: 1

• Explanation: The answer is "b", with a length of 1.

Approach:

To efficiently solve the problem, we can use the sliding window technique with the help of a set or a hash map to track characters in the current substring.

Key Idea:

• Use two pointers, left and right, to define the current window.

• Expand the window by moving the right pointer and add characters to the set or map.

• If a repeating character is found, shrink the window from the left until there are no repeats.

• Track the maximum length of the substring seen so far.

Java code Implementation:

import java.util.HashSet;

public class LongestSubstringWithoutRepeating {
    public int lengthOfLongestSubstring(String s) {
        // Initialize a set to track characters in the current substring
        HashSet<Character> set = new HashSet<>();

        // Initialize two pointers and a variable to store the max length
        int left = 0, maxLength = 0;

        // Expand the window using the right pointer
        for (int right = 0; right < s.length(); right++) {
            char current = s.charAt(right);

            // If the character is already in the set, shrink the window
            while (set.contains(current)) {
                set.remove(s.charAt(left));
                left++;
            }

            // Add the current character to the set
            set.add(current);

            // Update the max length of the substring
            maxLength = Math.max(maxLength, right - left + 1);
        }

        return maxLength;
    }

    public static void main(String[] args) {
        LongestSubstringWithoutRepeating solution = new LongestSubstringWithoutRepeating();
        System.out.println(solution.lengthOfLongestSubstring("abcabcbb")); // Output: 3
        System.out.println(solution.lengthOfLongestSubstring("bbbbb"));    // Output: 1

    }
}

Output: 3

Time Complexity:

• O(n): Each character is visited at most twice (once by right and once by left).

Space Complexity:

• O(k): Where k is the size of the character set. For English lowercase letters, k=26.

Example Walkthrough

Input: "abcabcbb"

• Initial state: left = 0, right = 0, set = {}, maxLength = 0

• Iteration 1: Add 'a', set = {a}, maxLength = 1

• Iteration 2: Add 'b', set = {a, b}, maxLength = 2

• Iteration 3: Add 'c', set = {a, b, c}, maxLength = 3

• Iteration 4: Duplicate 'a', remove characters until 'a' is removed, set = {b, c}, left = 1

• Continue similarly...

Result: maxLength = 3.

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