Two Pointers - Word Abbreviation

Estimated Time: 2 hours
Tech Stack: Java
Keywords: Data Structure - Algorithms
Experience Level: Beginner - Advanced
Category: TwoPointers
Skill: Parsing Numbers; Using Two Pointers on two different strings.

408. Valid Word Abbreviation

A string can be abbreviated by replacing any number of non-adjacent, non-empty substrings with their lengths. The lengths should not have leading zeros. For example, a string such as "substitution" could be abbreviated as (but not limited to):

• "s10n" ("s ubstitutio n")

• "sub4u4" ("sub stit u tion")

• "12" ("substitution")

• "su3i1u2on" ("su bst i t u ti on")

• "substitution" (no substrings replaced)

• The following are not valid abbreviations:

• "s55n" ("s ubsti tutio n", the replaced substrings are adjacent)

• "s010n" (has leading zeros)

• "s0ubstitution" (replaces an empty substring)

Given a string word and an abbreviation abbr, return whether the string matches the given abbreviation. A substring is a contiguous non-empty sequence of characters within a string.

Example 1:

Input: word = "internationalization", abbr = "i12iz4n" Output: true Explanation: The word "internationalization" can be abbreviated as "i12iz4n" ("i nternational iz atio n").

Example 2:

Input: word = "apple", abbr = "a2e" Output: false Explanation: The word "apple" cannot be abbreviated as "a2e".

Constraints:

1 <= word.length <= 20 word consists of only lowercase English letters. 1 <= abbr.length <= 10 abbr consists of lowercase English letters and digits. All the integers in abbr will fit in a 32-bit integer.

Solution:

package TwoPointers;

public class dAbbreviationValidator {

    public static boolean validWordAbbreviation(String word, String abbr) {
        int i = 0; // pointer for word
        int j = 0; // pointer for abbr

        while (i < word.length() && j < abbr.length()) {
            char current = abbr.charAt(j);

            // 🟠 Case 1: Digit in abbreviation
            if (Character.isDigit(current)) {
                if (current == '0') {
                    return false; // 🚫 Leading zero
                }

                int num = 0;
                while (j < abbr.length() && Character.isDigit(abbr.charAt(j))) {
                    num = num * 10 + (abbr.charAt(j) - '0');
                    j++;
                }

                i += num; // Skip characters in word
            }

            // 🟢 Case 2: Letter in abbreviation
            else {
                if (i >= word.length() || word.charAt(i) != current) {
                    return false; // 🚫 Mismatch
                }
                i++;
                j++;
            }
        }

        // ✅ Both pointers must reach the end
        return i == word.length() && j == abbr.length();
    }

    public static void main(String[] args) {
        System.out.println(validWordAbbreviation("internationalization", "i12iz4n")); // true
        System.out.println(validWordAbbreviation("apple", "a2e")); // false
        System.out.println(validWordAbbreviation("substitution", "s10n")); // true
        System.out.println(validWordAbbreviation("substitution", "s010n")); // false (leading zero)
        System.out.println(validWordAbbreviation("substitution", "s0ubstitution")); // false (empty substring)
    }
}

🔍 Class Header

package AdHoc;

🗣️ Say it out loud:

“This class is part of the TwoPointers package.”

📘 Explanation:

Organizes your code inside a named package — like putting a file in a folder.

public class AbbreviationValidator {

🗣️ Say it out loud:

“This is a public class named dAbbreviationValidator.”

📘 Explanation:

Defines the class where your logic and main method will live.

🧠 Method Declaration

public static boolean validWordAbbreviation(String word, String abbr) {

🗣️ Say it out loud:

“Define a public static method called validWordAbbreviation that takes a word and an abbreviation, and returns true or false.”

📘 Explanation:

This method checks if abbr is a valid abbreviation for word.

🧮 Pointers Initialization

int i = 0; // pointer for word
int j = 0; // pointer for abbr

🗣️ Say it out loud:

“Create two integer pointers i and j, both starting at zero.”

📘 Explanation:

i tracks position in word, j tracks position in abbr.

🔁 Loop Through Characters

while (i < word.length() && j < abbr.length()) {

🗣️ Say it out loud:

“While both pointers are within bounds of their strings, keep looping.”

📘 Explanation:

Loop continues as long as we haven’t finished parsing both strings.

🔎 Get the Current Abbreviation Character

char current = abbr.charAt(j);

🗣️ Say it out loud:

“Get the current character in abbr at position j.”

📘 Explanation:

Stores the character to decide if it’s a letter or digit.

🔠 If the Current Character is a Digit

if (Character.isDigit(current)) {

🗣️ Say it out loud:

“If the character is a digit...”

📘 Explanation:

Abbreviations can include numbers that tell us how many characters to skip in the word.

🛑 Check for Leading Zero

if (current == '0') {
    return false;
}

🗣️ Say it out loud:

“If the digit is zero, return false.”

📘 Explanation:

Leading zeros like 01, 007 are not valid — skip is not allowed to begin with zero.

🔢 Parse the Number

int num = 0;
while (j < abbr.length() && Character.isDigit(abbr.charAt(j))) {
    num = num * 10 + (abbr.charAt(j) - '0');
    j++;
}

🗣️ Say it out loud:

“While we’re still seeing digits, build the full number and move j forward.”

📘 Explanation:

This handles multi-digit numbers (e.g., "12"), and moves the j pointer past all digits.

🧪 Walkthrough 1: Parsing "15"

Let's say abbr = "a15b" and we're starting at index j = 1.

✅ Final result: num = 15

⏩ Skip Characters in Word

i += num;

🗣️ Say it out loud:

“Move the word pointer forward by that number.”

📘 Explanation:

You're skipping over num characters in the original word — as instructed by the abbreviation.

🔤 If It's a Letter

} else {
    if (i >= word.length() || word.charAt(i) != current) {
        return false;
    }
    i++;
    j++;
}

🗣️ Say it out loud:

“Otherwise, if it’s a letter, compare it with the current letter in word.”

📘 Explanation:

If the letters don’t match, it’s invalid. If they do, move both pointers forward.

✅ Final Check

return i == word.length() && j == abbr.length();

🗣️ Say it out loud:

“Return true only if both pointers reached the end of their strings.”

📘 Explanation:

We’re only done if both strings are fully parsed — no leftovers.

🧪 Test Code

public static void main(String[] args) {

🗣️ Say it out loud:

“Main method — this is where the program starts.”

📘 Explanation:

This is your test area to run the validator with different inputs.

System.out.println(validWordAbbreviation("internationalization", "i12iz4n")); // true
System.out.println(validWordAbbreviation("apple", "a2e")); // false
System.out.println(validWordAbbreviation("substitution", "s10n")); // true
System.out.println(validWordAbbreviation("substitution", "s010n")); // false
System.out.println(validWordAbbreviation("substitution", "s0ubstitution")); // false

🗣️ Say it out loud:

“Call the method with different test cases and print the results.”

📘 Explanation:

Validates both correct and incorrect abbreviations to confirm your logic works as expected.

✅ Output

true
false
true
false
false

✏️ Challenges in the Word Abbreviation Problem

🧠 You're doing something that few learners take the time to do — extracting wisdom from experience. Let’s now reflect on the challenges of the Word Abbreviation Validation problem.

🧠 Bonus Challenge: Manual character-by-character parsing

Parsing a number from a string without built-in parsing teaches you how characters and digits relate via ASCII math, which is an advanced beginner skill you now have!

✅ Summary

💡 Reflection Insight

This problem builds precision, pointer mastery, and string parsing fluency — a combo that makes you dangerous in string-heavy DSA challenges (like wildcard matching, regex parsing, or expression evaluation). You're building a Jedi toolkit, shinobi 🥷📜🔥