Palindrome Partitioning Using Backtracking in DSA Typescript - Build from Scratch

Palindrome Partitioning Using Backtracking

📖 Scenario: Imagine you have a word, and you want to split it into parts where each part reads the same forwards and backwards. This is called palindrome partitioning. For example, the word "noon" can be split into ["n", "oo", "n"] or ["noon"].

🎯 Goal: You will build a program that finds all the ways to split a given word into palindrome parts using backtracking. This helps understand how to explore all possible splits and check palindrome conditions step-by-step.

📋 What You'll Learn

Create a string variable with a specific word

Create a helper function to check if a substring is a palindrome

Use backtracking to find all palindrome partitions

Print all palindrome partitions as arrays of strings

💡 Why This Matters

🌍 Real World

Palindrome partitioning helps in text processing tasks like DNA sequence analysis, data compression, and pattern recognition where symmetrical patterns matter.

💼 Career

Understanding backtracking and palindrome checks is useful for software engineers working on algorithms, coding interviews, and solving complex recursive problems.

1

Create the input string

Create a string variable called input and set it to the value "aab".

DSA Typescript

# Create a string variable called input and set it to "aab"
// Your code here

Hint

Use const input: string = "aab"; to create the string.

2

Create a palindrome check function

Create a function called isPalindrome that takes a string s and returns true if s is a palindrome, otherwise false. Use a simple loop to compare characters from start and end.

DSA Typescript

const input: string = "aab";
// Create function isPalindrome(s: string): boolean
// Your code here

Hint

Use two pointers from start and end to check characters. Return false if mismatch found, else true.

3

Implement backtracking to find palindrome partitions

Create a function called backtrack that takes start: number, path: string[], and result: string[][]. Use a for loop from start to input.length. For each substring from start to i + 1, check if it is a palindrome using isPalindrome. If yes, add substring to path, call backtrack recursively with i + 1, then remove last element from path. If start equals input.length, push a copy of path to result.

DSA Typescript

const input: string = "aab";

function isPalindrome(s: string): boolean {
    let left = 0;
    let right = s.length - 1;
    while (left < right) {
        if (s[left] !== s[right]) {
            return false;
        }
        left++;
        right--;
    }
    return true;
}

// Create function backtrack(start: number, path: string[], result: string[][])
// Your code here

Hint

Use recursion and loop to explore all substring splits. Add palindrome substrings to path and backtrack.

4

Print all palindrome partitions

Create an empty array called result of type string[][]. Call backtrack with 0, an empty array, and result. Then print result using console.log(result).

DSA Typescript

const input: string = "aab";

function isPalindrome(s: string): boolean {
    let left = 0;
    let right = s.length - 1;
    while (left < right) {
        if (s[left] !== s[right]) {
            return false;
        }
        left++;
        right--;
    }
    return true;
}

function backtrack(start: number, path: string[], result: string[][]): void {
    if (start === input.length) {
        result.push([...path]);
        return;
    }
    for (let i = start; i < input.length; i++) {
        const substring = input.substring(start, i + 1);
        if (isPalindrome(substring)) {
            path.push(substring);
            backtrack(i + 1, path, result);
            path.pop();
        }
    }
}

// Create result array and call backtrack, then print result
// Your code here

Hint

Initialize an empty array, call backtrack starting at 0, then print the result array.

Practice

1. What is the main goal of palindrome partitioning using backtracking?

easy

A. To reverse the string and check if it is a palindrome

B. To find the longest palindrome substring in a string

C. To count the number of palindromes in a string

D. To split a string into all possible palindrome substrings

Solution

1. Step 1: Understand palindrome partitioning

   Palindrome partitioning means dividing a string into parts where each part is a palindrome.

2. Step 2: Role of backtracking

   Backtracking helps explore all possible ways to split the string into palindrome parts.

3. Final Answer:

   To split a string into all possible palindrome substrings -> Option D

4. Quick Check:

   Palindrome partitioning = all palindrome splits [OK]

Hint: Backtracking explores all palindrome splits step-by-step [OK]

Common Mistakes:

• Confusing palindrome partitioning with longest palindrome substring
• Thinking it only counts palindromes instead of listing partitions
• Assuming it just reverses the string

2. Which of the following TypeScript function signatures correctly represents a backtracking function for palindrome partitioning?

easy

A. function backtrack(s: number[], start: string, path: string[], result: string[][]): string

B. function backtrack(s: string, start: number, path: string, result: string[]): boolean

C. function backtrack(s: string, start: number, path: string[], result: string[][]): void

D. function backtrack(s: string[], start: number, path: string[], result: string[][]): number

Solution

1. Step 1: Analyze parameter types

   The string to partition is 's: string', start index is 'number', path is an array of strings, and result is an array of string arrays.

2. Step 2: Check return type

   Backtracking functions usually return void because they build results by modifying the result array.

3. Final Answer:

   function backtrack(s: string, start: number, path: string[], result: string[][]): void -> Option C

4. Quick Check:

   Correct signature = function backtrack(s: string, start: number, path: string[], result: string[][]): void [OK]

Hint: Backtracking returns void and uses arrays for path and result [OK]

Common Mistakes:

• Using wrong parameter types like number[] instead of string
• Returning boolean or number instead of void
• Passing path as string instead of string[]

3. What will be the output of the following TypeScript code snippet for palindrome partitioning of string "aab"?

function isPalindrome(s: string): boolean {
  return s === s.split('').reverse().join('');
}

function backtrack(s: string, start: number, path: string[], result: string[][]): void {
  if (start === s.length) {
    result.push([...path]);
    return;
  }
  for (let end = start + 1; end <= s.length; end++) {
    const substr = s.substring(start, end);
    if (isPalindrome(substr)) {
      path.push(substr);
      backtrack(s, end, path, result);
      path.pop();
    }
  }
}

const result: string[][] = [];
backtrack("aab", 0, [], result);
console.log(result);

medium

A. [["aa", "ab"]]

B. [["a", "a", "b"], ["aa", "b"]]

C. [["a", "ab"]]

D. [["aab"]]

Solution

1. Step 1: Identify palindrome substrings in "aab"

   "a" and "aa" are palindromes; "b" is a palindrome; "ab" and "aab" are not.

2. Step 2: Trace backtracking calls

   Partitions found are ["a", "a", "b"] and ["aa", "b"] added to result.

3. Final Answer:

   [["a", "a", "b"], ["aa", "b"]] -> Option B

4. Quick Check:

   All palindrome splits of "aab" = [["a", "a", "b"], ["aa", "b"]] [OK]

Hint: Check all palindrome substrings and backtrack all splits [OK]

Common Mistakes:

• Including non-palindrome substrings like "ab"
• Missing the "aa" substring partition
• Returning only one partition instead of all

4. Identify the bug in the following backtracking code for palindrome partitioning and select the fix:

function isPalindrome(s: string): boolean {
  return s === s.split('').reverse().join('');
}

function backtrack(s: string, start: number, path: string[], result: string[][]): void {
  if (start > s.length) {
    result.push([...path]);
    return;
  }
  for (let end = start + 1; end <= s.length; end++) {
    const substr = s.substring(start, end);
    if (isPalindrome(substr)) {
      path.push(substr);
      backtrack(s, end, path, result);
      path.pop();
    }
  }
}

medium

A. Change condition to if (start === s.length) to stop recursion correctly

B. Change substring to slice for correct substring extraction

C. Remove path.pop() to keep all partitions

D. Change for loop to end < s.length to avoid out of bounds

Solution

1. Step 1: Analyze base case condition

   Condition 'start > s.length' allows start to be s.length + 1, which is invalid index.

2. Step 2: Correct base case for recursion end

   It should be 'start === s.length' to indicate full string processed and add path to result.

3. Final Answer:

   Change condition to if (start === s.length) to stop recursion correctly -> Option A

4. Quick Check:

   Base case must be start === s.length [OK]

Hint: Base case triggers when start equals string length [OK]

Common Mistakes:

• Using > instead of === in base case
• Removing path.pop() causing wrong partitions
• Changing substring method unnecessarily

5. Given the string "racecar", which of the following is NOT a valid palindrome partitioning produced by a backtracking algorithm?

hard

A. ["rac", "ecar"]

B. ["racecar"]

C. ["r", "a", "c", "e", "c", "a", "r"]

D. ["r", "aceca", "r"]

Solution

1. Step 1: Check palindrome validity of each substring

   "rac" and "ecar" are not palindromes; others are single letters or full palindrome.

2. Step 2: Confirm valid partitions

   Partitions with only palindrome substrings are valid; those with non-palindromes are invalid.

3. Final Answer:

   ["rac", "ecar"] -> Option A

4. Quick Check:

   Non-palindrome substrings invalidate partition [OK]

Hint: Check each substring is palindrome to validate partition [OK]

Common Mistakes:

• Assuming any split is valid without palindrome check
• Confusing full string palindrome with partial substrings
• Ignoring single character substrings as palindromes