L17. Palindrome Partitioning | Leetcode | Recursion | C++ | Java

Main Ideas / Concepts

• Problem Definition (Palindrome Partitioning):

  • Given a string s, partition it into substrings such that every substring is a palindrome.
  • Return all possible palindrome partitions for the string.

• Palindrome Concept:

  • A palindrome reads the same forward and backward.

• Example Discussed:

  • For a string conceptually shaped like a a b b, the transcript examples indicate multiple valid partitions, such as:
    • [aa, bb]
    • [a, a, bb]
    • [aa, b, b]
    • [a, a, b, b]
  • Key lesson: you must consider partitions at multiple cut positions, but only those where each resulting substring is a palindrome are valid.

• Why Recursion Is Used:

  • Partitioning repeatedly (choosing a cut, then partitioning the remaining suffix) naturally forms a recursive tree.
  • The approach explores:
    1. where the first cut can be,
    2. then recursively where the next cut can be,
    3. until the end of the string.

• Backtracking Concept:

  • While building a current partition (“path”), once a branch completes, you must undo the last chosen substring so the next branch can try a different cut.

• Core Structure:

  • A recursion that maintains:
    • an index (current starting position),
    • a path (chosen substrings so far),
    • a result (all completed valid partitions).

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Methodology / Instructions (As Implied)

1) Recursive Function Design

• Function parameters / state:

  • index: starting position in s for the next substring to choose
  • path: list of palindrome substrings chosen so far
  • result (or rest): list of all complete partitions (list of lists of strings)

• Base case:

  • If index == n (where n = s.length):
    • the partition is complete
    • add the current path to result
    • return

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2) Transition / Recursion Step (Main Loop)

• For each candidate end position i from index to n - 1:
  • Consider the substring s[index..i]
  • Palindrome check:
    • Only proceed if s[index..i] is a palindrome
  • If it is a palindrome:
    • append s[index..i] to path
    • recursively call with index = i + 1 (next cut starts right after this substring)
    • after returning, backtrack:
      • remove the last appended substring from path (pop)

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3) Palindrome Checking Helper

• Use a helper function to determine whether s[l..r] is a palindrome:
  • Employ two pointers: l (left) and r (right)
  • Move inward while l < r
  • If s[l] != s[r], return false
  • If all checks pass, return true

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4) Output

• After exploring all recursion branches:
  • return result, which contains all valid palindrome partitions

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Speakers / Sources Featured

• Speaker: The video’s presenter/teacher (name not provided in the subtitles).
• Sponsor / source mentioned: Unacademy
  • Instructors mentioned for an Unacademy batch:
    • Deepp Gaur
    • Ria Bunse (spelling may differ from the transcript)
• Referenced platform/company: Flipkart (mentioned via an instructor’s employment reference in the sponsor segment).