Palindrome Partitioning LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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In this post, you will find the solution for the Palindrome Partitioning in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the Problem – Palindrome Partitioning– LeetCode Problem

Palindrome Partitioning– LeetCode Problem

Problem:

Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s.

A palindrome string is a string that reads the same backward as forward.

Example 1:

Input: s = "aab"
Output: [["a","a","b"],["aa","b"]]

Example 2:

Input: s = "a"
Output: [["a"]]

Constraints:

• 1 <= s.length <= 16
• s contains only lowercase English letters.

Palindrome Partitioning– LeetCode Solutions

Palindrome Partitioning Solution in C++:

class Solution {
 public:
  vector<vector<string>> partition(string s) {
    vector<vector<string>> ans;

    dfs(s, 0, {}, ans);

    return ans;
  }

 private:
  void dfs(const string& s, int start, vector<string>&& path,
           vector<vector<string>>& ans) {
    if (start == s.length()) {
      ans.push_back(path);
      return;
    }

    for (int i = start; i < s.length(); ++i)
      if (isPalindrome(s, start, i)) {
        path.push_back(s.substr(start, i - start + 1));
        dfs(s, i + 1, move(path), ans);
        path.pop_back();
      }
  }

  bool isPalindrome(const string& s, int l, int r) {
    while (l < r)
      if (s[l++] != s[r--])
        return false;
    return true;
  }
};

Palindrome Partitioning Solution in Java:

class Solution {
  public List<List<String>> partition(String s) {
    List<List<String>> ans = new ArrayList<>();

    dfs(s, 0, new ArrayList<>(), ans);

    return ans;
  }

  private void dfs(final String s, int start, List<String> path, List<List<String>> ans) {
    if (start == s.length()) {
      ans.add(new ArrayList<>(path));
      return;
    }

    for (int i = start; i < s.length(); ++i)
      if (isPalindrome(s, start, i)) {
        path.add(s.substring(start, i + 1));
        dfs(s, i + 1, path, ans);
        path.remove(path.size() - 1);
      }
  }

  private boolean isPalindrome(final String s, int l, int r) {
    while (l < r)
      if (s.charAt(l++) != s.charAt(r--))
        return false;
    return true;
  }
}

Palindrome Partitioning Solution in Python:

class Solution:
  def partition(self, s: str) -> List[List[str]]:
    ans = []

    def isPalindrome(s: str) -> bool:
      return s == s[::-1]

    def dfs(s: str, j: int, path: List[str], ans: List[List[str]]) -> None:
      if j == len(s):
        ans.append(path)
        return

      for i in range(j, len(s)):
        if isPalindrome(s[j: i + 1]):
          dfs(s, i + 1, path + [s[j: i + 1]], ans)

    dfs(s, 0, [], ans)

    return ans