Had an interview with the team that works on opeartor fustion.

Failed this first round interview.

I only gave very naive solution. Not really sad.

I am well prepared for kinds of outcomes.

Problem:

given two strings a and b with same length n, you can partition two strings at the same place for two strings into a1, a2 and b1 b2, ie. length of a1 is equal to length of b1 and length of a2 is equal to the length of b2. please give a algorithm to test if there is a partition that can form a concat string such at a1 + b2 or b1 + a2 is palindrome string

corresponding leetcode problem:

https://leetcode.com/problems/split-two-strings-to-make-palindrome/solutions/888885/c-java-greedy-o-n-o-1/

Certainly! Here’s a step-by-step algorithm to test if there is a partition that can form a palindrome by concatenating ( a1 + b2 ) or ( b1 + a2 ):

Algorithm

1. Input: Two strings a and b of length n.
2. Output: A boolean value indicating if there exists a partition point where ( a1 + b2 ) or ( b1 + a2 ) is a palindrome.

Steps:

1. Initialization:
   • Loop through each possible partition point i from 0 to n.
2. Check Partitions:
   • For each partition point i, split the strings into a1, a2, b1, and b2.
     • ( a1 = a[0:i] )
     • ( a2 = a[i:n] )
     • ( b1 = b[0:i] )
     • ( b2 = b[i:n] )
3. Check Palindromes:
   • Concatenate a1 + b2 and check if it’s a palindrome.
   • Concatenate b1 + a2 and check if it’s a palindrome.
   • If either concatenation is a palindrome, return True.
4. Return False:
   • If no valid palindrome is found after all partitions, return False.

Pseudocode

def is_palindrome(s):
    return s == s[::-1]

def can_form_palindrome(a, b):
    n = len(a)
    for i in range(n + 1):
        a1, a2 = a[:i], a[i:]
        b1, b2 = b[:i], b[i:]
        if is_palindrome(a1 + b2) or is_palindrome(b1 + a2):
            return True
    return False

# Example usage:
a = "yourstringa"
b = "yourstringb"
print(can_form_palindrome(a, b))

Explanation

• is_palindrome(s): This function checks if a given string s is a palindrome.
• can_form_palindrome(a, b): This function iterates through each possible partition point, creates the substrings a1, a2, b1, b2, and checks if either a1 + b2 or b1 + a2 forms a palindrome.

This algorithm runs in ( O(n^2) ) time complexity because it performs a linear check for palindromes for each of the ( n + 1 ) partition points.

Feel free to test this with your specific strings! If you need further refinements or have more questions, I’m here to help! 😊

Understood! We can develop a more efficient algorithm to check for a palindrome partition by leveraging additional data structures and dynamic programming principles. Here’s a refined and more efficient approach:

Efficient Algorithm

1. Input: Two strings a and b of length n.
2. Output: A boolean value indicating if there exists a partition point where ( a1 + b2 ) or ( b1 + a2 ) is a palindrome.

Steps:

1. Initialization:
   • Precompute palindromic substrings for both a and b using a 2D boolean array P[i][j] which indicates if the substring from index i to j is a palindrome.
2. Precompute Palindromic Substrings:
   • Use dynamic programming to fill in the 2D boolean arrays Pa and Pb for strings a and b, respectively.
3. Check Possible Partitions:
   • For each partition point i from 0 to n:
     • Check if a[:i] + b[i:] forms a palindrome by ensuring both substrings are palindromes.
     • Check if b[:i] + a[i:] forms a palindrome by ensuring both substrings are palindromes.
4. Return Result:
   • If a valid palindrome partition is found, return True.
   • If no valid palindrome partition is found after all partitions, return False.

Pseudocode

def is_palindrome(s, start, end):
    while start < end:
        if s[start] != s[end]:
            return False
        start += 1
        end -= 1
    return True

def can_form_palindrome(a, b):
    n = len(a)
    
    # Precompute palindromes for string a
    Pa = [[False] * n for _ in range(n)]
    for i in range(n):
        Pa[i][i] = True
    for length in range(2, n + 1):
        for i in range(n - length + 1):
            j = i + length - 1
            if a[i] == a[j] and (length == 2 or Pa[i + 1][j - 1]):
                Pa[i][j] = True

    # Precompute palindromes for string b
    Pb = [[False] * n for _ in range(n)]
    for i in range(n):
        Pb[i][i] = True
    for length in range(2, n + 1):
        for i in range(n - length + 1):
            j = i + length - 1
            if b[i] == b[j] and (length == 2 or Pb[i + 1][j - 1]):
                Pb[i][j] = True

    # Check for possible partitions
    for i in range(n + 1):
        if (i == 0 or Pa[0][i - 1]) and (i == n or Pb[i][n - 1]):
            return True
        if (i == 0 or Pb[0][i - 1]) and (i == n or Pa[i][n - 1]):
            return True

    return False

# Example usage:
a = "yourstringa"
b = "yourstringb"
print(can_form_palindrome(a, b))

Explanation

• is_palindrome(s, start, end): Helper function to check if the substring s[start:end] is a palindrome.
• can_form_palindrome(a, b): Main function to precompute palindromic substrings and check for valid partitions.
  • Precomputation: Use dynamic programming to fill in the 2D boolean arrays Pa and Pb indicating palindromic substrings for a and b.
  • Partition Check: For each partition point i, check if a[:i] + b[i:] or b[:i] + a[i:] forms a palindrome by ensuring both substrings are palindromic.

This approach reduces the time complexity by precomputing the palindromic substrings and performing efficient checks during partitioning.

Feel free to test this refined algorithm and let me know if you need further assistance or refinements! 😊

Simple greedy solution

class Solution {
public:
    bool checkPalindromeFormation(string a, string b) {
        return validate(a, b) || validate(b, a);
    }

    bool validate(string &a, string& b) {
       int l=0;
        int r = a.size() -1;
        while(l < r) {
            if(a[l] != b[r]) {
                return checkpalindrome(a, l, r) || checkpalindrome(b, l, r);
            }
            l++;
            r--;
        }
        return true;
    }

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

Explanation: https://leetcode.com/problems/split-two-strings-to-make-palindrome/solutions/891646/java-greedy-explanation-to-o-n-solution/