Leetcode 1616: Split Two Strings to Make Palindrome
You are given two strings, a and b, of the same length. You need to split both strings at the same index into two parts: one prefix and one suffix for each string. Then, check if concatenating one prefix from one string with the suffix of the other string forms a palindrome. Specifically, you need to check if either
aprefix + bsuffix or bprefix + asuffix forms a palindrome.Problem
Approach
Steps
Complexity
Input: The input consists of two strings, a and b, both of the same length.
Example: a = 'abc', b = 'cba'
Constraints:
• 1 <= a.length, b.length <= 10^5
• a.length == b.length
• a and b consist of lowercase English letters
Output: Return true if it is possible to form a palindrome string by concatenating either `aprefix + bsuffix` or `bprefix + asuffix`. Otherwise, return false.
Example: For a = 'abc' and b = 'cba', the output would be true.
Constraints:
• The output should be a boolean value, true or false.
Goal: Check if it is possible to form a palindrome by choosing a split index for both strings and combining the corresponding parts.
Steps:
• For each possible split point, split both strings into two parts: a prefix and a suffix.
• Check if either concatenation of `aprefix + bsuffix` or `bprefix + asuffix` forms a palindrome.
• Return true if either concatenation forms a palindrome, otherwise return false.
Goal: The input strings should have the same length and consist of lowercase letters.
Steps:
• Both strings a and b have the same length.
• The lengths of a and b can range from 1 to 100,000.
Assumptions:
• The two input strings are guaranteed to have the same length.
• You can split the strings at any valid index.
• Input: Input: a = 'abc', b = 'cba'
• Explanation: In this case, we can split both strings at index 1: aprefix = 'a', asuffix = 'bc', bprefix = 'c', and bsuffix = 'ba'. Concatenating `aprefix + bsuffix = 'a' + 'ba' = 'aba'`, which is a palindrome.
• Input: Input: a = 'ab', b = 'ba'
• Explanation: Here, splitting the strings at index 1: aprefix = 'a', asuffix = 'b', bprefix = 'b', and bsuffix = 'a'. Concatenating `aprefix + bsuffix = 'a' + 'a' = 'aa'`, which is a palindrome.
• Input: Input: a = 'abc', b = 'def'
• Explanation: In this case, no matter how we split the strings, no combination of prefix and suffix will form a palindrome.
Approach: The approach involves iterating through all possible split points, checking the formed concatenations for palindrome properties.
Observations:
• We need to check both concatenations (`aprefix + bsuffix` and `bprefix + asuffix`) for each possible split.
• If a combination forms a palindrome, we return true; otherwise, we return false.
Steps:
• Define a helper function `check` that checks if either concatenation forms a palindrome.
• Iterate through each possible split index of the strings a and b.
• For each index, check if either `aprefix + bsuffix` or `bprefix + asuffix` is a palindrome.
Empty Inputs:
• If either string is empty, the result should be false as there are no valid splits.
Large Inputs:
• For large strings, the approach needs to be efficient enough to handle up to 100,000 characters.
Special Values:
• Strings that are already palindromes can be split trivially.
Constraints:
• The algorithm needs to check possible splits efficiently.
bool checkPalindromeFormation(string a, string b) {
return check(a, b) || check(b, a);
}
bool check(string a, string b) {
int i = 0, j = a.size() -1;
while(i < j && a[i] == b[j])
i++, j--;
return isPalindrom(a, i, j) || isPalindrom(b, i, j);
}
bool isPalindrom(string a, int i , int j) {
while(i < j && a[i] == a[j])
i++, j--;
return i >= j;
}
1 : Function Definition
bool checkPalindromeFormation(string a, string b) {
Define the function `checkPalindromeFormation` that takes two strings `a` and `b` and returns true if a palindrome can be formed by rearranging them.
2 : Function Logic
return check(a, b) || check(b, a);
Call the `check` function for both possible combinations of the strings `a` and `b`. If either combination can form a palindrome, return true.
3 : Helper Function Definition
bool check(string a, string b) {
Define the helper function `check` that will determine if the strings `a` and `b` can form a palindrome by checking character matches from both ends.
4 : Variable Initialization
int i = 0, j = a.size() -1;
Initialize two pointers `i` at the start of string `a` and `j` at the end to check for palindrome conditions.
5 : While Loop
while(i < j && a[i] == b[j])
Use a while loop to compare characters of the two strings from both ends (`a[i]` with `b[j]`). Continue as long as they match.
6 : Pointer Adjustment
i++, j--;
If characters match, move the pointers towards the center (`i++` and `j--`) to check the next pair of characters.
7 : Return Statement
return isPalindrom(a, i, j) || isPalindrom(b, i, j);
Return the result of checking if either string from the updated pointers forms a palindrome using the helper function `isPalindrom`.
8 : Palindrome Check Function Definition
bool isPalindrom(string a, int i , int j) {
Define the function `isPalindrom` that checks whether the substring from index `i` to `j` in string `a` is a palindrome.
9 : While Loop
while(i < j && a[i] == a[j])
Use a while loop to compare characters from both ends of the substring in string `a` (`a[i]` with `a[j]`). Continue as long as they match.
10 : Pointer Adjustment
i++, j--;
If characters match, move the pointers towards the center (`i++` and `j--`) to check the next pair of characters.
11 : Return Statement
return i >= j;
If the pointers have crossed or met, it means the substring is a palindrome. Return true, else false.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The worst case occurs when we check palindrome properties at each split for strings of length n.
Best Case: O(1)
Worst Case: O(n)
Description: The space complexity is O(n) in the worst case for storing prefixes and suffixes.
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