Introduction
Palindrome checking is a common task in programming, as we've seen in a number of problems often discussed. In the scope of this work, however, they are essential as these are the markable sequences at the string level; palindromes are sequences that read the same from the front or back, such as 'madam'. There is a unique palindrome problem that we are going to solve here. Using C++ , two strings will be provided. The goal is to tell whenever the strings can be 'split' in such a way that when joined together, they can form palindromes. It can be viewed as enabling the construct of string splitting, concatenation and construction of palindromes.
Problem Statement
Let us define two strings, s1 and s2. Then let us outline the goal: the goal is to take each of these strings and cut them into two halves along any index and take one half from each string to create a new string by concatenating them. This concatenated result needs to be checked if it is a palindrome. In such a case, if there is at least one string that is true, return true; otherwise, return false.
Example
Let us take an example so as to illustrate the problem:
Program 1:
Example
#include <iostream>
#include <string>
bool isPalindrome(const std::string& str) {
int left = 0;
int right = str.size() - 1;
while (left < right) {
if (str[left] != str[right]) {
return false;
}
left++;
right--;
}
return true;
}
bool canFormPalindromeBySplitting(const std::string& s1, const std::string& s2) {
int n = s1.size();
for (int i = 0; i <= n; i++) {
// Split s1 and s2 at index i
std::string s1Prefix = s1.substr(0, i);
std::string s1Suffix = s1.substr(i);
std::string s2Prefix = s2.substr(0, i);
std::string s2Suffix = s2.substr(i);
// Concatenate parts and check if they form a palindrome
if (isPalindrome(s1Prefix + s2Suffix) || isPalindrome(s2Prefix + s1Suffix)) {
return true;
}
}
return false;
}
int main() {
std::string s1 = "abcba";
std::string s2 = "xyzzyx";
if (canFormPalindromeBySplitting(s1, s2)) {
std::cout << "Yes, a palindromic combination can be formed." << std::endl;
} else {
std::cout << "No, no palindromic combination can be formed." << std::endl;
}
return 0;
}
Output:
Output
Yes, a palindromic combination can be formed.
Explanation
- Function isPalindrome: This function helps check if the string with a single centre can be a palindrome by checking the string from the starting and the ending towards the centre of the string.
- Function canFormPalindromeBySplitting: This main function breaks the string into two parts and join joins them together to find if there are palindromic combinations through by checking the isPalindrome function.
- Main Function: Initializes the strings s1 and s2 and uses the function canFormPalindromeBySplitting to check whether the passed strings can form a valid palindrome.
- Time Complexity: O(n2), where n is the size of the string arrays. Go through the possible splits and check each risk for the single and all possible combinations of the allies.
- Space Complexity: O(n) because of the carving of the substrings in every round.
Complexity Analysis
Program 2:
Example
#include <iostream>
#include <string>
#include <unordered_set>
#include <vector>
// Function to check if a string is a palindrome
bool isPalindrome(const std::string& str) {
int left = 0;
int right = str.size() - 1;
while (left < right) {
if (str[left] != str[right]) {
return false;
}
left++;
right--;
}
return true;
}
// Helper function to split and concatenate
void generatePalindromicCombinations(const std::string& s1, const std::string& s2, int min_palindromic_combinations,
std::unordered_set<std::string>& palindromicCombinations,
int depth, int maxDepth) {
int n = s1.size();
int m = s2.size();
// Stop if the required number of palindromic combinations is reached
if (palindromicCombinations.size() >= min_palindromic_combinations) return;
// Split each string up to a certain depth
for (int i = 1; i < n && i < m && depth <= maxDepth; i++) {
std::string s1Prefix = s1.substr(0, i);
std::string s1Suffix = s1.substr(i);
std::string s2Prefix = s2.substr(0, i);
std::string s2Suffix = s2.substr(i);
// Concatenate in different orders and check for palindrome
std::vector<std::string> combinations = {
s1Prefix + s2Suffix,
s2Prefix + s1Suffix,
s1Suffix + s2Prefix,
s2Suffix + s1Prefix
};
for (const auto& comb : combinations) {
if (isPalindrome(comb)) {
palindromicCombinations.insert(comb);
}
}
// Recursive call to increase depth and explore further splits
generatePalindromicCombinations(s1Suffix, s2Suffix, min_palindromic_combinations, palindromicCombinations, depth + 1, maxDepth);
}
}
// Main function to check if the minimum number of palindromic combinations can be formed
bool canFormMultiplePalindromicPatterns(const std::string& s1, const std::string& s2, int min_palindromic_combinations) {
std::unordered_set<std::string> palindromicCombinations;
int maxDepth = 3; // Maximum recursion depth for splits (adjustable based on problem complexity)
// Generate palindromic combinations
generatePalindromicCombinations(s1, s2, min_palindromic_combinations, palindromicCombinations, 1, maxDepth);
// Check if we reached the required count
return palindromicCombinations.size() >= min_palindromic_combinations;
}
int main() {
std::string s1 = "abaxyz";
std::string s2 = "zyxba";
int min_palindromic_combinations = 3;
if (canFormMultiplePalindromicPatterns(s1, s2, min_palindromic_combinations)) {
std::cout << "Yes, at least " << min_palindromic_combinations << " palindromic combinations can be formed." << std::endl;
} else {
std::cout << "No, less than " << min_palindromic_combinations << " palindromic combinations can be formed." << std::endl;
}
return 0;
}
Output:
Output
Yes, at least 3 palindromic combinations can be formed.
Explanation:
- isPalindrome Function: Confirms whether the string that's been provided is the same when read from either direction.
- generatePalindromicCombinations Function: This procedure recursively cuts s1 and s2 at varying lengths, sequences the segments in all permutations, and then observes whether any palindrome has formed. Each set of boolean values makes sure that whenever a new palindrome is detected, it's added to ensure none were left behind. In situations where there is no need for additional palindromic pairs, the ordering terminates quickly.
- The function canFormMultiplePalindromicPatterns: This function sets the parameters for the recursive depth and the required number of combinations, invokes the recursive method, and, at last, verifies whether the intended count of palindromic patterns has been achieved.
- Time Complexity: The complexity increases with the recursive splits to O (n m D), with n and m being the length of s1 and s2, respectively, while D is the recursion depth in which splits occur.
- Space Complexity: The size of the space complexity for this study is O(P) due to the storage of unique combinations of the palindromic pattern combinations with respect to the number of fragments, P, of the global palindromic patterns identified in this work.
Complexity Analysis
Program 3:
Example
#include <iostream>
#include <string>
#include <unordered_set>
bool isPalindrome(const std::string& str) {
int left = 0, right = str.size() - 1;
while (left < right) {
if (str[left] != str[right]) {
return false;
}
left++;
right--;
}
return true;
}
bool checkPalindromeCombinations(const std::string& s1, const std::string& s2, int min_palindromic_combinations) {
int n = s1.size();
std::unordered_set<std::string> uniquePalindromes;
// Check combinations of s1 prefixes and s2 suffixes
for (int i = 0; i <= n; i++) {
std::string prefix = s1.substr(0, i);
std::string suffix = (i < n) ? s2.substr(i) : ""; // Avoid out-of-range
if (isPalindrome(prefix + suffix)) {
uniquePalindromes.insert(prefix + suffix);
if (uniquePalindromes.size() >= min_palindromic_combinations) return true;
}
}
// Check combinations of s2 prefixes and s1 suffixes
for (int i = 0; i <= n; i++) {
std::string prefix = s2.substr(0, i);
std::string suffix = (i < n) ? s1.substr(i) : ""; // Avoid out-of-range
if (isPalindrome(prefix + suffix)) {
uniquePalindromes.insert(prefix + suffix);
if (uniquePalindromes.size() >= min_palindromic_combinations) return true;
}
}
return uniquePalindromes.size() >= min_palindromic_combinations;
}
int main() {
std::string s1 = "abaxyz";
std::string s2 = "zyxba";
int min_palindromic_combinations = 3;
if (checkPalindromeCombinations(s1, s2, min_palindromic_combinations)) {
std::cout << "Yes, at least " << min_palindromic_combinations << " palindromic combinations can be formed." << std::endl;
} else {
std::cout << "No, less than " << min_palindromic_combinations << " palindromic combinations can be formed." << std::endl;
}
return 0;
}
Output:
Output
Yes, at least 3 palindromic combinations can be formed.
Explanation:
- Use two pointers, left and right, starting at the beginning of s1 and the end of s2 (or vice versa).
- Check if combining prefixes of s1 with suffixes of s2 forms a palindrome.
- Similarly, check suffixes of s1 combined with prefixes of s2.
- Move the pointers inward until a certain depth or stop once a minimum number of palindromic combinations is found.
Complexity: This reduces complexity by avoiding unnecessary recursive calls and substring operations.
Program 4:
If we need to check many palindromic substrings of s1 and s2, we can use dynamic programming to precompute palindromic substrings in both strings, making subsequent palindrome checks more efficient.
Example
#include <iostream>
#include <string>
#include <vector>
#include <unordered_set>
// Check if a given string is a palindrome
bool isPalindrome(const std::string& str) {
int left = 0, right = str.size() - 1;
while (left < right) {
if (str[left] != str[right]) {
return false;
}
left++;
right--;
}
return true;
}
// Compute a DP table to determine palindromic substrings in a string
std::vector<std::vector<bool>> computePalindromeDP(const std::string& s) {
int n = s.size();
std::vector<std::vector<bool>> dp(n, std::vector<bool>(n, false));
for (int i = 0; i < n; i++) dp[i][i] = true;
for (int i = 0; i < n - 1; i++) {
dp[i][i + 1] = (s[i] == s[i + 1]);
}
for (int len = 3; len <= n; len++) {
for (int i = 0; i <= n - len; i++) {
int j = i + len - 1;
dp[i][j] = (s[i] == s[j] && dp[i + 1][j - 1]);
}
}
return dp;
}
// Check if at least a certain number of palindromic combinations can be formed
bool canFormPalindromicPatternsDP(const std::string& s1, const std::string& s2, int min_palindromic_combinations) {
int n = s1.size();
auto dp1 = computePalindromeDP(s1);
auto dp2 = computePalindromeDP(s2);
std::unordered_set<std::string> palindromicCombinations;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
if (dp1[i][j]) {
std::string prefix = s1.substr(i, j - i + 1);
for (int k = 0; k < n; k++) {
std::string combined = prefix + s2.substr(k);
if (isPalindrome(combined)) {
palindromicCombinations.insert(combined);
if (palindromicCombinations.size() >= min_palindromic_combinations) return true;
}
}
}
}
}
return false;
}
int main() {
std::string s1 = "abaxyz";
std::string s2 = "zyxba";
int min_palindromic_combinations = 3;
if (canFormPalindromicPatternsDP(s1, s2, min_palindromic_combinations)) {
std::cout << "Yes, at least " << min_palindromic_combinations << " palindromic combinations can be formed." << std::endl;
} else {
std::cout << "No, less than " << min_palindromic_combinations << " palindromic combinations can be formed." << std::endl;
}
return 0;
}
Output:
Output
Yes, at least 3 palindromic combinations can be formed.
Explanation:
- Use dynamic programming to create a 2D table for each string, where dpi is true if the substring s[i:j+1] is a palindrome.
- With precomputed palindromic substrings, quickly check valid combinations of substrings from s1 and s2.
- This allows us to avoid repeatedly checking for palindromic properties and directly focus on combinations.
Complexity: This approach can reduce redundant checks for palindrome properties and improve efficiency for larger strings.
Conclusion:
An interesting problem of the construction of palindromic patterns by concatenation of the substrings of two given strings is that of string manipulation along with verification of its palindromicity and substring tasks that fit within resource constraints. Although a brute-force approach appears workable at this point, later on it would need to be addressed in order to optimize the solution towards searching within larger strings and higher complexity constraints.