Palindrome Pairs – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
Given a list of unique words, return all the pairs of the distinct indices (i, j) in the given list, so that the concatenation of the two words words[i] + words[j] is a palindrome.
Example 1:
Input: words = ["abcd","dcba","lls","s","sssll"] Output: [[0,1],[1,0],[3,2],[2,4]] Explanation: The palindromes are ["dcbaabcd","abcddcba","slls","llssssll"]
Example 2:
Input: words = ["bat","tab","cat"] Output: [[0,1],[1,0]] Explanation: The palindromes are ["battab","tabbat"]
Example 3:
Input: words = ["a",""] Output: [[0,1],[1,0]]
Constraints:
1 <= words.length <= 50000 <= words[i].length <= 300words[i]consists of lower-case English letters.
C++ Palindrome Pairs LeetCode Solution
class Solution {
public:
vector<vector<int>> palindromePairs(vector<string>& words) {
unordered_map<string, int> dict;
vector<vector<int>> ans;
// build dictionary
for(int i = 0; i < words.size(); i++) {
string key = words[i];
reverse(key.begin(), key.end());
dict[key] = i;
}
// edge case: if empty string "" exists, find all palindromes to become pairs ("", self)
if(dict.find("")!=dict.end()){
for(int i = 0; i < words.size(); i++){
if(i == dict[""]) continue;
if(isPalindrome(words[i])) ans.push_back({dict[""], i}); // 1) if self is palindrome, here ans covers concatenate("", self)
}
}
for(int i = 0; i < words.size(); i++) {
for(int j = 0; j < words[i].size(); j++) {
string left = words[i].substr(0, j);
string right = words[i].substr(j, words[i].size() - j);
if(dict.find(left) != dict.end() && isPalindrome(right) && dict[left] != i) {
ans.push_back({i, dict[left]}); // 2) when j = 0, left = "", right = self, so here covers concatenate(self, "")
}
if(dict.find(right) != dict.end() && isPalindrome(left) && dict[right] != i) {
ans.push_back({dict[right], i});
}
}
}
return ans;
}
bool isPalindrome(string str){
int i = 0;
int j = str.size() - 1;
while(i < j) {
if(str[i++] != str[j--]) return false;
}
return true;
}
};
Java Palindrome Pairs LeetCode Solution
class Solution {
public:
vector<vector<int>> palindromePairs(vector<string>& words) {
unordered_map<string, int> dict;
vector<vector<int>> ans;
// build dictionary
for(int i = 0; i < words.size(); i++) {
string key = words[i];
reverse(key.begin(), key.end());
dict[key] = i;
}
// edge case: if empty string "" exists, find all palindromes to become pairs ("", self)
if(dict.find("")!=dict.end()){
for(int i = 0; i < words.size(); i++){
if(i == dict[""]) continue;
if(isPalindrome(words[i])) ans.push_back({dict[""], i}); // 1) if self is palindrome, here ans covers concatenate("", self)
}
}
for(int i = 0; i < words.size(); i++) {
for(int j = 0; j < words[i].size(); j++) {
string left = words[i].substr(0, j);
string right = words[i].substr(j, words[i].size() - j);
if(dict.find(left) != dict.end() && isPalindrome(right) && dict[left] != i) {
ans.push_back({i, dict[left]}); // 2) when j = 0, left = "", right = self, so here covers concatenate(self, "")
}
if(dict.find(right) != dict.end() && isPalindrome(left) && dict[right] != i) {
ans.push_back({dict[right], i});
}
}
}
return ans;
}
bool isPalindrome(string str){
int i = 0;
int j = str.size() - 1;
while(i < j) {
if(str[i++] != str[j--]) return false;
}
return true;
}
};
Python 3 Palindrome Pairs LeetCode Solution
def is_palindrome(check):
return check == check[::-1]
words = {word: i for i, word in enumerate(words)}
valid_pals = []
for word, k in words.iteritems():
n = len(word)
for j in range(n+1):
pref = word[:j]
suf = word[j:]
if is_palindrome(pref):
back = suf[::-1]
if back != word and back in words:
valid_pals.append([words[back], k])
if j != n and is_palindrome(suf):
back = pref[::-1]
if back != word and back in words:
valid_pals.append([k, words[back]])
return valid_pals
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