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Home Code Solutions

Palindrome Pairs – LeetCode Solution

Palindrome Pairs - LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

BhautikBhalala by BhautikBhalala
September 6, 2022
Reading Time: 2 mins read
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Leetcode All Problems Solutions

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Table of Contents

  • Palindrome Pairs – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
  • C++ Palindrome Pairs LeetCode Solution
  • Java Palindrome Pairs LeetCode Solution
  • Python 3 Palindrome Pairs LeetCode Solution
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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 <= 5000
  • 0 <= words[i].length <= 300
  • words[i] consists of lower-case English letters.

C++ Palindrome Pairs LeetCode Solution


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 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


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 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


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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



 

Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264
Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195
Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140
Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96
Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66
Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40
Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31
Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21
Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16
Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9
Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7
Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4
Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3
Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1

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Tags: Cc++14full solutionGojavajava 15java 7java 8java8javascriptkotlinLeetCodeLeetCodeSolutionPalindrome Pairspypy 3Python 2python 3rubyrustSolution
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