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Palindrome Pairs

Certification: Advanced Level Accuracy: 33.33% Submissions: 3 Points: 15

Write a Java program to find all pairs of distinct indices i and j in the given list of words where the concatenation of the two words, i.e., words[i] + words[j] is a palindrome.

Example 1
    
  
  • Input: words = ["abcd", "dcba", "lls", "s", "sssll"]
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  • Output: [[0, 1], [1, 0], [3, 2], [2, 4]]
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  • Explanation:
    Step 1: words[0] + words[1] = "abcd" + "dcba" = "abcddcba" is a palindrome.
    Step 2: words[1] + words[0] = "dcba" + "abcd" = "dcbaabcd" is a palindrome.
    Step 3: words[3] + words[2] = "s" + "lls" = "slls" is a palindrome.
    Step 4: words[2] + words[4] = "lls" + "sssll" = "llssssll" is a palindrome.
    •
Example 2
    
  
  • Input: words = ["bat", "tab", "cat"]
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  • Output: [[0, 1], [1, 0]]
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  • Explanation:
    Step 1: words[0] + words[1] = "bat" + "tab" = "battab" is a palindrome.
    Step 2: words[1] + words[0] = "tab" + "bat" = "tabbat" is a palindrome.
    Step 3: words[0] + words[2] = "bat" + "cat" = "batcat" is not a palindrome.
    Step 4: words[2] + words[0] = "cat" + "bat" = "catbat" is not a palindrome.
    •
Constraints
    
  
  • 1 <= words.length <= 5000
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  • 0 <= words[i].length <= 300
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  • words[i] consists of lowercase English letters.
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  • All entries in words are distinct.
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  • Time Complexity: O(n * k²) where n is the number of words and k is the average length of words
    • 
  
  • Space Complexity: O(n * k)
    •
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Hint
Solution

Solution Hints

    
  
  • Use a trie data structure to efficiently search for palindrome pairs.
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  • For each word, check if its reverse exists in the list.
    • 
  
  • Handle empty strings as special cases.
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  • Consider partial matches where only part of the word needs to be palindromic.
    • 
  
  • Preprocess the list to efficiently look up words.
    •

Steps to solve by this approach:

 Step 1: Create a hashmap to store each word and its index for quick lookup.

 Step 2: For each word, check if the reversed version exists in the map.

 Step 3: Handle the empty string as a special case. If empty string exists and current word is a palindrome, they form palindrome pairs.

 Step 4: For each word, split it into prefix and suffix at every position.

 Step 5: If the prefix is a palindrome, check if the reversed suffix exists in the map.

 Step 6: If the suffix is a palindrome, check if the reversed prefix exists in the map.

 Step 7: Collect all valid pairs of indices where concatenating the words forms a palindrome.

Submitted Code :