LeetCode – Palindrome Pairs (Java)

Category: Others   May 11, 2014

Given a list of unique words. Find all pairs of distinct indices (i, j) in the given list, so that the concatenation of the two words, i.e. words[i] + words[j] is a palindrome.

Example 1:
Given words = ["bat", "tab", "cat"]
Return [[0, 1], [1, 0]]
The palindromes are ["battab", "tabbat"]

Java Solution

public List<List<Integer>> palindromePairs(String[] words) {
    List<List<Integer>> result = new ArrayList<>();
    if(words == null || words.length < 2){
        return result;
    }
 
    HashMap<String, Integer> map = new HashMap<>();
    for(int i=0; i<words.length; i++){
        map.put(words[i], i);
    }
 
    for(int i=0; i<words.length; i++){
        String s = words[i];
 
        for(int k=0; k<=s.length(); k++){
            String left = s.substring(0, k);
            String right= s.substring(k);
 
            //if left part is palindrome, find reversed right part
            if(isPalindrome(left)){
                String reversedRight = new StringBuilder(right).reverse().toString();
                if(map.containsKey(reversedRight) && map.get(reversedRight) != i){
                    ArrayList<Integer> l = new ArrayList<>();
                    l.add(map.get(reversedRight));
                    l.add(i);
                    result.add(l);
                }
            }
 
            //if right part is a palindrome, find reversed left part
            if(isPalindrome(right)){
                String reversedLeft = new StringBuilder(left).reverse().toString();
                if(map.containsKey(reversedLeft) 
                        && map.get(reversedLeft) != i 
                        && right.length() != 0){ 
                        //make sure right is not "", already handled in the if above
                    ArrayList<Integer> l = new ArrayList<>();
                    l.add(i);
                    l.add(map.get(reversedLeft));
                    result.add(l);
                }
            }
        }
    }
 
    return result;
}
 
public boolean isPalindrome(String s){
    int i=0;
    int j=s.length()-1;
 
    while(i<=j){
        if(s.charAt(i)!=s.charAt(j)){
            return false;
        }
        i++;
        j--;
    }
 
    return true;
}

Time complexity is O(n*k^2), where n is the number of words and k is the average length of each word.

Related posts:

  1. LeetCode – Palindrome Partitioning (Java)
  2. LeetCode – Substring with Concatenation of All Words (Java)
  3. LeetCode – Find K Pairs with Smallest Sums (Java)
  4. LeetCode – Shortest Word Distance II (Java)
Category >> Others  
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  • Satish

    Time: O(n^m)
    Space: O(m)
    n = nos of strings
    m = max len of string

  • Satish

    public static ArrayList<ArrayList> helperx(String[] strs) {
    if (strs == null || strs.length < 2)
    return null;

    ArrayList<ArrayList> result = new ArrayList();
    HashMap map = new HashMap();
    for (int i = 0; i < strs.length; i++) {
    String str = strs[i];
    String rev = reverse(str);
    if (map.containsKey(rev) || map.containsKey(rev.substring(0, rev.length() - 1))) {
    ArrayList list1 = new ArrayList();
    ArrayList list2 = new ArrayList();
    if (map.containsKey(rev)) {
    list1.add(i);
    list1.add(map.get(rev));
    list2.add(i);
    list2.add(map.get(rev));
    }
    else{
    list1.add(i);
    list1.add(map.get(rev.substring(0, rev.length() - 1)));
    list2.add(map.get(rev.substring(0, rev.length() - 1)));
    list2.add(i);
    }

    result.add(list1);
    result.add(list2);
    } else {
    map.put(str, i);
    }
    }

    return result;
    }

    public static String reverse(String str) {
    return str.length() > 0 ? str.substring(str.length()-1) + reverse(str.substring(0, str.length()-1)) : "";
    }

  • Bukary Kandeh

    public static List<List> PalindromePairs(String[] Words){
    List<List> finalList = new ArrayList<List>();

    for(int i=0; i<Words.length; i++){
    for(int j=i+1; j<Words.length; j++){
    String s1 = Words[i].concat(Words[j]);
    String s2 = Words[j].concat(Words[i]);
    String sRev1 = new StringBuilder(s1).reverse().toString();
    String sRev2 = new StringBuilder(s2).reverse().toString();

    if(s1.equals(sRev1)){
    List list = new ArrayList();
    list.add(Arrays.asList(Words).indexOf(Words[i]));
    list.add(Arrays.asList(Words).indexOf(Words[j]));
    finalList.add(list);
    }
    if(s2.equals(sRev2)){
    List list = new ArrayList();
    list.add(Arrays.asList(Words).indexOf(Words[j]));
    list.add(Arrays.asList(Words).indexOf(Words[i]));
    finalList.add(list);
    }
    }
    }
    return finalList;
    }

  • Bukary Kandeh


    Java, easy to understand and much shorter solutions:

    public static List<List> PalindromePairs(String[] Words){
    List<List> finalList = new ArrayList<List>();
    for(int i=0; i<Words.length; i++){
    for(int j=i+1; j<Words.length; j++){
    String s1 = Words[i].concat(Words[j]);
    String s2 = Words[j].concat(Words[i]);
    String sRev1 = new StringBuilder(s1).reverse().toString();
    String sRev2 = new StringBuilder(s2).reverse().toString();

    if(s1.equals(sRev1)){
    List list = new ArrayList();
    list.add(Arrays.asList(Words).indexOf(Words[i]));
    list.add(Arrays.asList(Words).indexOf(Words[j]));
    finalList.add(list);
    }
    if(s2.equals(sRev2)){
    List list = new ArrayList();
    list.add(Arrays.asList(Words).indexOf(Words[j]));
    list.add(Arrays.asList(Words).indexOf(Words[i]));
    finalList.add(list);
    }
    }
    }
    return finalList;
    }