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Commit9e5f18f

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890_FindAndReplacePattern890
1 parent30e2023 commit9e5f18f

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‎890_FindAndReplacePattern890.java‎

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/**
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* You have a list of words and a pattern, and you want to know which words in
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* words matches the pattern.
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*
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* A word matches the pattern if there exists a permutation of letters p so
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* that after replacing every letter x in the pattern with p(x), we get the
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* desired word.
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*
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* (Recall that a permutation of letters is a bijection from letters to
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* letters: every letter maps to another letter, and no two letters map to
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* the same letter.)
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*
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* Return a list of the words in words that match the given pattern.
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*
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* You may return the answer in any order.
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*
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* Example 1:
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* Input: words = ["abc","deq","mee","aqq","dkd","ccc"], pattern = "abb"
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* Output: ["mee","aqq"]
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* Explanation: "mee" matches the pattern because there is a permutation {a -> m, b -> e, ...}.
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* "ccc" does not match the pattern because {a -> c, b -> c, ...} is not a permutation,
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* since a and b map to the same letter.
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*
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* Note:
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* 1 <= words.length <= 50
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* 1 <= pattern.length = words[i].length <= 20
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*/
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publicclassFindAndReplacePattern890 {
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publicList<String>findAndReplacePattern(String[]words,Stringpattern) {
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List<String>res =newArrayList<>();
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char[]pat =pattern.toCharArray();
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intN =pattern.length();
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for (Stringword:words) {
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if (isPermutation(word.toCharArray(),pat,N)) {
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res.add(word);
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}
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}
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returnres;
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}
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publicbooleanisPermutation(char[]word,char[]pattern,intN) {
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Map<Character,Character>map =newHashMap<>();
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for (inti=0;i<N;i++) {
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if (map.containsKey(word[i])) {
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if (map.get(word[i]) !=pattern[i])returnfalse;
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}else {
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if (map.values().contains(pattern[i]))returnfalse;
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map.put(word[i],pattern[i]);
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}
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}
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returntrue;
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}
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publicList<String>findAndReplacePattern2(String[]words,Stringpattern) {
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List<String>res =newArrayList<>();
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intcode =encode(pattern);
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for (Stringw:words) {
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if (code ==encode(w)) {
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res.add(w);
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}
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}
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returnres;
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}
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publicintencode(Strings) {
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intres =0;
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char[]chars =s.toCharArray();
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charfirst =chars[0];
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Map<Integer,Integer>map =newHashMap<>();
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map.put(0,0);
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inti =1;
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for (intj=0;j<chars.length;j++) {
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intoffset =chars[j] -first;
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if (map.containsKey(offset)) {
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res +=map.get(offset) *j;
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}else {
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map.put(offset,i);
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res +=i *j;
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i++;
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}
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}
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returnres;
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}
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}

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