LeetCode – Group Shifted Strings (Java)

Given a string, we can "shift" each of its letter to its successive letter, for example: "abc" -> "bcd". We can keep "shifting" which forms the sequence: "abc" -> "bcd" -> ... -> "xyz".

Given a list of strings which contains only lowercase alphabets, group all strings that belong to the same shifting sequence, return:

[
  ["abc","bcd","xyz"],
  ["az","ba"],
  ["acef"],
  ["a","z"]
]

Java Solution

public List<List<String>> groupStrings(String[] strings) {
    List<List<String>> result = new ArrayList<List<String>>();
    HashMap<String, ArrayList<String>> map 
                    = new HashMap<String, ArrayList<String>>();
 
    for(String s: strings){
        char[] arr = s.toCharArray();
        if(arr.length>0){
            int diff = arr[0]-'a';
            for(int i=0; i<arr.length; i++){
                if(arr[i]-diff<'a'){
                   arr[i] = (char) (arr[i]-diff+26);
                }else{
                   arr[i] = (char) (arr[i]-diff); 
                }
 
            }
        }  
 
        String ns = new String(arr);
        if(map.containsKey(ns)){
            map.get(ns).add(s);
        }else{
            ArrayList<String> al = new ArrayList<String>();
            al.add(s);
            map.put(ns, al);
        }
    }
 
    for(Map.Entry<String, ArrayList<String>> entry: map.entrySet()){
        Collections.sort(entry.getValue());
    }
 
    result.addAll(map.values());
 
    return result;
}

Category >> Algorithms

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Kaushlendra Singh

var x = ["az","ba","af"] for(var i = 0 ;i < x[0].length-1 ; i++){ var diff = x[0].charCodeAt(i+1) - x[0].charCodeAt(i) if(diff i",diff,i) for(var j=1 ;j < x.length; j++){ tempDiff = x[j].charCodeAt(i+1) - x[j].charCodeAt(i) if(tempDiff ",x[j]) x.splice(j,1) } } } console.log(x)

LeetCode – Group Shifted Strings (Java)

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