Completed Hashing-1

GroupAnagrams.java

@@ -0,0 +1,26 @@
+class Solution {
+    public List<List<String>> groupAnagrams(String[] strs) {
+        // can also be solved using prime product technique
+        //TC - O(nk)
+        //SC - O(n)
+        if(strs.length == 0)
+            return new ArrayList<>();
+
+        Map<String, List<String>> map = new HashMap<>();
+        for(String str : strs) {
+            int[] freq = new int[26];
+            for(char ch : str.toCharArray()) {
+                freq[ch - 'a']++;
+            }
+            String word = Arrays.toString(freq);
+
+
+            if(!map.containsKey(word))
+                map.put(word, new ArrayList<>());
+            map.get(word).add(str);
+
+        }
+        return new ArrayList<>(map.values());
+    }
+
+}

isIsomorphic.java

@@ -0,0 +1,30 @@
+//TC is O(n) - n is length of string s
+//SC is O(n)
+class Solution {
+    public boolean isIsomorphic(String s, String t) {
+        if(s.length() != t.length())
+            return false;
+        Map<Character, Character> sMap = new HashMap<>();
+        Map<Character, Character> tMap = new HashMap<>();
+
+        for(int i = 0 ; i < s.length() ; i++) {
+            char sChar = s.charAt(i);
+            char tChar = t.charAt(i);
+            if(sMap.containsKey(sChar)) {
+                if(sMap.get(sChar) != tChar)
+                    return false;
+            }
+            else
+                sMap.put(sChar, tChar);
+
+            if(tMap.containsKey(tChar)) {
+                if(tMap.get(tChar) != sChar)
+                    return false;
+            }
+            else
+                tMap.put(tChar, sChar);
+
+        }
+        return true;
+    }
+}

wordPattern.java

@@ -0,0 +1,30 @@
+//TC is O(n) with n as the length of string pattern
+//SC is O(n) with 2 hashmaps combined with n being the length of strings
+class Solution {
+    public boolean wordPattern(String pattern, String s) {
+        Map<Character, String> map = new HashMap<>();
+        Map<String, Character> sMap = new HashMap<>();
+        String[] words = s.split("\\s+");
+        if(pattern.length() != words.length)
+            return false;
+        for(int i = 0 ; i < pattern.length() ; i++) {
+            char ch = pattern.charAt(i);
+            String word = words[i];
+            if(map.containsKey(ch)) {
+                if(!map.get(ch).equals(word)) {
+                    System.out.println("map's word " +map.get(ch));
+                    return false;
+                }
+            }
+            else
+                map.put(ch, word);
+            if(sMap.containsKey(word)) {
+                if(!sMap.get(word).equals(ch))
+                    return false;
+            }
+            else
+                sMap.put(word, ch);
+        }
+        return true;
+    }
+}