Create 767-Reorganize-String.java

Author: SharmaTushar1

java/767-Reorganize-String.java

@@ -0,0 +1,40 @@
+//See this comment for explanation https://leetcode.com/problems/reorganize-string/discuss/113440/Java-solution-PriorityQueue/211009
+
+class Solution {
+    public String reorganizeString(String s) {
+        HashMap<Character, Integer> map = new HashMap<>();
+        for (int i = 0; i<s.length(); i++) {
+            map.put(s.charAt(i), map.getOrDefault(s.charAt(i), 0)+1);
+        }
+        PriorityQueue<Map.Entry<Character, Integer>> pq = new PriorityQueue<>((a,b)->b.getValue()-a.getValue());
+        pq.addAll(map.entrySet());
+        
+        StringBuilder sb = new StringBuilder();
+        
+        while (!pq.isEmpty()) {
+            Map.Entry<Character, Integer> temp1 = pq.poll();
+            //if the character at sb's end is different from the max frequency character or the string is empty
+            if (sb.length()==0 || sb.charAt(sb.length()-1)!=temp1.getKey()) {
+                sb.append(temp1.getKey());
+                //update the value
+                temp1.setValue(temp1.getValue()-1);
+            } else { //the character is same
+                //hold the current character and look for the 2nd most frequent character
+                Map.Entry<Character, Integer> temp2 = pq.poll();
+                //if there is no temp2 i.e. the temp1 was the only character in the heap then there is no way to avoid adjacent duplicate values
+                if (temp2==null) 
+                    return "";
+                //else do the same thing as above
+                sb.append(temp2.getKey());
+                //update the value
+                temp2.setValue(temp2.getValue()-1);
+                //if still has some value left add again to the heap
+                if (temp2.getValue()!=0) 
+                    pq.offer(temp2);
+            }
+            if (temp1.getValue()!=0) 
+                pq.offer(temp1);
+        }
+        return sb.toString();
+    }
+}