One-Hundred-Seventy Commit: Add Rearrange String problem to Top K Elements section

Author: AAdewunmi

src/Top_K_Elements/Problem_11_Rearrange_String.java

@@ -0,0 +1,40 @@
+package Top_K_Elements;
+
+// Problem Statement: Rearrange String (hard)
+// LeetCode Question: 767. Reorganize String
+
+import java.util.HashMap;
+import java.util.Map;
+import java.util.PriorityQueue;
+
+public class Problem_11_Rearrange_String {
+    public String rearrangeString(String str) {
+        Map<Character, Integer> charFrequencyMap = new HashMap<>();
+        for (char chr : str.toCharArray())
+            charFrequencyMap.put(chr, charFrequencyMap.getOrDefault(chr, 0) + 1);
+
+        PriorityQueue<Map.Entry<Character, Integer>> maxHeap =
+                new PriorityQueue<Map.Entry<Character, Integer>>(
+                        (e1, e2) -> e2.getValue() - e1.getValue());
+
+        // add all characters to the max heap
+        maxHeap.addAll(charFrequencyMap.entrySet());
+
+        Map.Entry<Character, Integer> previousEntry = null;
+        StringBuilder resultString = new StringBuilder(str.length());
+        while (!maxHeap.isEmpty()) {
+            Map.Entry<Character, Integer> currentEntry = maxHeap.poll();
+            // add the previous entry back in the heap if its frequency is greater than zero
+            if (previousEntry != null && previousEntry.getValue() > 0)
+                maxHeap.offer(previousEntry);
+            // append the current character to the result string and decrement its count
+            resultString.append(currentEntry.getKey());
+            currentEntry.setValue(currentEntry.getValue() - 1);
+            previousEntry = currentEntry;
+        }
+
+        // if we were successful in appending all the characters to the result string,
+        // return it
+        return resultString.length() == str.length() ? resultString.toString() : "";
+    }
+}