LeetCode Solutions

			

class Solution {
 public:
  string reorganizeString(string s) {
    unordered_map<char, int> count;
    int maxFreq = 0;

    for (const char c : s)
      maxFreq = max(maxFreq, ++count[c]);

    if (maxFreq > (s.length() + 1) / 2)
      return "";

    string ans;
    priority_queue<pair<int, char>> maxHeap;  // (freq, c)
    int prevFreq = 0;
    char prevChar = '@';

    for (const auto& [c, freq] : count)
      maxHeap.emplace(freq, c);

    while (!maxHeap.empty()) {
      // Get the most freq letter
      const auto [freq, c] = maxHeap.top();
      maxHeap.pop();
      ans += c;
      // Add the previous letter back so that
      // Any two adjacent characters are not the same
      if (prevFreq > 0)
        maxHeap.emplace(prevFreq, prevChar);
      prevFreq = freq - 1;
      prevChar = c;
    }

    return ans;
  }
};

			

public class Solution {
  public String reorganizeString(String s) {
    Map<Character, Integer> count = new HashMap<>();
    int maxFreq = 0;

    for (final char c : s.toCharArray()) {
      count.merge(c, 1, Integer::sum);
      maxFreq = Math.max(maxFreq, count.get(c));
    }

    if (maxFreq > (s.length() + 1) / 2)
      return "";

    StringBuilder sb = new StringBuilder();
    // (freq, c)
    Queue<Pair<Integer, Character>> maxHeap =
        new PriorityQueue<>((a, b) -> b.getKey() - a.getKey());
    int prevFreq = 0;
    char prevChar = '@';

    for (final char c : count.keySet())
      maxHeap.offer(new Pair<>(count.get(c), c));

    while (!maxHeap.isEmpty()) {
      // Get the most freq letter
      final int freq = maxHeap.peek().getKey();
      final char c = maxHeap.poll().getValue();
      sb.append(c);
      // Add the previous letter back so that
      // Any two adjacent characters are not the same
      if (prevFreq > 0)
        maxHeap.offer(new Pair<>(prevFreq, prevChar));
      prevFreq = freq - 1;
      prevChar = c;
    }

    return sb.toString();
  }
}

			

class Solution:
  def reorganizeString(self, s: str) -> str:
    count = Counter(s)
    if max(count.values()) > (len(s) + 1) // 2:
      return ''

    ans = []
    maxHeap = [(-freq, c) for c, freq in count.items()]
    heapq.heapify(maxHeap)
    prevFreq = 0
    prevChar = '@'

    while maxHeap:
      # Get the most freq letter
      freq, c = heapq.heappop(maxHeap)
      ans.append(c)
      # Add the previous letter back so that
      # Any two adjacent characters are not the same
      if prevFreq < 0:
        heapq.heappush(maxHeap, (prevFreq, prevChar))
      prevFreq = freq + 1
      prevChar = c

    return ''.join(ans)