767. Reorganize String LeetCode Solution

In this guide, you will get 767. Reorganize String LeetCode Solution with the best time and space complexity. The solution to Reorganize String problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Reorganize String solution in C++
Reorganize String solution in Java
Reorganize String solution in Python
Additional Resources

767. Reorganize String LeetCode Solution image

Problem Statement of Reorganize String

Given a string s, rearrange the characters of s so that any two adjacent characters are not the same.
Return any possible rearrangement of s or return “” if not possible.

Example 1:
Input: s = “aab”
Output: “aba”
Example 2:
Input: s = “aaab”
Output: “”

Constraints:

1 <= s.length <= 500
s consists of lowercase English letters.

Complexity Analysis

Time Complexity: O(n\log 26) = O(n)
Space Complexity: O(n)

767. Reorganize String LeetCode Solution in C++

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 letter with the maximum frequency.
      const auto [freq, c] = maxHeap.top();
      maxHeap.pop();
      ans += c;
      // Add the previous letter back s.t. any two adjacent characters are not
      // the same.
      if (prevFreq > 0)
        maxHeap.emplace(prevFreq, prevChar);
      prevFreq = freq - 1;
      prevChar = c;
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

767. Reorganize String LeetCode Solution in Java

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

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

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

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

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

    while (!maxHeap.isEmpty()) {
      // Get the letter with the maximum frequency.
      final int freq = maxHeap.peek().getKey();
      final char c = maxHeap.poll().getValue();
      sb.append(c);
      // Add the previous letter back s.t. 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();
  }
}
// code provided by PROGIEZ

767. Reorganize String LeetCode Solution in Python

class Solution:
  def reorganizeString(self, s: str) -> str:
    count = collections.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 letter with the maximum frequency.
      freq, c = heapq.heappop(maxHeap)
      ans.append(c)
      # Add the previous letter back s.t. any two adjacent characters are not
      # the same.
      if prevFreq < 0:
        heapq.heappush(maxHeap, (prevFreq, prevChar))
      prevFreq = freq + 1
      prevChar = c

    return ''.join(ans)
 # code by PROGIEZ