Rearrange characters in a string such that no two adjacent are same

Given a string with repeated characters, task is rearrange characters in a string so that no two adjacent characters are same.

Note : It may be assumed that the string has only lowercase English alphabets.

Examples:

Input: aaabc 
Output: abaca 

Input: aaabb
Output: ababa 

Input: aa 
Output: Not Possible

Input: aaaabc 
Output: Not Possible

Asked In : Amazon Interview

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Prerequisite : priority_queue.

The idea is to put highest frequency character first (a greedy approach). We use a priority queue (Or Binary Max Heap) and put all characters and ordered by their frequencies (highest frequency character at root). We one by one take highest frequency character from the heap and add it to result. After we add, we decrease frequency of the character and we temporarily move this character out of priority queue so that it is not picked next time.

We have to follow the step to solve this problem, they are:
1. Build a Priority_queue or max_heap, pq that stores characters and their frequencies.
…… Priority_queue or max_heap is built on the bases of frequency of character.
2. Create a temporary Key that will used as the previous visited element ( previous element in resultant string. Initialize it { char = ‘#’ , freq = ‘-1’ }
3. While pq is not empty.
….. Pop an element and add it to result.
….. Decrease frequency of the popped element by ‘1’
….. Push the previous element back into the priority_queue if it’s frequency > ‘0’
….. Make the current element as previous element for the next iteration.
4. If length of resultant string and original, print “not possible”. Else print result.

Below is the implementation of above idea

C++

// C++ program to rearrange characters in a string 
// so that no two adjacent characters are same. 
#include<bits/stdc++.h> 
using namespace std; 
  
const int MAX_CHAR = 26; 
  
struct Key 
{ 
    int freq; // store frequency of character 
    char ch; 
  
    // function for priority_queue to store Key 
    // according to freq 
    bool operator<(const Key &k) const
    { 
        return freq < k.freq; 
    } 
}; 
  
// Function to rearrange character of a string 
// so that no char repeat twice 
void rearrangeString(string str) 
{ 
    int n = str.length(); 
  
    // Store frequencies of all characters in string 
    int count[MAX_CHAR] = {0}; 
    for (int i = 0 ; i < n ; i++) 
        count[str[i]-'a']++; 
  
    // Insert all characters with their frequencies 
    // into a priority_queue 
    priority_queue< Key > pq; 
    for (char c = 'a' ; c <= 'z' ; c++) 
        if (count[c-'a']) 
            pq.push( Key { count[c-'a'], c} ); 
  
    // 'str' that will store resultant value 
    str = "" ; 
  
    // work as the previous visited element 
    // initial previous element be. ( '#' and 
    // it's frequency '-1' ) 
    Key prev {-1, '#'} ; 
  
    // traverse queue 
    while (!pq.empty()) 
    { 
        // pop top element from queue and add it 
        // to string. 
        Key k = pq.top(); 
        pq.pop(); 
        str = str + k.ch; 
  
        // IF frequency of previous character is less 
        // than zero that means it is useless, we 
        // need not to push it 
        if (prev.freq > 0) 
            pq.push(prev); 
  
        // make current character as the previous 'char' 
        // decrease frequency by 'one' 
        (k.freq)--; 
        prev = k; 
    } 
  
    // If length of the resultant string and original 
    // string is not same then string is not valid 
    if (n != str.length()) 
        cout << " Not valid String " << endl; 
  
    else // valid string 
        cout << str << endl; 
} 
  
// Driver program to test above function 
int main() 
{ 
    string str = "bbbaa" ; 
    rearrangeString(str); 
    return 0; 
} 

Java

// Java program to rearrange characters in a string 
// so that no two adjacent characters are same. 
import java.io.*; 
import java.util.*; 
  
class KeyComparator implements Comparator<Key> { 
  
      // Overriding compare()method of Comparator 
      public int compare(Key k1, Key k2) 
      { 
             if (k1.freq < k2.freq) 
                 return 1; 
             else if (k1.freq > k2.freq) 
                 return -1; 
             return 0; 
      } 
} 
  
class Key { 
      int freq; // store frequency of character 
      char ch; 
      Key(int val, char c)  
      { 
          freq = val;  
          ch = c; 
      } 
} 
  
class GFG { 
      static int MAX_CHAR = 26;  
  
      // Function to rearrange character of a string 
      // so that no char repeat twice 
      static void rearrangeString(String str) 
      { 
             int n = str.length(); 
  
             // Store frequencies of all characters in string 
             int[] count = new int[MAX_CHAR]; 
               
             for (int i = 0; i < n; i++) 
                  count[str.charAt(i) - 'a']++; 
  
              // Insert all characters with their frequencies 
              // into a priority_queue  
              PriorityQueue<Key> pq = new PriorityQueue<>(new 
                                                          KeyComparator()); 
              for (char c = 'a'; c <= 'z'; c++) { 
                   int val = c - 'a'; 
                   if (count[val] > 0) 
                       pq.add(new Key(count[val], c)); 
              } 
                   
              // 'str' that will store resultant value 
              str = "" ; 
  
              // work as the previous visited element 
              // initial previous element be. ( '#' and 
              // it's frequency '-1' ) 
              Key prev = new Key(-1, '#'); 
  
              // traverse queue 
              while (pq.size() != 0) { 
                
                     // pop top element from queue and add it 
                     // to string. 
                     Key k = pq.peek(); 
                     pq.poll(); 
                     str = str + k.ch; 
  
                     // If frequency of previous character is less 
                     // than zero that means it is useless, we 
                     // need not to push it  
                     if (prev.freq > 0) 
                         pq.add(prev); 
  
                     // make current character as the previous 'char' 
                     // decrease frequency by 'one' 
                     (k.freq)--; 
                      prev = k; 
              } 
  
              // If length of the resultant string and original 
              // string is not same then string is not valid 
              if (n != str.length()) 
                  System.out.println(" Not valid String "); 
              else
                  System.out.println(str); 
      } 
  
      // Driver program to test above function  
      public static void main(String args[]) 
      { 
             String str = "bbbaa" ; 
             rearrangeString(str); 
      } 
} 
  
// This code is contributed by rachana soma 

Output:

babab

Time complexity : O(n log n)

This article is contributed by Nishant Singh . If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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