Given a string with repeated characters, the task is to 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
Prerequisite : priority_queue.
The idea is to put the 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 the highest frequency character from the heap and add it to result. After we add, we decrease the 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 the frequency of character.
2. Create a temporary Key that will be used as the previously visited element (the previous element in the resultant string. Initialize it { char = ‘#’ , freq = ‘-1’ }
3. While pq is not empty.
….. Pop an element and add it to the 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 the previous element for the next iteration.
4. If the length of the resultant string and original string is not equal, 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 twicevoid 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) pq.push(Key{ count, 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 functionint 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(nlog(n))
Another approach :
Another approach is to fill all the even positions of the result string first, with the highest frequency character. If there are still some even positions remaining, fill them first. Once even positions are done, then fill the odd positions. This way, we can ensure that no two adjacent characters are the same.
C++14
#include <bits/stdc++.h>using namespace std;char getMaxCountChar(const vector<int>& count){ int max = 0; char ch; for (int i = 0; i < 26; i++) { if (count[i] > max) { max = count[i]; ch = 'a' + i; } } return ch;}string rearrangeString(string S){ int n = S.size(); if (!n) return ""; vector<int> count(26, 0); for (auto ch : S) count[ch - 'a']++; char ch_max = getMaxCountChar(count); int maxCount = count[ch_max - 'a']; // check if the result is possible or not if (maxCount > (n + 1) / 2) return ""; string res(n, ' '); int ind = 0; // filling the most frequently occuring char in the even // indices while (maxCount) { res[ind] = ch_max; ind = ind + 2; maxCount--; } count[ch_max - 'a'] = 0; // now filling the other Chars, first filling the even // positions and then the odd positions for (int i = 0; i < 26; i++) { while (count[i] > 0) { ind = (ind >= n) ? 1 : ind; res[ind] = 'a' + i; ind += 2; count[i]--; } } return res;}// Driver program to test above functionint main(){ string str = "bbbaa"; string res = rearrangeString(str); if (res == "") cout << "Not valid string" << endl; else cout << res << endl; return 0;} |
Output
Time complexity : O(n)
Space complexity : O(n+26) where 26 is the size of the vocabulary.
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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