Given a string with lowercase repeated characters, the task is to rearrange characters in a string so that no two adjacent characters are the same. If it is not possible to do so, then print “Not possible”.
Examples:
Input: aaabc
Output: abacaInput: aaabb
Output: ababaInput: aa
Output: Not PossibleInput: aaaabc
Output: Not Possible
Asked In: Amazon Interview
To solve the problem using this approach follow the below idea:
The idea is to put the highest frequency character first (a greedy approach). Use a priority queue (Or Binary Max Heap) and put all characters and ordered by their frequencies (highest frequency character at root). One by one take the highest frequency character from the heap and add it to result. After adding it, just decrease the frequency of the character and then temporarily move this character out of priority queue so that it is not picked again next time.
Follow the given steps to solve the problem:
- Build a Priority_queue or max_heap, pq that stores characters with their frequencies.
- Priority_queue or max_heap is built on the basis of the frequency of character.
- 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’ }
- While pq is not empty.
- Pop an element and add it to the result.
- Decrease the frequency of the popped element by ‘1’
- Push the previous element back into the priority_queue if its frequency is greater than zero.
- Make the current element as the previous element for the next iteration.
- If the length of the resultant string and the original string is not equal, then print “not possible”, else print the resultant string.
Below is the implementation of the above approach:
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++) { int val = c - 'a'; if (count[val]) { pq.push(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{ -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 possible " << endl; else // valid string cout << str << endl;}// Driver's codeint main(){ string str = "bbbaa"; // Function call 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 possible "); else System.out.println(str); } // Driver's code public static void main(String args[]) { String str = "bbbaa"; // Function call rearrangeString(str); }}// This code is contributed by rachana soma |
Python3
# Python program to rearrange characters in a string# so that no two adjacent characters are same.from heapq import heappush, heappopfrom collections import Counter# A key class for readabilityclass Key: def __init__(self, character: str, freq: int) -> None: self.character = character self.freq = freq def __lt__(self, other: "Key") -> bool: return self.freq > other.freq# Function to rearrange character of a string# so that no char repeat twicedef rearrangeString(str: str): n = len(str) # Creating a frequency hashmap count = dict() for i in str: count[ord(i)] = count.get(ord(i), 0) + 1 pq = [] for c in range(97, 123): if count.get(c, 0): heappush(pq, Key(chr(c), count)) # null character for default previous checking prev = Key('#', -1) str = "" while pq: key = heappop(pq) str += key.character # Since one character is already added key.freq -= 1 # We avoid inserting if the frequency drops to 0 if prev.freq > 0: heappush(pq, prev) prev = key if len(str) != n: print("Not possible") else: print(str)# Driver's Codeif __name__ == "__main__": string = "bbbaa" # Function call rearrangeString(string) # This code is contributed by kraanzu. |
C#
// C# codeusing System; public class GFG { // Maximum number of characters const int MAX_CHAR = 26; // Structure to store character and its // frequency in the string public struct Key { public int freq; public char ch; // Function to compare two Key values public int CompareTo(Key other) { return freq - other.freq; } }; // Function to rearrange character of a string // so that no char repeat twice public 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[i] - 'a']++; // Insert all characters with their frequencies // into a priority_queue PriorityQueue<Key> pq = new PriorityQueue<Key>(); for (int c = 'a'; c <= 'z'; c++) { int val = c - 'a'; if (count[val] > 0) pq.Enqueue(new Key(count[val], (char)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.Count != 0) { // pop top element from queue and add it // to string. Key k = pq.Dequeue(); 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.Enqueue(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) Console.WriteLine(" Not possible "); else // valid string Console.WriteLine(str); } // Driver's code public static void Main() { String str = "bbbaa"; // Function call rearrangeString(str); } }// This code is contributed by ishankhandelwals. |
Javascript
// JS codeconst MAX_CHAR = 26;const rearrangeString = (str) => { let N = str.length; // Store frequencies of all characters in string let count = Array(MAX_CHAR).fill(0); for (let i = 0; i < N; i++) { count[str.charCodeAt(i) - 'a'.charCodeAt(0)]++; } // Insert all characters with their frequencies // into a priority_queue let pq = []; for (let c = 'a'; c <= 'z'; c++) { let val = c.charCodeAt(0) - 'a'.charCodeAt(0); if (count[val]) { pq.push({ freq: count[val], ch: c }); } } // 'str' that will store resultant value str = ""; // work as the previous visited element // initial previous element be. ( '#' and // it's frequency '-1' ) let prev = { freq: -1, ch: '#' }; // traverse queue while (pq.length > 0) { // pop top element from queue and add it // to string. let k = pq.shift(); 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) { console.log(" Not possible "); } else { // valid string console.log(str); }};// This code is contributed by ishankhandelwals. |
Output
babab
Time complexity : O(N log(N))
Auxiliary Space: O(N), Extra space is used to store the resultant string
Rearrange characters in a String such that no two adjacent characters are same using Frequency:
To solve the problem using this approach follow the below idea:
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, it can be ensured that no two adjacent characters are the same.
Follow the given steps to solve the problem:
- Calculate the frequencies of every character in the input string
- If a character with a maximum frequency has a frequency greater than (n + 1) / 2, then return an empty string, as it is not possible to construct a string
- Now fill the even index positions with the maximum frequency character, if some even positions are remaining then first fill them with remaining characters
- Then fill odd index positions with the remaining characters
- Return the constructed string
Below is the implementation of the above approach:
C
// C code for the above approach#include <stdio.h>#include <stdlib.h>#include <string.h>char getMaxCountChar(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;}char* rearrangeString(char S[]){ int N = strlen(S); if (N == 0) return ""; int count[26] = { 0 }; for (int i = 0; i < N; i++) count[S[i] - 'a']++; char ch_max = getMaxCountChar(count); int maxCount = count[ch_max - 'a']; char* res = malloc(N * sizeof(int)); int ind = 0; // check if the result is possible or not if (maxCount > (N + 1) / 2) return res; // filling the most frequently occurring 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's codeint main(){ char str[] = { "bbbaa" }; // Function call char* res = rearrangeString(str); if (res == "") printf("Not possible \n"); else printf("%s \n", res); return 0;} |
C++14
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;char getMaxCountChar(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 == 0) 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 occurring 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's codeint main(){ string str = "bbbaa"; // Function call string res = rearrangeString(str); if (res == "") cout << "Not possible" << endl; else cout << res << endl; return 0;} |
Java
// Java code for the above approachimport java.io.*;class GFG { static char getMaxCountChar(int[] count) { int max = 0; char ch = 0; for (int i = 0; i < 26; i++) { if (count[i] > max) { max = count[i]; ch = (char)((int)'a' + i); } } return ch; } static String rearrangeString(String S) { int N = S.length(); if (N == 0) return ""; int[] count = new int[26]; for (int i = 0; i < 26; i++) { count[i] = 0; } for (char ch : S.toCharArray()) { count[(int)ch - (int)'a']++; } char ch_max = getMaxCountChar(count); int maxCount = count[(int)ch_max - (int)'a']; // check if the result is possible or not if (maxCount > (N + 1) / 2) return ""; String res = ""; for (int i = 0; i < N; i++) { res += ' '; } int ind = 0; // filling the most frequently occurring char in the // even indices while (maxCount > 0) { res = res.substring(0, ind) + ch_max + res.substring(ind + 1); ind = ind + 2; maxCount--; } count[(int)ch_max - (int)'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 = res.substring(0, ind) + (char)((int)'a' + i) + res.substring(ind + 1); ind += 2; count[i]--; } } return res; } // Driver's Code public static void main(String args[]) { String str = "bbbaa"; // Function call String res = rearrangeString(str); if (res == "") System.out.println("Not possible"); else System.out.println(res); }}// This code is contributed by shinjanpatra |
Python3
# Python program for rearranging characters in a string such# that no two adjacent are same# Function to find the char with maximum frequency in the given# stringdef getMaxCountChar(count): maxCount = 0 for i in range(26): if count[i] > maxCount: maxCount = count[i] maxChar = chr(i + ord('a')) return maxCount, maxChar# Main function for rearranging the charactersdef rearrangeString(S): N = len(S) # if length of string is None return False if not N: return False # create a hashmap for the alphabets count = [0] * 26 for char in S: count[ord(char) - ord('a')] += 1 maxCount, maxChar = getMaxCountChar(count) # if the char with maximum frequency is more than the half of the # total length of the string than return False if maxCount > (N + 1) // 2: return False # create a list for storing the result res = [None] * N ind = 0 # place all occurrences of the char with maximum frequency in # even positions while maxCount: res[ind] = maxChar ind += 2 maxCount -= 1 # replace the count of the char with maximum frequency to zero # as all the maxChar are already placed in the result count[ord(maxChar) - ord('a')] = 0 # place all other char in the result starting from remaining even # positions and then place in the odd positions for i in range(26): while count[i] > 0: if ind >= N: ind = 1 res[ind] = chr(i + ord('a')) ind += 2 count[i] -= 1 # convert the result list to string and return return ''.join(res)# Driver Codeif __name__ == '__main__': str = 'bbbaa' # Function call res = rearrangeString(str) if res: print(res) else: print('Not possible')# This code is contributed by Manish Thapa |
C#
// C# codeusing System; public class GFG { // Maximum number of characters const int MAX_CHAR = 26; // Structure to store character and its // frequency in the string public struct Key { public int freq; public char ch; // Function to compare two Key values public int CompareTo(Key other) { return freq - other.freq; } }; // Function to rearrange character of a string // so that no char repeat twice public 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[i] - 'a']++; // Insert all characters with their frequencies // into a priority_queue PriorityQueue<Key> pq = new PriorityQueue<Key>(); for (int c = 'a'; c <= 'z'; c++) { int val = c - 'a'; if (count[val] > 0) pq.Enqueue(new Key(count[val], (char)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.Count != 0) { // pop top element from queue and add it // to string. Key k = pq.Dequeue(); 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.Enqueue(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) Console.WriteLine(" Not possible "); else // valid string Console.WriteLine(str); } // Driver's code public static void Main() { String str = "bbbaa"; // Function call rearrangeString(str); } } |
Javascript
// JavaScript program for rearranging characters in a string such// that no two adjacent are same// Function to find the char with maximum frequency in the given// stringfunction getMaxCountChar(count){let maxCount = 0let maxCharfor(let i = 0; i < 26; i++){ if(count[i] > maxCount){ maxCount = count[i] maxChar = String.fromCharCode(i + ('a').charCodeAt(0)) }}return [maxCount, maxChar]}// Main function for rearranging the charactersfunction rearrangeString(S){let n = S.length// if length of string is None return falseif(!n) return false // create a hashmap for the alphabetslet count = new Array(26).fill(0)for(let char of S) count[char.charCodeAt(0) - ('a').charCodeAt(0)] += 1let [maxCount, maxChar] = getMaxCountChar(count)// if the char with maximum frequency is more than the half of the// total length of the string than return falseif(maxCount > Math.floor((n + 1) / 2)) return false// create a list for storing the resultlet res = new Array(n)let ind = 0// place all occurrences of the char with maximum frequency in// even positionswhile(maxCount){ res[ind] = maxChar ind += 2 maxCount -= 1} // replace the count of the char with maximum frequency to zero// as all the maxChar are already placed in the resultcount[maxChar.charCodeAt(0) - 'a'.charCodeAt(0)] = 0// place all other char in the result starting from remaining even// positions and then place in the odd positionsfor(let i = 0; i < 26; i++){ while(count[i] > 0) { if(ind >= n) ind = 1 res[ind] = String.fromCharCode(i + ('a').charCodeAt(0)) ind += 2 count[i] -= 1 }}// convert the result list to string and returnreturn res.join('')}// Driver Codelet str = 'bbbaa'let res = rearrangeString(str)if(res) document.write(res)else document.write('Not valid string')// This code is contributed by shinjanpatra |
Output
babab
Time complexity: O(N)
Auxiliary Space: O(N+26), as there are 26 lowercase english characters
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