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

Given a string str with repeated characters, the task is to rearrange the characters in a string such that no two adjacent characters are the same. If it is possible then print Yes else print No.

Examples:

Input: str = “geeksforgeeks”
Output: Yes
“egeksforegeks” is one such arrangement.

Input: str = “bbbbb”
Output: No

Approach: The idea is to store the frequency of each character in an unordered_map and compare maximum frequency of character with the difference of string length and maximum frequency number. If the maximum frequency is less than the difference then it can be arranged otherwise not.

1. Let we start putting all the character having maximum frequency alternatively. Then at minimum, we need (max_freq-1) spaces between them to solve the question so that they are not adjacent to each other.
2. But we have  (length of the string – max_freq) spaces left. So, (length of the string – max_freq) should be at least (max_freq-1) to arrange such that no two char are the same.
3. So, it goes this way: (max_freq-1) <= (length of the string – max_freq)

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``#include ``using` `namespace` `std;` `// Function that returns true if it is possible``// to rearrange the characters of the string``// such that no two consecutive characters are same``int` `isPossible(string str)``{` `    ``// To store the frequency of``    ``// each of the character``    ``unordered_map<``char``, ``int``> freq;` `    ``// To store the maximum frequency so far``    ``int` `max_freq = 0;``    ``for` `(``int` `j = 0; j < (str.length()); j++) {``        ``freq[str[j]]++;``        ``if` `(freq[str[j]] > max_freq)``            ``max_freq = freq[str[j]];``    ``}` `    ``// If possible``    ``if` `(max_freq <= (str.length() - max_freq + 1))``        ``return` `true``;``    ``return` `false``;``}` `// Driver code``int` `main()``{``    ``string str = ``"geeksforgeeks"``;` `    ``if` `(isPossible(str))``        ``cout << ``"Yes"``;``    ``else``        ``cout << ``"No"``;` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG {` `    ``// Function that returns true if it is possible``    ``// to rearrange the characters of the string``    ``// such that no two consecutive characters are same``    ``static` `boolean` `isPossible(``char``[] str)``    ``{` `        ``// To store the frequency of``        ``// each of the character``        ``Map freq = ``new` `HashMap<>();` `        ``// To store the maximum frequency so far``        ``int` `max_freq = ``0``;``        ``for` `(``int` `j = ``0``; j < (str.length); j++) {``            ``if` `(freq.containsKey(str[j])) {``                ``freq.put(str[j], freq.get(str[j]) + ``1``);``                ``if` `(freq.get(str[j]) > max_freq)``                    ``max_freq = freq.get(str[j]);``            ``}``            ``else` `{``                ``freq.put(str[j], ``1``);``                ``if` `(freq.get(str[j]) > max_freq)``                    ``max_freq = freq.get(str[j]);``            ``}``        ``}` `        ``// If possible``        ``if` `(max_freq <= (str.length - max_freq + ``1``))``            ``return` `true``;``        ``return` `false``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String str = ``"geeksforgeeks"``;` `        ``if` `(isPossible(str.toCharArray()))``            ``System.out.println(``"Yes"``);``        ``else``            ``System.out.println(``"No"``);``    ``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation of the approach` `# Function that returns true if it is possible``# to rearrange the characters of the String``# such that no two consecutive characters are same``def` `isPossible(``Str``):` `    ``# To store the frequency of``    ``# each of the character``    ``freq ``=` `dict``()` `    ``# To store the maximum frequency so far``    ``max_freq ``=` `0``    ``for` `j ``in` `range``(``len``(``Str``)):``        ``freq[``Str``[j]] ``=` `freq.get(``Str``[j], ``0``) ``+` `1``        ``if` `(freq[``Str``[j]] > max_freq):``            ``max_freq ``=` `freq[``Str``[j]]` `    ``# If possible``    ``if` `(max_freq <``=` `(``len``(``Str``) ``-` `max_freq ``+` `1``)):``        ``return` `True``    ``return` `False` `# Driver code``Str` `=` `"geeksforgeeks"` `if` `(isPossible(``Str``)):``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)` `# This code is contributed by Mohit Kumar`

## C#

 `// C# implementation of the approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG {` `    ``// Function that returns true if it is possible``    ``// to rearrange the characters of the string``    ``// such that no two consecutive characters are same``    ``static` `Boolean isPossible(``char``[] str)``    ``{` `        ``// To store the frequency of``        ``// each of the character``        ``Dictionary<``char``, ``int``> freq = ``new` `Dictionary<``char``, ``int``>();` `        ``// To store the maximum frequency so far``        ``int` `max_freq = 0;``        ``for` `(``int` `j = 0; j < (str.Length); j++) {``            ``if` `(freq.ContainsKey(str[j])) {``                ``var` `v = freq[str[j]] + 1;``                ``freq.Remove(str[j]);``                ``freq.Add(str[j], v);``                ``if` `(freq[str[j]] > max_freq)``                    ``max_freq = freq[str[j]];``            ``}``            ``else` `{``                ``freq.Add(str[j], 1);``                ``if` `(freq[str[j]] > max_freq)``                    ``max_freq = freq[str[j]];``            ``}``        ``}` `        ``// If possible``        ``if` `(max_freq <= (str.Length - max_freq + 1))``            ``return` `true``;``        ``return` `false``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``String str = ``"geeksforgeeks"``;` `        ``if` `(isPossible(str.ToCharArray()))``            ``Console.WriteLine(``"Yes"``);``        ``else``            ``Console.WriteLine(``"No"``);``    ``}``}` `// This code is contributed by Princi Singh`

## Javascript

 ``

Output:

`Yes`

Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(1), as we are using frequency map.

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