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 <bits/stdc++.h>
#include <time.h>
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<Character, Integer> 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

<script>

    // Javascript 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
    function isPossible(str)
    {

        // To store the frequency of
        // each of the character
        let freq = new Map();

        // To store the maximum frequency so far
        let max_freq = 0;
        for (let j = 0; j < (str.length); j++) {
            if (freq.has(str[j])) {
                freq.set(str[j], freq.get(str[j]) + 1);
                if (freq.get(str[j]) > max_freq)
                    max_freq = freq.get(str[j]);
            }
            else {
                freq.set(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

     let str = "geeksforgeeks";

        if (isPossible(str.split('')))
            document.write("Yes");
        else
           document.write("No");

// This code is contributed by rrrtnx.
</script>

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.