Check if there exists a permutation of given string which doesn’t contain any monotonous substring

Given a string S of lowercase English alphabets, the task is to check if there exists an arrangement of string S such that it doesn’t contain any monotonous substring. 

A monotonous substring has the following properties: 

• Length of such substring is 2.
• Both the characters are consecutive, For example – “ab”, “cd”, “dc”, “zy” etc.

Examples:

Input: S = “abcd” 
Output: Yes 
Explanation:
String S can be rearranged into “cadb” or “bdac”

Input: string = “aab” 
Output: No 
Explanation:
Every arrangement of the string contains a monotonous substring. 

Approach: The idea is to group the characters into two different buckets, where one bucket contains the characters which are at the even places and another bucket contains the characters which are at the odd places. Finally, checking for the concatenation point of both groups is not a monotonous substring.

Below is the implementation of the above approach: 

C++

// C++ implementation such that there
// are no monotonous
// string in given string

#include <bits/stdc++.h>

using namespace std;

// Function to check a string doesn't
// contains a monotonous substring
bool check(string s)
{
    bool ok = true;

    // Loop to iterate over the string
    // and check that it doesn't contains
    // the monotonous substring
    for (int i = 0; i + 1 < s.size(); ++i)
        ok &= (abs(s[i] - s[i + 1]) != 1);
    return ok;
}

// Function to check that there exist
// a arrangement of string such that
// it doesn't contains monotonous substring
string monotonousString(string s)
{
    string odd = "", even = "";

    // Loop to group the characters
    // of the string into two buckets
    for (int i = 0; i < s.size(); ++i) {
        if (s[i] % 2 == 0)
            odd += s[i];
        else
            even += s[i];
    }

    // Sorting the two buckets
    sort(odd.begin(), odd.end());
    sort(even.begin(), even.end());

    // Condition to check if the
    // concatenation point doesn't
    // contains the monotonous string
    if (check(odd + even))
        return "Yes";
    else if (check(even + odd))
        return "Yes";
    return "No";
}

// Driver Code
int main()
{
    string str = "abcd";
    string ans;
    ans = monotonousString(str);
    cout << ans << endl;
    return 0;
}

Java

// Java implementation such that there
// are no monotonous
// string in given string
import java.io.*;
import java.util.*;

class GFG{

// Function to check a string doesn't
// contains a monotonous substring
static boolean check(String s)
{
    boolean ok = true;

    // Loop to iterate over the string
    // and check that it doesn't contains
    // the monotonous substring
    for (int i = 0; i + 1 < s.length(); ++i)
        ok &= (Math.abs(s.charAt(i) -
                        s.charAt(i + 1)) != 1);
    return ok;
}

// Function to check that there exist
// a arrangement of string such that
// it doesn't contains monotonous substring
static String monotonousString(String s)
{
    String odd = "", even = "";

    // Loop to group the characters
    // of the string into two buckets
    for (int i = 0; i < s.length(); ++i)
    {
          char c=s.charAt(i);
        if (c % 2 == 0)
            odd += c;
        else
            even += c;
    }

    // Sorting the two buckets
    char oddArray[] = odd.toCharArray();
    Arrays.sort(oddArray);
    odd = new String(oddArray);

    char evenArray[] = even.toCharArray();
    Arrays.sort(evenArray);
    even = new String(evenArray);

    // Condition to check if the
    // concatenation point doesn't
    // contains the monotonous string
    if (check(odd + even))
        return "Yes";
    else if (check(even + odd))
        return "Yes";
    return "No";
}

// Driver Code
public static void main(String []args)
{
    String str = "abcd";
    String ans;
    ans = monotonousString(str);
    System.out.println( ans);
}
}

// This code is contributed by ChitraNayal

Python3

# Python3 implementation such that there
# are no monotonous string in given string

# Function to check a string doesn't
# contains a monotonous substring
def check(s):

    ok = True

    # Loop to iterate over the string
    # and check that it doesn't contains
    # the monotonous substring
    for i in range(0, len(s) - 1, 1):
        ok = (ok & (abs(ord(s[i]) -
                        ord(s[i + 1])) != 1))
    return ok

# Function to check that there exist
# a arrangement of string such that
# it doesn't contains monotonous substring
def monotonousString(s):

    odd = ""
    even = ""

    # Loop to group the characters
    # of the string into two buckets
    for i in range(len(s)):

        if (ord(s[i]) % 2 == 0):
            odd += s[i]
        else:
            even += s[i]

    # Sorting the two buckets
    odd = list(odd)
    odd.sort(reverse = False)
    odd = str(odd)

    even = list(even)
    even.sort(reverse = False)
    even = str(even)

    # Condition to check if the
    # concatenation point doesn't
    # contains the monotonous string
    if (check(odd + even)):
        return "Yes"
    elif (check(even + odd)):
        return "Yes"

    return "No"

# Driver Code
if __name__ == '__main__':

    str1 = "abcd"
    ans = monotonousString(str1)

    print(ans)

# This code is contributed by Samarth

C#

// C# implementation such that there
// are no monotonous
// string in given string
using System;

class GFG{

// Function to check a string doesn't
// contains a monotonous substring
static bool check(string s)
{
    bool ok = true;

    // Loop to iterate over the string
    // and check that it doesn't contains
    // the monotonous substring
    for(int i = 0; i + 1 < s.Length; ++i)
        ok &= (Math.Abs(s[i] -
                        s[i + 1]) != 1);

    return ok;
}

// Function to check that there exist
// a arrangement of string such that
// it doesn't contains monotonous substring
static string monotonousString(string s)
{
    string odd = "", even = "";

    // Loop to group the characters
    // of the string into two buckets
    for(int i = 0; i < s.Length; ++i)
    {
        if (s[i] % 2 == 0)
            odd += s[i];
        else
            even += s[i];
    }

    // Sorting the two buckets
    char []oddArray = odd.ToCharArray();
    Array.Sort(oddArray);
    odd = new String(oddArray);

    char []evenArray = even.ToCharArray();
    Array.Sort(evenArray);
    even = new String(evenArray);

    // Condition to check if the
    // concatenation point doesn't
    // contains the monotonous string
    if (check(odd + even))
        return "Yes";

    else if (check(even + odd))
        return "Yes";

    return "No";
}

// Driver Code
public static void Main(string []args)
{
    string str = "abcd";
    string ans;

    ans = monotonousString(str);

    Console.Write(ans);
}
}

// This code is contributed by rutvik_56

Javascript

<script>

// Javascript implementation such that there
// are no monotonous
// string in given string

// Function to check a string doesn't
// contains a monotonous substring
function check(s)
{
    var ok = true;

    // Loop to iterate over the string
    // and check that it doesn't contains
    // the monotonous substring
    for (var i = 0; i + 1 < s.length; ++i)
        ok &= (Math.abs(s[i] - s[i + 1]) != 1);
    return ok;
}

// Function to check that there exist
// a arrangement of string such that
// it doesn't contains monotonous substring
function monotonousString(s)
{
    var odd = "", even = "";

    // Loop to group the characters
    // of the string into two buckets
    for (var i = 0; i < s.length; ++i) {
        if (s[i] % 2 == 0)
            odd += s[i];
        else
            even += s[i];
    }

    // Sorting the two buckets
    odd  = odd.split('').sort().join('');
    even = even.split('').sort().join('');

    // Condition to check if the
    // concatenation point doesn't
    // contains the monotonous string
    if (check(odd + even))
        return "Yes";
    else if (check(even + odd))
        return "Yes";
    return "No";
}

// Driver Code
var str = "abcd";
var ans;
ans = monotonousString(str);
document.write( ans );


</script>

Output:

Yes

Time Complexity : O(|S|*log|S|) ,where |S| is size of string.
Auxiliary Space: O(1)