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

• Difficulty Level : Medium
• Last Updated : 02 Dec, 2022

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 os 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 ` `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

 ``

Output:

`Yes`

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

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