Related Articles

# Check if string can be rearranged so that every Odd length Substring is Palindrome

• Last Updated : 31 Oct, 2020

Given a string S. The task is to check whether it is possible to rearrange the string such that every substring of odd length is a palindrome.
Examples:

Input: S = “oiooi”
Output: YES
The string can be rearranged as “oioio”
Input: S = “yuyuo”
Output: NO

Approach:

• The very first observation is if all the characters of the string are the same then every substring of odd length is a palindrome, and we do not need to rearrange them.
• The second observation is if the number of distinct characters is more than 2 then it is impossible to rearrange.
• Now if the number of distinct characters are exactly 2 then to get all odd length substring to be a palindrome, the difference of their count must be less than or equal to 1, and if this satisfies then we rearrange the string in an alternate manner means • for i <— ( 1 to n – 1 ). Where n is the length of the string.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of``// the above approach``#include ``using` `namespace` `std;` `// Function to check is it``// possible to rearrange the string``// such that every odd length``// substring is palindrome``bool` `IsPossible(string s)``{` `    ``// Length of the string``    ``int` `n = s.length();` `    ``// To count number of distinct``    ``// character in string``    ``set<``char``> count;` `    ``// To count frequency of``    ``// each character``    ``map<``char``, ``int``> map;` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Inserting into set``        ``count.insert(s[i]);` `        ``// Incrementing the frequency``        ``map[s[i]] += 1;``    ``}` `    ``// All characters in``    ``// the string are same``    ``if` `(count.size() == 1) {``        ``return` `true``;``    ``}` `    ``// There are more than 2 different``    ``// character in string``    ``if` `(count.size() > 2) {``        ``return` `false``;``    ``}` `    ``// Currently there is 2 different``    ``// character in string``    ``auto` `it = count.begin();` `    ``// Get the frequencies of the``    ``// characters that present``    ``// in string``    ``int` `x = 0, y = 0;``    ``x = map[*it];` `    ``it++;``    ``y = map[*it];` `    ``// Difference between their``    ``// count is less than or``    ``// equal to 1``    ``if` `(``abs``(x - y) <= 1) {``        ``return` `true``;``    ``}` `    ``return` `false``;``}` `// Driver code``int` `main()``{``    ``string s = ``"aaaddad"``;` `    ``if` `(IsPossible(s))``        ``cout << ``"YES\n"``;``    ``else``        ``cout << ``"NO\n"``;` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{` `    ``// Function to check is it``    ``// possible to rearrange the string``    ``// such that every odd length``    ``// substring is palindrome``    ``static` `boolean` `IsPossible(String s)``    ``{` `        ``// Length of the string``        ``int` `n = s.length();` `        ``// To count number of distinct``        ``// character in string``        ``HashSet count = ``new` `HashSet<>();` `        ``// To count frequency of``        ``// each character``        ``HashMap map = ``new` `HashMap<>();` `        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{` `            ``// Inserting into set``            ``count.add(s.charAt(i));` `            ``// Incrementing the frequency``            ``map.put(s.charAt(i), map.get(s.charAt(i)) ==``                    ``null` `? ``1` `: map.get(s.charAt(i)) + ``1``);``        ``}` `        ``// All characters in``        ``// the string are same``        ``if` `(count.size() == ``1``)``            ``return` `true``;` `        ``// There are more than 2 different``        ``// character in string``        ``if` `(count.size() > ``2``)``            ``return` `false``;` `        ``String newString = count.toArray().toString();` `        ``// Currently there is 2 different``        ``// character in string``        ``int` `j = ``0``;``        ``char` `it = newString.charAt(j);` `        ``// Get the frequencies of the``        ``// characters that present``        ``// in string``        ``int` `x = ``0``, y = ``0``;``        ``x = map.get(it) == ``null` `? ``0` `: map.get(it);``        ``j++;` `        ``it = newString.charAt(j);``        ``y = map.get(it) == ``null` `? ``0` `: map.get(it);` `        ``// Difference between their``        ``// count is less than or``        ``// equal to 1``        ``if` `(Math.abs(x - y) <= ``1``)``            ``return` `true``;``        ``return` `false``;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String s = ``"aaaddad"``;``        ``if` `(IsPossible(s))``            ``System.out.println(``"YES"``);``        ``else``            ``System.out.println(``"NO"``);``    ``}``}` `// This code is contributed by``// sanjeev2552`

## Python3

 `# Python3 implementation of``# the above approach` `# Function to check is it``# possible to rearrange the string``# such that every odd length``# substring is palindrome``def` `IsPossible(s) :` `    ``# Length of the string``    ``n ``=` `len``(s);` `    ``# To count number of distinct``    ``# character in string``    ``count ``=` `set``();``    ` `    ``# To count frequency of``    ``# each character``    ``map` `=` `dict``.fromkeys(s, ``0``);` `    ``for` `i ``in` `range``(n) :` `        ``# Inserting into set``        ``count.add(s[i]);` `        ``# Incrementing the frequency``        ``map``[s[i]] ``+``=` `1``;` `    ``# All characters in``    ``# the string are same``    ``if` `(``len``(count) ``=``=` `1``) :``        ``return` `True``;` `    ``# There are more than 2 different``    ``# character in string``    ``if` `(``len``(count) > ``2``) :``        ``return` `False``;``        ` `    ``# Currently there is 2 different``    ``# character in string``    ``j ``=` `0``    ``it ``=` `list``(count)[j];` `    ``# Get the frequencies of the``    ``# characters that present``    ``# in string``    ``x ``=` `0``; y ``=` `0``;``    ``x ``=` `map``[it];``    ` `    ``j ``+``=` `1``    ``it ``=` `list``(count)[j];``    ``y ``=` `map``[it];` `    ``# Difference between their``    ``# count is less than or``    ``# equal to 1``    ``if` `(``abs``(x ``-` `y) <``=` `1``) :``        ``return` `True``;` `    ``return` `False``;` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``s ``=` `"aaaddad"``;` `    ``if` `(IsPossible(s)) :``        ``print``(``"YES"``);``    ``else` `:``        ``print``(``"NO"``);``        ` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the``// above approach``using` `System;``using` `System.Collections.Generic;``class` `GFG{` `// Function to check is it``// possible to rearrange the string``// such that every odd length``// substring is palindrome``static` `bool` `IsPossible(String s)``{` `  ``// Length of the string``  ``int` `n = s.Length;` `  ``// To count number of distinct``  ``// character in string``  ``HashSet<``char``> count =``          ``new` `HashSet<``char``>();` `  ``// To count frequency of``  ``// each character``  ``Dictionary<``char``,``             ``int``> map = ``new` `Dictionary<``char``,``                                       ``int``>();` `  ``for` `(``int` `i = 0; i < n; i++)``  ``{``    ``// Inserting into set``    ``count.Add(s[i]);` `    ``// Incrementing the frequency``    ``if``(map.ContainsKey(s[i]))``      ``map[s[i]] = map[s[i]] + 1;``    ``else``      ``map.Add(s[i], 1);``  ``}` `  ``// All characters in``  ``// the string are same``  ``if` `(count.Count == 1)``    ``return` `true``;` `  ``// There are more than 2 different``  ``// character in string``  ``if` `(count.Count > 2)``    ``return` `false``;` `  ``String newString = count.ToString();` `  ``// Currently there is 2 different``  ``// character in string``  ``int` `j = 0;``  ``char` `it = newString[j];` `  ``// Get the frequencies of the``  ``// characters that present``  ``// in string``  ``int` `x = 0, y = 0;``  ``x = !map.ContainsKey(it) ?``       ``0 : map[it];``  ``j++;` `  ``it = newString[j];``  ``y = !map.ContainsKey(it) ?``       ``0 : map[it];` `  ``// Difference between their``  ``// count is less than or``  ``// equal to 1``  ``if` `(Math.Abs(x - y) <= 1)``    ``return` `true``;``  ``return` `false``;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``  ``String s = ``"aaaddad"``;``  ``if` `(IsPossible(s))``    ``Console.WriteLine(``"YES"``);``  ``else``    ``Console.WriteLine(``"NO"``);``}``}` `// This code is contributed by 29AjayKumar`
Output:
```YES

```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up