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

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• The very first observation is if all the characters of the string are same then every substring of odd length is a palindrome and we do not need to rearrange them.
• Second observation is if the number of distinct characters are 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 satisfy then we rearrange the string in alternate manner means for i <— ( 1 to n – 1 ). Where n is the length of the string.

Below is the implementation of 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; ` `} `

## 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 `

Output:

```YES
```

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