# Minimum length of substring whose rotation generates a palindromic substring

Given a string str, the task is to find the minimum length of substring required to rotate that generates a palindromic substring from the given string.

Examples:

Input: str = “abcbd”
Output: 0
Explanation: No palindromic substring can be generated there is no repeating character in the string.

Input: str = “abcdeba”
Output: 3
Explanation: Rotate substring “deb” to convert the given string to abcbeda with a palindromic substring “bcb”.

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

Approach:

• If no character repeats in the string, then no palindromic substring can be generated.
• For every repeating character, check if the index of its previous occurrence is within one or two indices from the current index. If so, then a palindromic substring already exists.
• Otherwise, calculate the length of (current index – index of previous occurrence – 1).
• Return the minimum of all such lengths as the answer

Below is the implementation of the above approach:

## C++

 `// C++ Program to find the minimum ` `// length of substring whose rotation ` `// generates a palindromic substring ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to return the ` `// minimum lenth of substring ` `int` `count_min_length(string s) ` `{ ` ` `  `    ``// Store the index of ` `    ``// previous occurrence ` `    ``// of the character ` `    ``int` `hash; ` ` `  `    ``// Variable to store ` `    ``// the maximum length ` `    ``// of substring ` `    ``int` `ans = INT_MAX; ` ` `  `    ``for` `(``int` `i = 0; i < 26; i++) ` `        ``hash[i] = -1; ` ` `  `    ``for` `(``int` `i = 0; i < s.size(); i++) { ` `        ``// If the current character ` `        ``// hasn't appeared yet ` `        ``if` `(hash[s[i] - ``'a'``] == -1) ` `            ``hash[s[i] - ``'a'``] = i; ` `        ``else` `{ ` `            ``// If the character has occured ` `            ``// within one or two previous ` `            ``// index, a palindromic substring ` `            ``// already exists ` `            ``if` `(hash[s[i] - ``'a'``] == i - 1 ` `                ``|| hash[s[i] - ``'a'``] == i - 2) ` `                ``return` `0; ` ` `  `            ``// Update the maximum ` `            ``ans = min(ans, ` `                      ``i - hash[s[i] - ``'a'``] - 1); ` ` `  `            ``// Replace the previous ` `            ``// index of the character by ` `            ``// the current index ` `            ``hash[s[i] - ``'a'``] = i; ` `        ``} ` `    ``} ` ` `  `    ``// If character appeared ` `    ``// at least twice ` `    ``if` `(ans == INT_MAX) ` `        ``return` `-1; ` ` `  `    ``return` `ans; ` `} ` `// Driver Code ` `int` `main() ` `{ ` `    ``string str = ``"abcdeba"``; ` `    ``cout << count_min_length(str); ` `} `

## Java

 `// Java Program to find the minimum ` `// length of substring whose rotation ` `// generates a palindromic substring ` `import` `java.util.*; ` `import` `java.lang.*; ` `class` `GFG{ ` ` `  `// Function to return the ` `// minimum lenth of substring ` `static` `int` `count_min_length(String s) ` `{ ` ` `  `    ``// Store the index of ` `    ``// previous occurrence ` `    ``// of the character ` `    ``int``[] hash = ``new` `int``[``26``]; ` ` `  `    ``// Variable to store ` `    ``// the maximum length ` `    ``// of substring ` `    ``int` `ans = Integer.MAX_VALUE; ` ` `  `    ``for` `(``int` `i = ``0``; i < ``26``; i++) ` `        ``hash[i] = -``1``; ` ` `  `    ``for` `(``int` `i = ``0``; i < s.length(); i++)  ` `    ``{ ` `        ``// If the current character ` `        ``// hasn't appeared yet ` `        ``if` `(hash[s.charAt(i) - ``'a'``] == -``1``) ` `            ``hash[s.charAt(i) - ``'a'``] = i; ` `        ``else`  `        ``{ ` `            ``// If the character has occured ` `            ``// within one or two previous ` `            ``// index, a palindromic substring ` `            ``// already exists ` `            ``if` `(hash[s.charAt(i) - ``'a'``] == i - ``1` `||  ` `                ``hash[s.charAt(i) - ``'a'``] == i - ``2``) ` `                ``return` `0``; ` ` `  `            ``// Update the maximum ` `            ``ans = Math.min(ans,  ` `                       ``i - hash[s.charAt(i) - ``'a'``] - ``1``); ` ` `  `            ``// Replace the previous ` `            ``// index of the character by ` `            ``// the current index ` `            ``hash[s.charAt(i) - ``'a'``] = i; ` `        ``} ` `    ``} ` ` `  `    ``// If character appeared ` `    ``// at least twice ` `    ``if` `(ans == Integer.MAX_VALUE) ` `        ``return` `-``1``; ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``String str = ``"abcdeba"``; ` ` `  `    ``System.out.println(count_min_length(str)); ` `} ` `} ` ` `  `// This code is contributed by offbeat `

## Python3

 `# Python3 program to find the minimum  ` `# length of substring whose rotation  ` `# generates a palindromic substring  ` `import` `sys ` ` `  `INT_MAX ``=` `sys.maxsize; ` ` `  `# Function to return the  ` `# minimum lenth of substring  ` `def` `count_min_length(s):  ` ` `  `    ``# Store the index of  ` `    ``# previous occurrence  ` `    ``# of the character  ` `    ``hash` `=` `[``0``] ``*` `26``;  ` ` `  `    ``# Variable to store  ` `    ``# the maximum length  ` `    ``# of substring  ` `    ``ans ``=` `sys.maxsize;  ` ` `  `    ``for` `i ``in` `range``(``26``): ` `        ``hash``[i] ``=` `-``1``;  ` ` `  `    ``for` `i ``in` `range``(``len``(s)): ` `         `  `        ``# If the current character  ` `        ``# hasn't appeared yet  ` `        ``if` `(``hash``[``ord``(s[i]) ``-` `ord``(``'a'``)] ``=``=` `-``1``): ` `            ``hash``[``ord``(s[i]) ``-` `ord``(``'a'``)] ``=` `i;  ` `        ``else` `: ` `             `  `            ``# If the character has occured  ` `            ``# within one or two previous  ` `            ``# index, a palindromic substring  ` `            ``# already exists  ` `            ``if` `(``hash``[``ord``(s[i]) ``-` `ord``(``'a'``)] ``=``=` `i ``-` `1` `or`  `                ``hash``[``ord``(s[i]) ``-` `ord``(``'a'``)] ``=``=` `i ``-` `2``) : ` `                ``return` `0``;  ` ` `  `            ``# Update the maximum  ` `            ``ans ``=` `min``(ans, i ``-` `hash``[``ord``(s[i]) ``-`  `                                    ``ord``(``'a'``)] ``-` `1``);  ` ` `  `            ``# Replace the previous  ` `            ``# index of the character by  ` `            ``# the current index  ` `            ``hash``[``ord``(s[i]) ``-` `ord``(``'a'``)] ``=` `i;  ` ` `  `    ``# If character appeared  ` `    ``# at least twice  ` `    ``if` `(ans ``=``=` `INT_MAX): ` `        ``return` `-``1``;  ` ` `  `    ``return` `ans;  ` ` `  `# Driver Code  ` `if` `__name__ ``=``=` `"__main__"``:  ` ` `  `    ``string ``=` `"abcdeba"``;  ` `     `  `    ``print``(count_min_length(string));  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# Program to find the minimum ` `// length of substring whose rotation ` `// generates a palindromic substring ` `using` `System; ` `class` `GFG{ ` ` `  `// Function to return the ` `// minimum lenth of substring ` `static` `int` `count_min_length(``string` `s) ` `{ ` ` `  `    ``// Store the index of ` `    ``// previous occurrence ` `    ``// of the character ` `    ``int``[] hash = ``new` `int``; ` ` `  `    ``// Variable to store ` `    ``// the maximum length ` `    ``// of substring ` `    ``int` `ans = ``int``.MaxValue; ` ` `  `    ``for` `(``int` `i = 0; i < 26; i++) ` `        ``hash[i] = -1; ` ` `  `    ``for` `(``int` `i = 0; i < s.Length; i++)  ` `    ``{ ` `        ``// If the current character ` `        ``// hasn't appeared yet ` `        ``if` `(hash[s[i] - ``'a'``] == -1) ` `            ``hash[s[i] - ``'a'``] = i; ` `        ``else` `        ``{ ` `            ``// If the character has occured ` `            ``// within one or two previous ` `            ``// index, a palindromic substring ` `            ``// already exists ` `            ``if` `(hash[s[i] - ``'a'``] == i - 1 ||  ` `                ``hash[s[i] - ``'a'``] == i - 2) ` `                ``return` `0; ` ` `  `            ``// Update the maximum ` `            ``ans = Math.Min(ans,  ` `                      ``i - hash[s[i] - ``'a'``] - 1); ` ` `  `            ``// Replace the previous ` `            ``// index of the character by ` `            ``// the current index ` `            ``hash[s[i] - ``'a'``] = i; ` `        ``} ` `    ``} ` ` `  `    ``// If character appeared ` `    ``// at least twice ` `    ``if` `(ans == ``int``.MaxValue) ` `        ``return` `-1; ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(``string``[] args) ` `{ ` `    ``string` `str = ``"abcdeba"``; ` ` `  `    ``Console.WriteLine(count_min_length(str)); ` `} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```3
```

Time Complexity: O(N) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : offbeat, AnkitRai01, nidhi_biet