Minimum length of substring whose rotation generates a palindromic substring

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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 repeated 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”.

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 the 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 <bits/stdc++.h>
using namespace std;
 
// Function to return the
// minimum length of substring
int count_min_length(string s)
{
 
    // Store the index of
    // previous occurrence
    // of the character
    int hash[26];
 
    // 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 occurred
            // 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 length 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 occurred
            // 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 length 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 occurred 
            # 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 length 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 = 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 occurred
            // 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

Javascript

<script>
      // JavaScript Program to find the minimum
      // length of substring whose rotation
      // generates a palindromic substring
      // Function to return the
      // minimum length of substring
      function count_min_length(s) {
        // Store the index of
        // previous occurrence
        // of the character
        var hash = new Array(26).fill(0);
 
        // Variable to store
        // the maximum length
        // of substring
        var ans = 2147483648;
 
        for (var i = 0; i < 26; i++) 
            hash[i] = -1;
 
        for (var i = 0; i < s.length; i++) {
          // If the current character
          // hasn't appeared yet
          if (hash[s[i].charCodeAt(0) - "a".charCodeAt(0)] == -1)
            hash[s[i].charCodeAt(0) - "a".charCodeAt(0)] = i;
          else {
            // If the character has occurred
            // within one or two previous
            // index, a palindromic substring
            // already exists
            if (
              hash[s[i].charCodeAt(0) - "a".charCodeAt(0)] == i - 1 ||
              hash[s[i].charCodeAt(0) - "a".charCodeAt(0)] == i - 2
            )
              return 0;
 
            // Update the maximum
            ans = Math.min(
              ans,
              i - hash[s[i].charCodeAt(0) - "a".charCodeAt(0)] - 1
            );
 
            // Replace the previous
            // index of the character by
            // the current index
            hash[s[i].charCodeAt(0) - "a".charCodeAt(0)] = i;
          }
        }
 
        // If character appeared
        // at least twice
        if (ans === 2147483648) return -1;
 
        return ans;
      }
 
      // Driver code
      var str = "abcdeba";
 
      document.write(count_min_length(str));
</script>

Output:

Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time
Auxiliary Space: O(26), as we are using extra space for hash.

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