旋转产生回文子串的子串的最小长度

在给定字符串str的情况下，任务是找到旋转所需的最小子字符串长度，以从给定字符串生成回文子字符串。

例子：

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”.

方法：

• 如果字符串没有字符重复，则不会生成回文子字符串。
• 对于每个重复字符，请检查其先前出现的索引是否在当前索引的一两个索引之内。如果是这样，则回文子串已经存在。
• 否则，计算(当前索引–上一次出现的索引– 1)的长度。
• 返回所有此类长度中的最小值作为答案

下面是上述方法的实现：

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[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 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[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 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

输出：

3

时间复杂度： O(N)

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