其旋转产生回文子串的子串的最小长度

给定一个字符串str ，任务是找到从给定字符串生成回文子字符串所需的最小子字符串长度。

例子：

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”. 

方法：

• 如果字符串没有字符重复，则不能生成回文子串。
• 对于每个重复字符，检查其前一次出现的索引是否在当前索引的一个或两个索引之内。如果是这样，那么回文子串已经存在。
• 否则，计算(current index – 上一次出现的 index – 1)的长度。
• 返回所有这些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 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 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 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 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 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 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 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 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

Javascript

      // 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 occured
            // 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));

输出：

3

时间复杂度： O(N)