回文子串查询

给定一个字符串和对给定输入字符串的子字符串的几个查询，以检查子字符串是否为回文。

例子 ：

Suppose our input string is “abaaabaaaba” and the queries- [0, 10], [5, 8], [2, 5], [5, 9]

We have to tell that the substring having the starting and ending indices as above is a palindrome or not.

[0, 10] → Substring is “abaaabaaaba” which is a palindrome.
[5, 8] → Substring is “baaa” which is not a palindrome.
[2, 5] → Substring is “aaab” which is not a palindrome.
[5, 9] → Substring is “baaab” which is a palindrome.

让我们假设有 Q 个这样的查询需要回答，N 是我们输入字符串的长度。有以下两种方法可以回答这些查询

方法 1（幼稚）
我们逐一遍历查询的所有子串并检查所考虑的子串是否是回文。

由于有 Q 个查询，并且每个查询可能需要 O(N) 个最坏情况下的时间来回答，因此该方法在最坏情况下需要 O(QN) 个时间。尽管这是一种就地/空间高效的算法，但仍有一种更有效的方法可以做到这一点。

方法二（累积哈希）
这个想法类似于 Rabin Karp字符串匹配。我们使用字符串哈希。我们要做的就是，我们计算出原始字符串以及两个arrays-前缀[]和后缀[]反转字符串字符串的累积哈希值。

如何计算累积哈希值？

假设我们的字符串是 str[]，那么用于填充我们的 prefix[] 数组的累积散列函数是-

prefix[0] = 0
prefix[i] = str[0] + str[1] * 101 + str[2] * 101² + …… + str[i-1] * 101^i-1

For example, take the string- “abaaabxyaba”

prefix[0] = 0
prefix[1] = 97 (ASCII Value of ‘a’ is 97)
prefix[2] = 97 + 98 * 101
prefix[3] = 97 + 98 * 101 + 97 * 101²
………………………
………………………
prefix[11] = 97 + 98 * 101 + 97 * 101² + ……..+ 97 * 101¹⁰

现在以这种方式存储的原因是我们可以使用 – 在 O(1) 时间内轻松找到任何子字符串的哈希值 –

hash(L, R) = prefix[R+1] – prefix[L]

例如，hash (1, 5) = hash (“baaab”) = prefix[6] – prefix[1] = 98 * 101 + 97 * 101 ² + 97 * 101 ³ + 97 * 101 ⁴ + 98 * 101 ⁵ = 1040184646587 [我们稍后将使用这个奇怪的值来解释发生了什么]。

与此类似，我们将 suffix[] 数组填充为-

suffix[0] = 0
suffix[i] = str[n-1] + str[n-2] * 10¹ + str[n-3] * 101² + …… + str[n-i] * 101^i-1

For example, take the string- “abaaabxyaba”

suffix[0] = 0
suffix[1] = 97 (ASCII Value of ‘a’ is 97)
suffix[2] = 97 + 98 * 101
suffix[3] = 97 + 98 * 101 + 97 * 101²
………………………
………………………
suffix[11] = 97 + 98 * 101 + 97 * 101² + ……..+ 97 * 101¹⁰

Now the reason to store in that way is that we can easily find the reverse hash value of any substring in O(1) time using

reverse_hash(L, R) = hash (R, L) = suffix[n-L] – suffix[n-R-1] 

where n = length of string.

For “abaaabxyaba”, n = 11
reverse_hash(1, 5) = reverse_hash(“baaab”) = hash(“baaab”) [Reversing “baaab” gives “baaab”]

hash(“baaab”) = suffix[11-1] – suffix[11-5-1] = suffix[10] – suffix[5] = 98 * 101⁵ + 97 * 101⁶ + 97 * 101⁷ + 97 * 101⁸ + 98 * 101⁹ = 108242031437886501387

Now there doesn’t seem to be any relationship between these two weird integers – 1040184646587 and 108242031437886501387

Think again. Is there any relation between these two massive integers ?

Yes, there is and this observation is the core of this program/article.

1040184646587 * 101⁴ = 108242031437886501387

Try thinking about this and you will find that any substring starting at index- L and ending at index- R (both inclusive) will be a palindrome if

(prefix[R + 1] – prefix[L]) / (101^L) = (suffix [n – L] – suffix [n – R- 1] ) / (101^{n – R – 1})

The rest part is just implementation.

The function computerPowers() in the program computes the powers of 101 using dynamic programming.

Overflow Issues:
As, we can see that the hash values and the reverse hash values can become huge for even the small strings of length – 8. Since C and C++ doesn’t provide support for such large numbers, so it will cause overflows. To avoid this we will take modulo of a prime (a prime number is chosen for some specific mathematical reasons). We choose the biggest possible prime which fits in an integer value. The best such value is 1000000007. Hence all the operations are done modulo 1000000007.

However, Java and Python has no such issues and can be implemented without the modulo operator.

The fundamental modulo operations which are used extensively in the program are listed below.
1) Addition-

(a + b) %M = (a %M + b % M) % M
(a + b + c) % M = (a % M + b % M + c % M) % M
(a + b + c + d) % M = (a % M + b % M + c % M+ d% M) % M
…. ….. ….. ……
…. ….. ….. ……

2) Multiplication-

(a * b) % M = (a * b) % M
(a * b * c) % M = ((a * b) % M * c % M) % M
(a * b * c * d) % M = ((((a * b) % M * c) % M) * d) % M
…. ….. ….. ……
…. ….. ….. ……

This property is used by modPow() function which computes power of a number modulo M
3) Mixture of addition and multiplication-

(a * x + b * y + c) % M = ( (a * x) % M +(b * y) % M+ c % M ) % M

4) Subtraction-

(a – b) % M = (a % M – b % M + M) % M [Correct]
(a – b) % M = (a % M – b % M) % M [Wrong]

5) Division-

(a / b) % M = (a * MMI(b)) % M

Where MMI() is a function to calculate Modulo Multiplicative Inverse. In our program this is implemented by the function- findMMI().

C++

/* A C++ program to answer queries to check whether 
the substrings are palindrome or not efficiently */
#include 
using namespace std;
  
#define p 101
#define MOD 1000000007
  
// Structure to represent a query. A query consists
// of (L, R) and we have to answer whether the substring
// from index-L to R is a palindrome or not
struct Query {
    int L, R;
};
  
// A function to check if a string str is palindrome
// in the ranfe L to R
bool isPalindrome(string str, int L, int R)
{
    // Keep comparing characters while they are same
    while (R > L)
        if (str[L++] != str[R--])
            return (false);
    return (true);
}
  
// A Function to find pow (base, exponent) % MOD
// in log (exponent) time
unsigned long long int modPow(
    unsigned long long int base,
    unsigned long long int exponent)
{
    if (exponent == 0)
        return 1;
    if (exponent == 1)
        return base;
  
    unsigned long long int temp = modPow(base, exponent / 2);
  
    if (exponent % 2 == 0)
        return (temp % MOD * temp % MOD) % MOD;
    else
        return (((temp % MOD * temp % MOD) % MOD)
                * base % MOD)
               % MOD;
}
  
// A Function to calculate Modulo Multiplicative Inverse of 'n'
unsigned long long int findMMI(unsigned long long int n)
{
    return modPow(n, MOD - 2);
}
  
// A Function to calculate the prefix hash
void computePrefixHash(
    string str, int n,
    unsigned long long int prefix[],
    unsigned long long int power[])
{
    prefix[0] = 0;
    prefix[1] = str[0];
  
    for (int i = 2; i <= n; i++)
        prefix[i] = (prefix[i - 1] % MOD
                     + (str[i - 1] % MOD
                        * power[i - 1] % MOD)
                           % MOD)
                    % MOD;
  
    return;
}
  
// A Function to calculate the suffix hash
// Suffix hash is nothing but the prefix hash of
// the reversed string
void computeSuffixHash(
    string str, int n,
    unsigned long long int suffix[],
    unsigned long long int power[])
{
    suffix[0] = 0;
    suffix[1] = str[n - 1];
  
    for (int i = n - 2, j = 2; i >= 0 && j <= n; i--, j++)
        suffix[j] = (suffix[j - 1] % MOD
                     + (str[i] % MOD
                        * power[j - 1] % MOD)
                           % MOD)
                    % MOD;
    return;
}
  
// A Function to answer the Queries
void queryResults(string str, Query q[], int m, int n,
                  unsigned long long int prefix[],
                  unsigned long long int suffix[],
                  unsigned long long int power[])
{
    for (int i = 0; i <= m - 1; i++) {
        int L = q[i].L;
        int R = q[i].R;
  
        // Hash Value of Substring [L, R]
        unsigned long long hash_LR
            = ((prefix[R + 1] - prefix[L] + MOD) % MOD
               * findMMI(power[L]) % MOD)
              % MOD;
  
        // Reverse Hash Value of Substring [L, R]
        unsigned long long reverse_hash_LR
            = ((suffix[n - L] - suffix[n - R - 1] + MOD) % MOD
               * findMMI(power[n - R - 1]) % MOD)
              % MOD;
  
        // If both are equal then
        // the substring is a palindrome
        if (hash_LR == reverse_hash_LR) {
            if (isPalindrome(str, L, R) == true)
                printf("The Substring [%d %d] is a "
                       "palindrome\n",
                       L, R);
            else
                printf("The Substring [%d %d] is not a "
                       "palindrome\n",
                       L, R);
        }
  
        else
            printf("The Substring [%d %d] is not a "
                   "palindrome\n",
                   L, R);
    }
  
    return;
}
  
// A Dynamic Programming Based Approach to compute the
// powers of 101
void computePowers(unsigned long long int power[], int n)
{
    // 101^0 = 1
    power[0] = 1;
  
    for (int i = 1; i <= n; i++)
        power[i] = (power[i - 1] % MOD * p % MOD) % MOD;
  
    return;
}
  
/* Driver program to test above function */
int main()
{
    string str = "abaaabaaaba";
    int n = str.length();
  
    // A Table to store the powers of 101
    unsigned long long int power[n + 1];
  
    computePowers(power, n);
  
    // Arrays to hold prefix and suffix hash values
    unsigned long long int prefix[n + 1], suffix[n + 1];
  
    // Compute Prefix Hash and Suffix Hash Arrays
    computePrefixHash(str, n, prefix, power);
    computeSuffixHash(str, n, suffix, power);
  
    Query q[] = { { 0, 10 }, { 5, 8 }, { 2, 5 }, { 5, 9 } };
    int m = sizeof(q) / sizeof(q[0]);
  
    queryResults(str, q, m, n, prefix, suffix, power);
    return (0);
}

Java

/* A Java program to answer queries to check whether 
the substrings are palindrome or not efficiently */
  
public class GFG {
  
    static int p = 101;
    static int MOD = 1000000007;
  
    // Structure to represent a query. A query consists
    // of (L, R) and we have to answer whether the substring
    // from index-L to R is a palindrome or not
    static class Query {
  
        int L, R;
  
        public Query(int L, int R)
        {
            this.L = L;
            this.R = R;
        }
    };
  
    // A function to check if a string str is palindrome
    // in the ranfe L to R
    static boolean isPalindrome(String str, int L, int R)
    {
        // Keep comparing characters while they are same
        while (R > L) {
            if (str.charAt(L++) != str.charAt(R--)) {
                return (false);
            }
        }
        return (true);
    }
  
    // A Function to find pow (base, exponent) % MOD
    // in log (exponent) time
    static int modPow(int base, int exponent)
    {
        if (exponent == 0) {
            return 1;
        }
        if (exponent == 1) {
            return base;
        }
  
        int temp = modPow(base, exponent / 2);
  
        if (exponent % 2 == 0) {
            return (temp % MOD * temp % MOD) % MOD;
        }
        else {
            return (((temp % MOD * temp % MOD) % MOD)
                    * base % MOD)
                % MOD;
        }
    }
  
    // A Function to calculate
    // Modulo Multiplicative Inverse of 'n'
    static int findMMI(int n)
    {
        return modPow(n, MOD - 2);
    }
  
    // A Function to calculate the prefix hash
    static void computePrefixHash(String str, int n,
                                  int prefix[], int power[])
    {
        prefix[0] = 0;
        prefix[1] = str.charAt(0);
  
        for (int i = 2; i <= n; i++) {
            prefix[i] = (prefix[i - 1] % MOD
                         + (str.charAt(i - 1) % MOD
                            * power[i - 1] % MOD)
                               % MOD)
                        % MOD;
        }
  
        return;
    }
  
    // A Function to calculate the suffix hash
    // Suffix hash is nothing but the prefix hash of
    // the reversed string
    static void computeSuffixHash(String str, int n,
                                  int suffix[], int power[])
    {
        suffix[0] = 0;
        suffix[1] = str.charAt(n - 1);
  
        for (int i = n - 2, j = 2; i >= 0 && j <= n; i--, j++) {
            suffix[j] = (suffix[j - 1] % MOD
                         + (str.charAt(i) % MOD
                            * power[j - 1] % MOD)
                               % MOD)
                        % MOD;
        }
        return;
    }
  
    // A Function to answer the Queries
    static void queryResults(
        String str, Query q[], int m, int n,
        int prefix[], int suffix[], int power[])
    {
        for (int i = 0; i <= m - 1; i++) {
            int L = q[i].L;
            int R = q[i].R;
  
            // Hash Value of Substring [L, R]
            long hash_LR
                = ((prefix[R + 1] - prefix[L] + MOD) % MOD
                   * findMMI(power[L]) % MOD)
                  % MOD;
  
            // Reverse Hash Value of Substring [L, R]
            long reverse_hash_LR
                = ((suffix[n - L] - suffix[n - R - 1] + MOD) % MOD
                   * findMMI(power[n - R - 1]) % MOD)
                  % MOD;
  
            // If both are equal then the substring is a palindrome
            if (hash_LR == reverse_hash_LR) {
                if (isPalindrome(str, L, R) == true) {
                    System.out.printf("The Substring [%d %d] is a "
                                          + "palindrome\n",
                                      L, R);
                }
                else {
                    System.out.printf("The Substring [%d %d] is not a "
                                          + "palindrome\n",
                                      L, R);
                }
            }
            else {
                System.out.printf("The Substring [%d %d] is not a "
                                      + "palindrome\n",
                                  L, R);
            }
        }
  
        return;
    }
  
    // A Dynamic Programming Based Approach to compute the
    // powers of 101
    static void computePowers(int power[], int n)
    {
        // 101^0 = 1
        power[0] = 1;
  
        for (int i = 1; i <= n; i++) {
            power[i] = (power[i - 1] % MOD * p % MOD) % MOD;
        }
  
        return;
    }
  
    /* Driver code */
    public static void main(String[] args)
    {
        String str = "abaaabaaaba";
        int n = str.length();
  
        // A Table to store the powers of 101
        int[] power = new int[n + 1];
  
        computePowers(power, n);
  
        // Arrays to hold prefix and suffix hash values
        int[] prefix = new int[n + 1];
        int[] suffix = new int[n + 1];
  
        // Compute Prefix Hash and Suffix Hash Arrays
        computePrefixHash(str, n, prefix, power);
        computeSuffixHash(str, n, suffix, power);
  
        Query q[] = { new Query(0, 10), new Query(5, 8),
                      new Query(2, 5), new Query(5, 9) };
        int m = q.length;
  
        queryResults(str, q, m, n, prefix, suffix, power);
    }
}
  
// This code is contributed by Princi Singh

C#

/* A C# program to answer queries to check whether 
the substrings are palindrome or not efficiently */
using System;
  
class GFG {
  
    static int p = 101;
    static int MOD = 1000000007;
  
    // Structure to represent a query. A query consists
    // of (L, R) and we have to answer whether the substring
    // from index-L to R is a palindrome or not
    public class Query {
  
        public int L, R;
  
        public Query(int L, int R)
        {
            this.L = L;
            this.R = R;
        }
    };
  
    // A function to check if a string str is palindrome
    // in the ranfe L to R
    static Boolean isPalindrome(String str, int L, int R)
    {
        // Keep comparing characters while they are same
        while (R > L) {
            if (str[L++] != str[R--]) {
                return (false);
            }
        }
        return (true);
    }
  
    // A Function to find pow (base, exponent) % MOD
    // in log (exponent) time
    static int modPow(int Base, int exponent)
    {
        if (exponent == 0) {
            return 1;
        }
        if (exponent == 1) {
            return Base;
        }
  
        int temp = modPow(Base, exponent / 2);
  
        if (exponent % 2 == 0) {
            return (temp % MOD * temp % MOD) % MOD;
        }
        else {
            return (((temp % MOD * temp % MOD) % MOD) * Base % MOD) % MOD;
        }
    }
  
    // A Function to calculate Modulo Multiplicative Inverse of 'n'
    static int findMMI(int n)
    {
        return modPow(n, MOD - 2);
    }
  
    // A Function to calculate the prefix hash
    static void computePrefixHash(String str, int n,
                                  int[] prefix, int[] power)
    {
        prefix[0] = 0;
        prefix[1] = str[0];
  
        for (int i = 2; i <= n; i++) {
            prefix[i] = (prefix[i - 1] % MOD
                         + (str[i - 1] % MOD * power[i - 1] % MOD) % MOD)
                        % MOD;
        }
  
        return;
    }
  
    // A Function to calculate the suffix hash
    // Suffix hash is nothing but the prefix hash of
    // the reversed string
    static void computeSuffixHash(String str, int n,
                                  int[] suffix, int[] power)
    {
        suffix[0] = 0;
        suffix[1] = str[n - 1];
  
        for (int i = n - 2, j = 2; i >= 0 && j <= n; i--, j++) {
            suffix[j] = (suffix[j - 1] % MOD
                         + (str[i] % MOD * power[j - 1] % MOD) % MOD)
                        % MOD;
        }
        return;
    }
  
    // A Function to answer the Queries
    static void queryResults(String str, Query[] q, int m, int n,
                             int[] prefix, int[] suffix, int[] power)
    {
        for (int i = 0; i <= m - 1; i++) {
            int L = q[i].L;
            int R = q[i].R;
  
            // Hash Value of Substring [L, R]
            long hash_LR
                = ((prefix[R + 1] - prefix[L] + MOD) % MOD
                   * findMMI(power[L]) % MOD)
                  % MOD;
  
            // Reverse Hash Value of Substring [L, R]
            long reverse_hash_LR
                = ((suffix[n - L] - suffix[n - R - 1] + MOD) % MOD
                   * findMMI(power[n - R - 1]) % MOD)
                  % MOD;
  
            // If both are equal then the substring is a palindrome
            if (hash_LR == reverse_hash_LR) {
                if (isPalindrome(str, L, R) == true) {
                    Console.Write("The Substring [{0} {1}] is a "
                                      + "palindrome\n",
                                  L, R);
                }
                else {
                    Console.Write("The Substring [{0} {1}] is not a "
                                      + "palindrome\n",
                                  L, R);
                }
            }
            else {
                Console.Write("The Substring [{0} {1}] is not a "
                                  + "palindrome\n",
                              L, R);
            }
        }
  
        return;
    }
  
    // A Dynamic Programming Based Approach to compute the
    // powers of 101
    static void computePowers(int[] power, int n)
    {
        // 101^0 = 1
        power[0] = 1;
  
        for (int i = 1; i <= n; i++) {
            power[i] = (power[i - 1] % MOD * p % MOD) % MOD;
        }
  
        return;
    }
  
    /* Driver code */
    public static void Main(String[] args)
    {
        String str = "abaaabaaaba";
        int n = str.Length;
  
        // A Table to store the powers of 101
        int[] power = new int[n + 1];
  
        computePowers(power, n);
  
        // Arrays to hold prefix and suffix hash values
        int[] prefix = new int[n + 1];
        int[] suffix = new int[n + 1];
  
        // Compute Prefix Hash and Suffix Hash Arrays
        computePrefixHash(str, n, prefix, power);
        computeSuffixHash(str, n, suffix, power);
  
        Query[] q = { new Query(0, 10), new Query(5, 8),
                      new Query(2, 5), new Query(5, 9) };
        int m = q.Length;
  
        queryResults(str, q, m, n, prefix, suffix, power);
    }
}
  
// This code is contributed by Rajput-Ji

Output:

The Substring [0 10] is a palindrome
The Substring [5 8] is not a palindrome
The Substring [2 5] is not a palindrome
The Substring [5 9] is a palindrome

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