Online algorithm for checking palindrome in a stream

Last Updated : 11 Feb, 2025

Given a stream of characters (characters are received one by one), write a function that prints ‘Yes’ if a character makes the complete string palindrome, else prints ‘No’.

Examples:

Input: str[] = "abcba"
Output: a Yes   // "a" is palindrome
        b No    // "ab" is not palindrome
        c No    // "abc" is not palindrome
        b No    // "abcb" is not palindrome
        a Yes   // "abcba" is palindrome
Input: str[] = "aabaacaabaa"
Output:  a Yes   // "a" is palindrome
         a Yes   // "aa" is palindrome
         b No    // "aab" is not palindrome 
         a No    // "aaba" is not palindrome  
         a Yes   // "aabaa" is palindrome  
         c No    // "aabaac" is not palindrome  
         a No    // "aabaaca" is not palindrome  
         a No    // "aabaacaa" is not palindrome  
         b No    // "aabaacaab" is not palindrome  
         a No    // "aabaacaaba" is not palindrome  
         a Yes   // "aabaacaabaa" is palindrome

Let input string be str[0..n-1].

A Simple Solution is to do following for every character str[i] in input string. Check if substring str[0…i] is palindrome, then print yes, else print no.

A Better Solution is to use the idea of Rolling Hash used in Rabin Karp algorithm. The idea is to keep track of reverse of first half and second half (we also use first half and reverse of second half) for every index. Below is complete algorithm.

  1) The first character is always a palindrome, so print yes for 
     first character.
  2) Initialize reverse of first half as "a" and second half as "b".  
     Let the hash value of first half reverse be 'firstr' and that of 
     second half be 'second'.
  3) Iterate through string starting from second character, do following
      for every character str[i], i.e., i varies from 1 to n-1.
     a) If 'firstr' and 'second' are same, then character by character 
        check the substring ending with current character and print 
        "Yes" if palindrome.
        Note that if hash values match, then strings need not be same.
        For example, hash values of "ab" and "ba" are same, but strings
        are different. That is why we check complete string after hash.
     b) Update 'firstr' and 'second' for next iteration.  
           If 'i' is even, then add next character to the beginning of 
                           'firstr' and end of second half and update 
                            hash values.
           If 'i' is odd,  then keep 'firstr' as it is, remove leading 
                           character from second and append a next 
                           character at end.

Let us see all steps for example

string “abcba” Initial values of ‘firstr’ and ‘second’

firstr’ = hash(“a”), ‘second’ = hash(“b”) Start from second character, i.e., i = 1

Compare ‘firstr’ and ‘second’, they don’t match, so print no.
Calculate hash values for next iteration, i.e., i = 2
Since i is odd, ‘firstr’ is not changed and ‘second’ becomes hash(“c”) i = 2

Compare ‘firstr’ and ‘second’, they don’t match, so print no.
Calculate hash values for next iteration, i.e., i = 3
Since i is even, ‘firstr’ becomes hash(“ba”) and ‘second’ becomes hash(“cb”) i = 3

Compare ‘first’ and ‘second’, they don’t match, so print no.
Calculate hash values for next iteration, i.e., i = 4
Since i is odd, ‘firstr’ is not changed and ‘second’ becomes hash(“ba”) i = 4

‘firstr’ and ‘second’ match, compare the whole strings, they match, so print yes
We don’t need to calculate next hash values as this is last index The idea of using rolling hashes is, next hash value can be calculated from previous in O(1) time by just doing some constant number of arithmetic operations. Below are the implementations of above approach.

C++

// C++ program for online algorithm for palindrome checking #include<bits/stdc++.h> using namespace std;     // d is the number of characters in input alphabet #define d 256     // q is a prime number used for evaluating Rabin Karp's Rolling hash #define q 103     void checkPalindromes(string str) {     // Length of input string     int N = str.size();         // A single character is always a palindrome     cout << str[0] << " Yes" << endl;         // Return if string has only one character     if (N == 1) return;         // Initialize first half reverse and second half for      // as firstr and second characters     int firstr  = str[0] % q;     int second = str[1] % q;         int h = 1, i, j;         // Now check for palindromes from second character     // onward     for (i=1; i<N; i++)     {         // If the hash values of 'firstr' and 'second'          // match, then only check individual characters         if (firstr == second)         {             /* Check if str[0..i] is palindrome using                simple character by character match */             for (j = 0; j < i/2; j++)             {                 if (str[j] != str[i-j])                     break;             }             (j == i/2)?  cout << str[i] << " Yes" << endl: cout << str[i] << " No" << endl;         }         else cout << str[i] << " No" << endl;             // Calculate hash values for next iteration.         // Don't calculate hash for next characters if         // this is the last character of string         if (i != N-1)         {             // If i is even (next i is odd)              if (i%2 == 0)             {                 // Add next character after first half at beginning                  // of 'firstr'                 h = (h*d) % q;                 firstr  = (firstr + h*str[i/2])%q;                                     // Add next character after second half at the end                 // of second half.                 second = (second*d + str[i+1])%q;             }             else             {                 // If next i is odd (next i is even) then we                 // need not to change firstr, we need to remove                 // first character of second and append a                 // character to it.                 second = (d*(second + q - str[(i+1)/2]*h)%q                           + str[i+1])%q;             }         }     } }     /* Driver program to test above function */ int main() {     string txt = "aabaacaabaa";     checkPalindromes(txt);     return 0; }  // The code is contributed by Nidhi goel.

// C program for online algorithm for palindrome checking  #include<stdio.h>  #include<string.h>     // d is the number of characters in input alphabet  #define d 256     // q is a prime number used for evaluating Rabin Karp's Rolling hash  #define q 103     void checkPalindromes(char str[])  {      // Length of input string      int N = strlen(str);         // A single character is always a palindrome      printf("%c Yes\n", str[0]);         // Return if string has only one character      if (N == 1) return;         // Initialize first half reverse and second half for       // as firstr and second characters      int firstr  = str[0] % q;      int second = str[1] % q;         int h = 1, i, j;         // Now check for palindromes from second character      // onward      for (i=1; i<N; i++)      {          // If the hash values of 'firstr' and 'second'           // match, then only check individual characters          if (firstr == second)          {              /* Check if str[0..i] is palindrome using                 simple character by character match */             for (j = 0; j < i/2; j++)              {                  if (str[j] != str[i-j])                      break;              }              (j == i/2)?  printf("%c Yes\n", str[i]):              printf("%c No\n", str[i]);          }          else printf("%c No\n", str[i]);             // Calculate hash values for next iteration.          // Don't calculate hash for next characters if          // this is the last character of string          if (i != N-1)          {              // If i is even (next i is odd)               if (i%2 == 0)              {                  // Add next character after first half at beginning                   // of 'firstr'                  h = (h*d) % q;                  firstr  = (firstr + h*str[i/2])%q;                                     // Add next character after second half at the end                  // of second half.                  second = (second*d + str[i+1])%q;              }              else             {                  // If next i is odd (next i is even) then we                  // need not to change firstr, we need to remove                  // first character of second and append a                  // character to it.                  second = (d*(second + q - str[(i+1)/2]*h)%q                            + str[i+1])%q;              }          }      }  }     /* Driver program to test above function */ int main()  {      char *txt = "aabaacaabaa";      checkPalindromes(txt);      getchar();      return 0;  }

Java

// Java program for online algorithm for // palindrome checking public class GFG  {           // d is the number of characters in      // input alphabet     static final int d = 256;           // q is a prime number used for      // evaluating Rabin Karp's Rolling hash     static final int q = 103;           static void checkPalindromes(String str)     {         // Length of input string         int N = str.length();               // A single character is always a palindrome         System.out.println(str.charAt(0)+" Yes");               // Return if string has only one character         if (N == 1) return;               // Initialize first half reverse and second          // half for as firstr and second characters         int firstr  = str.charAt(0) % q;         int second = str.charAt(1) % q;               int h = 1, i, j;               // Now check for palindromes from second          // character onward         for (i = 1; i < N; i++)         {             // If the hash values of 'firstr' and             // 'second' match, then only check              // individual characters             if (firstr == second)             {                 /* Check if str[0..i] is palindrome                 using simple character by character                   match */                 for (j = 0; j < i/2; j++)                 {                     if (str.charAt(j) != str.charAt(i                                                 - j))                         break;                 }                 System.out.println((j == i/2) ?                    str.charAt(i) + " Yes": str.charAt(i)+                   " No");             }             else System.out.println(str.charAt(i)+ " No");                   // Calculate hash values for next iteration.             // Don't calculate hash for next characters             // if this is the last character of string             if (i != N - 1)             {                 // If i is even (next i is odd)                  if (i % 2 == 0)                 {                     // Add next character after first                      // half at beginning of 'firstr'                     h = (h * d) % q;                     firstr  = (firstr + h *str.charAt(i /                                                   2)) % q;                                           // Add next character after second                      // half at the end of second half.                     second = (second * d + str.charAt(i +                                                  1)) % q;                 }                 else                 {                     // If next i is odd (next i is even)                     // then we need not to change firstr,                     // we need to remove first character                     // of second and append a character                     // to it.                     second = (d * (second + q - str.charAt(                              (i + 1) / 2) * h) % q +                                str.charAt(i + 1)) % q;                 }             }         }     }           /* Driver program to test above function */     public static void main(String args[])     {         String txt = "aabaacaabaa";         checkPalindromes(txt);     } } // This code is contributed by Sumit Ghosh

Python

# Python program Online algorithm for checking palindrome # in a stream  # d is the number of characters in input alphabet d = 256  # q is a prime number used for evaluating Rabin Karp's # Rolling hash q = 103  def checkPalindromes(string):      # Length of input string     N = len(string)      # A single character is always a palindrome     print string[0] + " Yes"      # Return if string has only one character     if N == 1:         return      # Initialize first half reverse and second half for     # as firstr and second characters     firstr = ord(string[0]) % q     second = ord(string[1]) % q      h = 1     i = 0     j = 0      # Now check for palindromes from second character     # onward     for i in xrange(1,N):          # If the hash values of 'firstr' and 'second'         # match, then only check individual characters         if firstr == second:              # Check if str[0..i] is palindrome using             # simple character by character match             for j in xrange(0,i/2):                 if string[j] != string[i-j]:                     break             j += 1             if j == i/2:                 print string[i] + " Yes"             else:                 print string[i] + " No"         else:             print string[i] + " No"          # Calculate hash values for next iteration.         # Don't calculate hash for next characters if         # this is the last character of string         if i != N-1:              # If i is even (next i is odd)             if i % 2 == 0:                  # Add next character after first half at                 # beginning of 'firstr'                 h = (h*d) % q                 firstr = (firstr + h*ord(string[i/2]))%q                  # Add next character after second half at                 # the end of second half.                 second = (second*d + ord(string[i+1]))%q             else:                 # If next i is odd (next i is even) then we                 # need not to change firstr, we need to remove                 # first character of second and append a                 # character to it.                 second = (d*(second + q - ord(string[(i+1)/2])*h)%q                             + ord(string[i+1]))%q  # Driver program txt = "aabaacaabaa" checkPalindromes(txt) # This code is contributed by Bhavya Jain

// C# program for online algorithm for  // palindrome checking  using System;  class GFG { // d is the number of characters   // in input alphabet  public const int d = 256;  // q is a prime number used for  // evaluating Rabin Karp's Rolling hash  public const int q = 103;  public static void checkPalindromes(string str) {     // Length of input string      int N = str.Length;      // A single character is always     // a palindrome      Console.WriteLine(str[0] + " Yes");      // Return if string has only      // one character      if (N == 1)     {         return;     }      // Initialize first half reverse and second      // half for as firstr and second characters      int firstr = str[0] % q;     int second = str[1] % q;      int h = 1, i, j;      // Now check for palindromes from      // second character onward      for (i = 1; i < N; i++)     {         // If the hash values of 'firstr'          // and 'second' match, then only          // check individual characters          if (firstr == second)         {             /* Check if str[0..i] is palindrome              using simple character by character              match */             for (j = 0; j < i / 2; j++)             {                 if (str[j] != str[i - j])                 {                     break;                 }             }             Console.WriteLine((j == i / 2) ? str[i] +                               " Yes": str[i] + " No");         }         else         {             Console.WriteLine(str[i] + " No");         }          // Calculate hash values for next iteration.          // Don't calculate hash for next characters          // if this is the last character of string          if (i != N - 1)         {             // If i is even (next i is odd)              if (i % 2 == 0)             {                 // Add next character after first                  // half at beginning of 'firstr'                  h = (h * d) % q;                 firstr = (firstr + h * str[i / 2]) % q;                  // Add next character after second                  // half at the end of second half.                  second = (second * d + str[i + 1]) % q;             }             else             {                 // If next i is odd (next i is even)                  // then we need not to change firstr,                  // we need to remove first character                  // of second and append a character                  // to it.                  second = (d * (second + q - str[(i + 1) / 2] *                                    h) % q + str[i + 1]) % q;             }         }     } }  // Driver Code public static void Main(string[] args) {     string txt = "aabaacaabaa";     checkPalindromes(txt); } }  // This code is contributed by Shrikant13

JavaScript

// JavaScript program to check palindromic prefixes  // Function for checking palindromic prefixes function checkPalindromes(str) {     // d is the number of characters in input alphabet     var d = 256;          // q is a prime number used for evaluating Rabin-Karp's Rolling hash     var q = 103;          var n = str.length;          console.log(str.charAt(0) + " Yes");          if (n == 1) {         return;     }      // Initialize first half reverse and second half for as firstr and second characters     var firstr = str.charCodeAt(0) % q;     var second = str.charCodeAt(1) % q;      var h = 1, i, j;          // Now check for palindromes from second character onward     for (var i = 1; i < n; i++) {         // If the hash values of 'firstr' and 'second' match, then check individual characters         if (firstr === second) {             // Check if str[0..i] is palindrome using simple character-by-character match             for (j = 0; j < i / 2; j++) {                 if (str.charAt(j) !== str.charAt(i - j)) {                     break;                 }             }             console.log((j === Math.floor(i / 2)) ? str.charAt(i) + " Yes" : str.charAt(i) + " No");         } else {             console.log(str.charAt(i) + " No");         }          // Calculate hash values for the next iteration.         if (i !== n - 1) {             // If i is even (next i is odd)             if (i % 2 === 0) {                 // Add next character after first half at beginning of 'firstr'                 h = (h * d) % q;                 firstr = (firstr + h * str.charCodeAt(i / 2)) % q;                                  // Add next character after second half at the end of second half.                 second = (second * d + str.charCodeAt(i + 1)) % q;             } else {                 // If next i is odd (next i is even)                 // then we need not change firstr,                 // we need to remove first character of second and append a character to it.                 second = (d * (second + q - str.charCodeAt((i + 1) / 2) * h) % q +                           str.charCodeAt(i + 1)) % q;             }         }     } }  // Driver code var txt = "aabaacaabaa"; checkPalindromes(txt);

PHP

<?php  // PHP program to delete elements from array.  // Function for deleting k elements function checkPalindromes($str) {     // d is the number of characters in the input alphabet     $d = 256;      // q is a prime number used for evaluating Rabin Karp's Rolling hash     $q = 103;      $n = strlen($str);      echo $str[0] . " Yes\n";      if ($n == 1) {         return;     }      // Initialize first half reverse and second     // half for as firstr and second characters     $firstr = ord($str[0]) % $q;     $second = ord($str[1]) % $q;      $h = 1;     $i = 0;     $j = 0;      // Now check for palindromes from second     // character onward     for ($i = 1; $i < $n; $i++) {         // If the hash values of 'firstr' and 'second'         // match, then only check individual characters         if ($firstr == $second) {             // Check if str[0..i] is palindrome             // using simple character by character match             for ($j = 0; $j < $i / 2; $j++) {                 if ($str[$j] != $str[$i - $j]) {                     break;                 }             }              if ($j == $i / 2) {                 echo $str[$i] . " Yes\n";             } else {                 echo $str[$i] . " No\n";             }         } else {             echo $str[$i] . " No\n";         }          // Calculate hash values for the next iteration.         // Don't calculate the hash for the next characters         // if this is the last character of the string         if ($i != $n - 1) {             // If i is even (next i is odd)             if ($i % 2 == 0) {                 // Add the next character after the first                 // half at the beginning of 'firstr'                 $h = ($h * $d) % $q;                 $firstr = ($firstr + $h * ord($str[$i / 2])) % $q;                  // Add the next character after the second                 // half at the end of the second half.                 $second = ($second * $d + ord($str[$i + 1])) % $q;             } else {                 // If next i is odd (next i is even)                 // then we need not change firstr,                 // we need to remove the first character                 // of the second and append a character to it.                 $second = ($d * ($second + $q - ord($str[($i + 1) / 2]) * $h) % $q                             + ord($str[$i + 1])) % $q;             }         }     } }  // Driver code $txt = "aabaacaabaa"; checkPalindromes($txt);   ?>

Output

a Yes a Yes b No a No a Yes c No a No a No b No a No a Yes

The worst case time complexity of the above solution remains O(n*n), but in general, it works much better than simple approach as we avoid complete substring comparison most of the time by first comparing hash values.

The worst case occurs for input strings with all same characters like “aaaaaa”.

Space Complexity: O(1)
The algorithm uses a fixed number of variables (which is independent of the size of the input string). Hence, the space complexity of the algorithm is O(1).

Palindrome Substring Queries

kartik

Improve

Article Tags :

Practice Tags :

Online algorithm for checking palindrome in a stream

Similar Reads

Check Palindrome by Different Language

Easy Problems on Palindrome