Count special palindromes in a String

Last Updated : 04 Aug, 2022

Given a String s, count all special palindromic substrings of size greater than 1. A Substring is called special palindromic substring if all the characters in the substring are same or only the middle character is different for odd length. Example "aabaa" and "aaa" are special palindromic substrings and "abcba" is not special palindromic substring.

Examples :

Input :  str = " abab" Output : 2 All Special Palindromic  substring are: "aba", "bab"  Input : str = "aabbb" Output : 4 All Special substring are: "aa", "bb", "bbb", "bb"

Simple Solution is that we simply generate all substrings one-by-one and count how many substring are Special Palindromic substring. This solution takes O(n³) time.

Efficient Solution

There are 2 Cases :

Case 1: All Palindromic substrings have same character :
We can handle this case by simply counting the same continuous character and using formula K*(K+1)/2 (total number of substring possible : Here K is count of Continuous same char).

 Lets Str = "aaabba"  Traverse string from left to right and Count of same char   "aaabba"  =  3, 2, 1    for "aaa" : total substring possible are    'aa' 'aa', 'aaa', 'a', 'a', 'a'  : 3(3+1)/2 = 6     "bb" : 'b', 'b', 'bb' : 2(2+1)/2 = 3    'a'  : 'a' : 1(1+1)/2 = 1

Case 2: We can handle this case by storing count of same character in another temporary array called "sameChar[n]" of size n. and pick each character one-by-one and check its previous and forward character are equal or not if equal then there are min_between( sameChar[previous], sameChar[forward] ) substring possible.

Let's Str = "aabaaab"  Count of smiler char from left to right :  that we will store in Temporary array "sameChar"   Str         =    " a a b a a a b " sameChar[]  =      2 2 1 3 3 3 1 According to the problem statement middle character is different: so we have only left with char "b" at index :2 ( index from 0 to n-1) substring : "aabaaa" so only two substring are possible : "aabaa", "aba" that is min (smilerChar[index-1], smilerChar[index+1] ) that is 2.

Below is the implementation of above idea

C++

// C++ program to count special Palindromic substring #include <bits/stdc++.h> using namespace std;  // Function to count special Palindromic substring int CountSpecialPalindrome(string str) {     int n = str.length();      // store count of special Palindromic substring     int result = 0;      // it will store the count of continues same char     int sameChar[n] = { 0 };      int i = 0;      // traverse string character from left to right     while (i < n) {          // store same character count         int sameCharCount = 1;          int j = i + 1;          // count similar character         while (str[i] == str[j] && j < n)             sameCharCount++, j++;          // Case : 1         // so total number of substring that we can         // generate are : K *( K + 1 ) / 2         // here K is sameCharCount         result += (sameCharCount * (sameCharCount + 1) / 2);          // store current same char count in sameChar[]         // array         sameChar[i] = sameCharCount;          // increment i         i = j;     }      // Case 2: Count all odd length Special Palindromic     // substring     for (int j = 1; j < n; j++)     {         // if current character is equal to previous         // one then we assign Previous same character         // count to current one         if (str[j] == str[j - 1])             sameChar[j] = sameChar[j - 1];          // case 2: odd length         if (j > 0 && j < (n - 1) &&             (str[j - 1] == str[j + 1] &&              str[j] != str[j - 1]))             result += min(sameChar[j - 1],                           sameChar[j + 1]);     }      // subtract all single length substring     return result - n; }  // driver program to test above fun int main() {     string str = "abccba";     cout << CountSpecialPalindrome(str) << endl;     return 0; }

Java

// Java program to count special // Palindromic substring import java.io.*; import java.util.*; import java.lang.*;  class GFG {      // Function to count special // Palindromic substring public static int CountSpecialPalindrome(String str) {     int n = str.length();      // store count of special     // Palindromic substring     int result = 0;      // it will store the count      // of continues same char     int[] sameChar = new int[n];     for(int v = 0; v < n; v++)     sameChar[v] = 0;      int i = 0;      // traverse string character     // from left to right     while (i < n)      {          // store same character count         int sameCharCount = 1;          int j = i + 1;          // count similar character         while (j < n &&              str.charAt(i) == str.charAt(j))         {             sameCharCount++;             j++;         }          // Case : 1         // so total number of          // substring that we can         // generate are : K *( K + 1 ) / 2         // here K is sameCharCount         result += (sameCharCount *                    (sameCharCount + 1) / 2);          // store current same char          // count in sameChar[] array         sameChar[i] = sameCharCount;          // increment i         i = j;     }      // Case 2: Count all odd length     //           Special Palindromic     //           substring     for (int j = 1; j < n; j++)     {         // if current character is          // equal to previous one          // then we assign Previous          // same character count to         // current one         if (str.charAt(j) == str.charAt(j - 1))             sameChar[j] = sameChar[j - 1];          // case 2: odd length         if (j > 0 && j < (n - 1) &&         (str.charAt(j - 1) == str.charAt(j + 1) &&              str.charAt(j) != str.charAt(j - 1)))             result += Math.min(sameChar[j - 1],                                sameChar[j + 1]);     }      // subtract all single      // length substring     return result - n; }  // Driver code public static void main(String args[]) {     String str = "abccba";     System.out.print(CountSpecialPalindrome(str)); } }  // This code is contributed // by Akanksha Rai(Abby_akku)

Python3

# Python3 program to count special # Palindromic substring  # Function to count special  # Palindromic substring def CountSpecialPalindrome(str):     n = len(str);      # store count of special     # Palindromic substring     result = 0;      # it will store the count      # of continues same char     sameChar=[0] * n;      i = 0;      # traverse string character      # from left to right     while (i < n):          # store same character count         sameCharCount = 1;          j = i + 1;          # count smiler character         while (j < n):             if(str[i] != str[j]):                 break;             sameCharCount += 1;             j += 1;                  # Case : 1         # so total number of substring          # that we can generate are :         # K *( K + 1 ) / 2         # here K is sameCharCount         result += int(sameCharCount *                       (sameCharCount + 1) / 2);          # store current same char          # count in sameChar[] array         sameChar[i] = sameCharCount;          # increment i         i = j;      # Case 2: Count all odd length      # Special Palindromic substring     for j in range(1, n):                  # if current character is equal          # to previous one then we assign          # Previous same character count          # to current one         if (str[j] == str[j - 1]):             sameChar[j] = sameChar[j - 1];          # case 2: odd length         if (j > 0 and j < (n - 1) and             (str[j - 1] == str[j + 1] and              str[j] != str[j - 1])):             result += (sameChar[j - 1]                     if(sameChar[j - 1] < sameChar[j + 1])                      else sameChar[j + 1]);      # subtract all single     # length substring     return result-n;  # Driver Code str = "abccba"; print(CountSpecialPalindrome(str));  # This code is contributed by mits.

// C# program to count special // Palindromic substring using System;  class GFG {      // Function to count special // Palindromic substring public static int CountSpecialPalindrome(String str) {     int n = str.Length;      // store count of special     // Palindromic substring     int result = 0;      // it will store the count      // of continues same char     int[] sameChar = new int[n];     for(int v = 0; v < n; v++)     sameChar[v] = 0;      int i = 0;      // traverse string character     // from left to right     while (i < n)      {          // store same character count         int sameCharCount = 1;          int j = i + 1;          // count smiler character         while (j < n &&                 str[i] == str[j])         {             sameCharCount++;             j++;         }          // Case : 1         // so total number of          // substring that we can         // generate are : K *( K + 1 ) / 2         // here K is sameCharCount         result += (sameCharCount *                    (sameCharCount + 1) / 2);          // store current same char          // count in sameChar[] array         sameChar[i] = sameCharCount;          // increment i         i = j;     }      // Case 2: Count all odd length     //         Special Palindromic     //         substring     for (int j = 1; j < n; j++)     {         // if current character is          // equal to previous one          // then we assign Previous          // same character count to         // current one         if (str[j] == str[j - 1])             sameChar[j] = sameChar[j - 1];          // case 2: odd length         if (j > 0 && j < (n - 1) &&            (str[j - 1] == str[j + 1] &&             str[j] != str[j - 1]))             result += Math.Min(sameChar[j - 1],                               sameChar[j + 1]);     }      // subtract all single      // length substring     return result - n; }  // Driver code public static void Main() {     String str = "abccba";     Console.Write(CountSpecialPalindrome(str)); } }  // This code is contributed by mits.

PHP

<?php // PHP program to count special // Palindromic substring  // Function to count special  // Palindromic substring function CountSpecialPalindrome($str) {     $n = strlen($str);      // store count of special     // Palindromic substring     $result = 0;      // it will store the count      // of continues same char     $sameChar=array_fill(0, $n, 0);      $i = 0;      // traverse string character      // from left to right     while ($i < $n)      {          // store same character count         $sameCharCount = 1;          $j = $i + 1;          // count smiler character         while ($j < $n)         {             if($str[$i] != $str[$j])             break;             $sameCharCount++;             $j++;         }                  // Case : 1         // so total number of substring          // that we can generate are :         // K *( K + 1 ) / 2         // here K is sameCharCount         $result += (int)($sameCharCount *                          ($sameCharCount + 1) / 2);          // store current same char          // count in sameChar[] array         $sameChar[$i] = $sameCharCount;          // increment i         $i = $j;     }      // Case 2: Count all odd length      // Special Palindromic substring     for ($j = 1; $j < $n; $j++)     {         // if current character is equal          // to previous one then we assign          // Previous same character count          // to current one         if ($str[$j] == $str[$j - 1])             $sameChar[$j] = $sameChar[$j - 1];          // case 2: odd length         if ($j > 0 && $j < ($n - 1) &&             ($str[$j - 1] == $str[$j + 1] &&                 $str[$j] != $str[$j - 1]))             $result += $sameChar[$j - 1] <                         $sameChar[$j + 1] ?                         $sameChar[$j - 1] :                         $sameChar[$j + 1];     }      // subtract all single     // length substring     return $result - $n; }  // Driver Code $str = "abccba"; echo CountSpecialPalindrome($str);  // This code is contributed by mits. ?>

JavaScript

<script>       // JavaScript program to count special Palindromic substring        // Function to count special Palindromic substring       function CountSpecialPalindrome(str) {         var n = str.length;          // store count of special Palindromic substring         var result = 0;          // it will store the count of continues same char         var sameChar = [...Array(n)];          var i = 0;          // traverse string character from left to right         while (i < n) {           // store same character count           var sameCharCount = 1;            var j = i + 1;            // count smiler character           while (str[i] == str[j] && j < n) sameCharCount++, j++;            // Case : 1           // so total number of substring that we can           // generate are : K *( K + 1 ) / 2           // here K is sameCharCount           result += (sameCharCount * (sameCharCount + 1)) / 2;            // store current same char count in sameChar[]           // array           sameChar[i] = sameCharCount;            // increment i           i = j;         }          // Case 2: Count all odd length Special Palindromic         // substring         for (var j = 1; j < n; j++) {           // if current character is equal to previous           // one then we assign Previous same character           // count to current one           if (str[j] == str[j - 1]) sameChar[j] = sameChar[j - 1];            // case 2: odd length           if (             j > 0 &&             j < n - 1 &&             str[j - 1] == str[j + 1] &&             str[j] != str[j - 1]           )             result += Math.min(sameChar[j - 1], sameChar[j + 1]);         }          // subtract all single length substring         return result - n;       }        // driver program to test above fun       var str = "abccba";       document.write(CountSpecialPalindrome(str) + "<br>");     </script>

Output

Time Complexity : O(n)
Auxiliary Space : O(n)

Count special palindromes in a String

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Count special palindromes in a String

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