Count special palindromes in a String

14 Mar, 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++
Java
Python3
C#
PHP
Javascript
// C++ program to count special Palindromic substring#include <bits/stdc++.h>using namespace std; // Function to count special Palindromic substringint 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 funint main(){    string str = "abccba";    cout << CountSpecialPalindrome(str) << endl;    return 0;}
// Java program to count special// Palindromic substringimport java.io.*;import java.util.*;import java.lang.*; class GFG{     // Function to count special// Palindromic substringpublic 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 codepublic static void main(String args[]){    String str = "abccba";    System.out.print(CountSpecialPalindrome(str));}} // This code is contributed// by Akanksha Rai(Abby_akku)
# Python3 program to count special# Palindromic substring # Function to count special# Palindromic substringdef 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 Codestr = "abccba";print(CountSpecialPalindrome(str)); # This code is contributed by mits.
// C# program to count special// Palindromic substringusing System; class GFG{     // Function to count special// Palindromic substringpublic 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 codepublic static void Main(){    String str = "abccba";    Console.Write(CountSpecialPalindrome(str));}} // This code is contributed by mits.
<?php// PHP program to count special// Palindromic substring // Function to count special// Palindromic substringfunction 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.?>
<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 :  
1
Time Complexity : O(n) Auxiliary Space : O(n) 
Mithun Kumar
Akanksha_Rai
Wagner da Silva Fontes
rdtank
varshagumber28
sumitgumber28
palindrome
Strings
Strings
palindrome
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