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 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 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 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.
?>
<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>
Time Complexity : O(n)
Auxiliary Space : O(n)
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