Given a string, the task is to count all palindrome substring in a given string. Length of palindrome substring is greater than or equal to 2.
Examples: Input : str = "abaab" Output: 3 Explanation : All palindrome substring are : "aba", "aa", "baab" Input : str = "abbaeae" Output: 4 Explanation : All palindrome substring are : "bb", "abba", "aea", "eae":
We have discussed a dynamic programming based solution in below post. Count All Palindrome Sub-Strings in a String | Set 1 The solution discussed here is extension of Longest Palindromic Substring problem.
The idea is for each character in the given input string, we consider it as midpoint of a palindrome and expand it in both directions to find all palindromes of even and odd lengths. We use hashmap to keep track of all the distinct palindromes of length greater than 1 and return map size which have count of all possible palindrome substrings.
Implementation:
// C++ program to count all distinct palindromic // substrings of a string. #include <bits/stdc++.h> using namespace std;
// Returns total number of palindrome substring of // length greater than equal to 2 int countPalindromes(string s)
{ unordered_map<string, int > m;
for ( int i = 0; i < s.length(); i++) {
// check for odd length palindromes
for ( int j = 0; j <= i; j++) {
if (!s[i + j])
break ;
if (s[i - j] == s[i + j]) {
// check for palindromes of length
// greater than 1
if ((i + j + 1) - (i - j) > 1)
m[s.substr(i - j,
(i + j + 1) - (i - j))]++;
} else
break ;
}
// check for even length palindromes
for ( int j = 0; j <= i; j++) {
if (!s[i + j + 1])
break ;
if (s[i - j] == s[i + j + 1]) {
// check for palindromes of length
// greater than 1
if ((i + j + 2) - (i - j) > 1)
m[s.substr(i - j,
(i + j + 2) - (i - j))]++;
} else
break ;
}
}
return m.size();
} // Driver code int main()
{ string s = "abbaeae" ;
cout << countPalindromes(s) << endl;
return 0;
} |
// Java Program to count palindrome substring // in a string public class PalindromeSubstring {
// Method which return count palindrome substring
static int countPS(String str){
String temp = "" ;
StringBuffer stf;
int count = 0 ;
// Iterate the loop twice
for ( int i = 0 ; i < str.length(); i++) {
for ( int j = i + 1 ; j <= str.length(); j++) {
// Get each substring
temp = str.substring(i, j);
// If length is greater than or equal to two
// Check for palindrome
if (temp.length() >= 2 ) {
// Use StringBuffer class to reverse the string
stf = new StringBuffer(temp);
stf.reverse();
// Compare substring with reverse of substring
if (stf.toString().compareTo(temp) == 0 )
count++;
}
}
}
// return the count
return count;
}
// Driver Code
public static void main(String args[]) throws Exception {
// Declare and initialize the string
String str = "abbaeae" ;
// Call the method
System.out.println(countPS(str));
}
} // This code is contributes by hungundji
|
# Python program to count all distinct palindromic # substrings of a string. # Returns total number of palindrome substring of # length greater than equal to 2 def countPalindromes(s):
m = {}
for i in range ( len (s)):
# check for odd length palindromes
for j in range (i + 1 ):
if (i + j > = len (s)):
break
if (s[i - j] = = s[i + j]):
# check for palindromes of length
# greater than 1
if ((i + j + 1 ) - (i - j) > 1 ):
if (s[i - j:(i + j + 1 )] in m):
m[s[i - j:(i + j + 1 )]] + = 1
else :
m[s[i - j:(i + j + 1 )]] = 1
else :
break
# check for even length palindromes
for j in range (i + 1 ):
if (i + j + 1 > = len (s)):
break
if (s[i - j] = = s[i + j + 1 ]):
# check for palindromes of length
# greater than 1
if ((i + j + 2 ) - (i - j) > 1 ):
if (s[i - j:i + j + 2 ] in m):
m[s[i - j:(i + j + 2 )]] + = 1
else :
m[s[i - j:(i + j + 2 )]] = 1
else :
break
return len (m)
# Driver code s = "abbaeae"
print (countPalindromes(s))
# This code is contributed by shinjanpatra |
// C# Program to count palindrome substring // in a string using System;
class GFG
{ // Method which return count palindrome substring
static int countPS(String str)
{
String temp = "" ;
String stf;
int count = 0;
// Iterate the loop twice
for ( int i = 0; i < str.Length; i++)
{
for ( int j = i + 1;
j <= str.Length; j++)
{
// Get each substring
temp = str.Substring(i, j-i);
// If length is greater than or equal to two
// Check for palindrome
if (temp.Length >= 2)
{
// Use StringBuffer class to reverse
// the string
stf = temp;
stf = reverse(temp);
// Compare substring with reverse of substring
if (stf.ToString().CompareTo(temp) == 0)
count++;
}
}
}
// return the count
return count;
}
static String reverse(String input)
{
char [] a = input.ToCharArray();
int l, r = 0;
r = a.Length - 1;
for (l = 0; l < r; l++, r--)
{
// Swap values of l and r
char temp = a[l];
a[l] = a[r];
a[r] = temp;
}
return String.Join( "" ,a);
}
// Driver Code
public static void Main(String []args)
{
// Declare and initialize the string
String str = "abbaeae" ;
// Call the method
Console.WriteLine(countPS(str));
}
} // This code is contributed by Rajput-Ji |
<script> // JavaScript program to count all distinct palindromic // substrings of a string. // Returns total number of palindrome substring of // length greater than equal to 2 function countPalindromes(s)
{ let m = new Map();
for (let i = 0; i < s.length; i++) {
// check for odd length palindromes
for (let j = 0; j <= i; j++) {
if (!s[i + j])
break ;
if (s[i - j] == s[i + j]) {
// check for palindromes of length
// greater than 1
if ((i + j + 1) - (i - j) > 1){
if (m.has(s.substr(i - j,(i + j + 1) - (i - j)))){
m.set(s.substr(i - j,(i + j + 1) - (i - j)),m.get(s.substr(i - j,(i + j + 1) - (i - j))) + 1)
}
else m.set(s.substr(i - j,(i + j + 1) - (i - j)),1)
}
} else
break ;
}
// check for even length palindromes
for (let j = 0; j <= i; j++) {
if (!s[i + j + 1])
break ;
if (s[i - j] == s[i + j + 1]) {
// check for palindromes of length
// greater than 1
if ((i + j + 2) - (i - j) > 1){
if (m.has(s.substr(i - j,(i + j + 2) - (i - j)))){
m.set(s.substr(i - j,(i + j + 2) - (i - j)),m.get(s.substr(i - j,(i + j + 2) - (i - j))) + 1)
}
else m.set(s.substr(i - j,(i + j + 2) - (i - j)),1)
}
} else
break ;
}
}
return m.size;
} // Driver code let s = "abbaeae"
document.write(countPalindromes(s), "</br>" )
// This code is contributed by shinjanpatra </script> |
4
Time complexity O(n2) Auxiliary Space O(n)
In the Approach discussed above we are using a unordered map which is taking O(n) space but we can do the same thing without using any extra space.Instead of making a hashmap we can directly count the palindromic substring.
Implementation:
#include <bits/stdc++.h> using namespace std;
int countPalindromes(string s)
{ int n = s.size();
int count = 0;
for ( int i = 0; i < s.size(); i++) {
int left = i - 1;
int right = i + 1;
while (left >= 0 and right < n) {
if (s[left] == s[right])
count++;
else
break ;
left--;
right++;
}
}
for ( int i = 0; i < s.size(); i++) {
int left = i;
int right = i + 1;
while (left >= 0 and right < n) {
if (s[left] == s[right])
count++;
else
break ;
left--;
right++;
}
}
return count;
} int main()
{ string s = "abbaeae" ;
cout << countPalindromes(s) << endl;
return 0;
} // This code is contributed by Arpit Jain |
class GFG {
public static int countPalindromes(String s){
int count = 0 ;
for ( int i= 0 ; i<s.length(); i++){
//Taking every node as a center and expanding it from there to check if it is a palindrome or not.
//This will make sure that the length of the string to be checked > 2.
count += expandFromCenter(s, i- 1 , i+ 1 ) + expandFromCenter(s, i , i+ 1 );
}
return count;
}
public static int expandFromCenter(String s, int left, int right){
int count = 0 ;
//As we expand from the center and find if the character matches, we increase the count.
while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)){
count++;
left--;
right++;
}
return count;
}
public static void main (String[] args) {
String s = "abbaeae" ;
System.out.println(countPalindromes(s));
}
} |
# Python program to count all distinct palindromic # substrings of a string. # Returns total number of palindrome substring of # length greater than equal to 2 def countPalindromes(s):
n = len (s)
count = 0
for i in range ( 0 , n):
left = i - 1
right = i + 1
while left > = 0 and right < n:
if s[left] = = s[right]:
count + = 1
else :
break
left - = 1
right + = 1
for i in range ( 0 , n):
left = i
right = i + 1
while left > = 0 and right < n:
if s[left] = = s[right]:
count + = 1
else :
break
left - = 1
right + = 1
return count
# Driver code
s = "abbaeae"
print (countPalindromes(s))
# This code is contributed by Arpit Jain |
using System;
public class GFG{
public static int countPalindromes( string s){
int count = 0;
for ( int i = 0; i < s.Length; i++)
{
// Taking every node as a center and expanding
// it from there to check if it is a palindrome or not.
// This will make sure that the length of the string to be checked > 2.
count += expandFromCenter(s, i - 1, i + 1) + expandFromCenter(s, i , i + 1);
}
return count;
}
public static int expandFromCenter( string s, int left, int right){
int count = 0;
// As we expand from the center and find
// if the character matches, we increase the count.
while (left >= 0 && right < s.Length && s[left] == s[right]){
count++;
left--;
right++;
}
return count;
}
public static void Main ( string [] args) {
string s = "abbaeae" ;
Console.WriteLine(countPalindromes(s));
}
} // This code is contributed by Shubham singh |
<script> function countPalindromes(s)
{ n = s.length;
count = 0;
for (i = 0; i < s.length; i++) {
left = i - 1;
right = i + 1;
while (left >= 0 && right < n) {
if (s[left] == s[right])
count++;
else
break ;
left--;
right++;
}
}
for (i = 0; i < s.length; i++) {
left = i;
right = i + 1;
while (left >= 0 && right < n) {
if (s[left] == s[right])
count++;
else
break ;
left--;
right++;
}
}
return count;
} s = "abbaeae" ;
document.write(countPalindromes(s)); // This code is contributed by Shubham Singh </script> |
4
Time complexity is O(n^2) in worst case because we are running one while loop inside a for loop because in while loop we are expanding to both directions so that’s why it’s O(n/2) in worst case when all characters are same eg : “aaaaaaaa” .
But we are not using any extra space , Space Complexity is O(1) .