Count all palindromic Substrings for each character in a given String

Last Updated : 05 Apr, 2023

Given a string S of length n, for each character S[i], the task is to find the number of palindromic substrings of length K such that no substring should contain S[i], the task is to return an array A of length n, where A[i] is the count of palindromic substrings of length K which does not include the character S[i].

Examples:

Input: S = "aababba", K = 2
Output : A[] = [1 1 2 2 1 1 2]
Explanation: A[0] = 1 => removing char s[0] = 'a' only one palindromic substring of length 2 is possible i.e. "bb"
A[1] = 1 => removing char s[1] = 'a' only one palindromic substring of length 2 is possible i.e. "bb"
A[2] = 2 => removing char s[2] = 'b' two palindromic substring of length 2 are possible i.e. "aa" and "bb"
A[3] = 2 => removing char s[3] = 'a' two palindromic substring of length 2 are possible i.e. "aa" and "bb"
A[4] = 1 => removing char s[4] = 'b' only one palindromic substring of length 2 is possible i.e. "aa"
A[5] = 1 => removing char s[5] = 'b' only one palindromic substring of length 2 is possible i.e. "aa"
A[6] = 2 => removing char s[6] = 'a' two palindromic substring of length 2 are possible i.e. "aa" and "bb"

Input: S = "abbab", K = 2
Output: A[] = [1 0 0 1 1]
Explanation: A[0] = 1 => removing char s[0] = 'a' only one palindromic substring of length 2 is possible i.e. "bb"
A[1] = 1 => removing char s[1] = 'b' no palindromic substring of length 2 is possible
A[2] = 2 => removing char s[2] = 'b' no palindromic substring of length 2 is possible
A[3] = 2 => removing char s[3] = 'a' only one palindromic substring of length 2 is possible i.e. "bb"
A[4] = 1 => removing char s[4] = 'b' only one palindromic substring of length 2 is possible i.e. "bb"

Approach: To solve the problem follow the below steps:

For character S[i] of the string, we need to find palindromic substrings of length K such that the substrings don't include S[i]. This approach requires one loop to iterate the string and two separate inner for loops because we don't have to include the character S[i] so we need to skip the entire range for which S[i] is included in any substring.
Run a loop to iterate the string from i = 0 to i < length of string.
The first inner loop will run from j = 0 to j ? i - K, then the last substring before S[i] will start from i - K and end at i - 1 which won't include S[i].
The second inner loop starts from j = i + 1 to n - 1.
For each substring, check if it is a palindrome, then increment the counter for that character S[i].

Below are the steps for the above approach:

Create an array A[] to store the count of the number of palindromic substrings for S[i].
Run a loop to iterate the string from i = 0 to i < S.length.
Initialize a counter-variable count = 0.
The first inner loop runs from j = 0 to j = i - K and generates a substring from j of length K to check if it is a palindrome, incrementing the counter.
The second inner loop runs from j = i+1 to j = l - 1 and generates a substring from j of length k to check if it is a palindrome, and increment the counter.
When both inner loop ends, update A[i] = count
Return the array A[].

Below is the implementation for the above approach:

C++

// C++ program for the above approach #include <bits/stdc++.h> using namespace std;  // function to check if a string is palindrome bool isPalin(string s, int l) {     for (int i = 0; i < l; i++) {         if (s[i] != s[l - i - 1])             return false;     }     return true; }  // function to count the number of palindromic // substrings for each character of a string void findCount(string s, int l, int k, int A[]) {     for (int i = 0; i < s.length(); i++) {         int count = 0;          // loop 1         for (int j = 0; j <= i - k; j++) {             string sub = s.substr(j, k);             if (isPalin(sub, k))                 count++;         }          // loop 2         for (int j = i + 1; j < l; j++) {             string sub = s.substr(j, k);             if (isPalin(sub, k))                 count++;         }          A[i] = count;     } }  // Driver function int main() {     string s = "aababba"; // given string     int l = 7; // length of string     int k = 2; // length of substring      // array to store the count     int A[s.length()];      // function call     findCount(s, l, k, A);      cout << "Count of palindromic substrings for "          << "each character of string s is: [ ";     for (int i = 0; i < s.length(); i++)         cout << A[i] << " ";     cout << "]" << '\n'; }

Java

// Java code implementation  import java.io.*;  class Main {      // function to check if a string is palindrome     static boolean isPalin(String s, int l)     {         for (int i = 0; i < l; i++) {             if (s.charAt(i) != s.charAt(l - i - 1))                 return false;         }         return true;     }      // function to count the number of palindromic     // substrings for each character of a string     static void findCount(String s, int l, int k, int[] A)     {         for (int i = 0; i < s.length(); i++) {             int count = 0;              // loop 1             for (int j = 0; j <= i - k; j++) {                 if (j + k <= s.length()) {                     String sub = s.substring(j, j + k);                     if (isPalin(sub, k))                         count++;                 }             }              // loop 2             for (int j = i + 1; j < l; j++) {                 if (j + k <= s.length()) {                     String sub = s.substring(j, j + k);                     if (isPalin(sub, k))                         count++;                 }             }              A[i] = count;         }     }      public static void main(String[] args)     {         String s = "aababba"; // given string         int l = 7; // length of string         int k = 2; // length of substring          // array to store the count         int[] A = new int[s.length()];          // function call         findCount(s, l, k, A);          System.out.print(             "Count of palindromic substrings for each character of string s is: [ ");         for (int i = 0; i < s.length(); i++)             System.out.print(A[i] + " ");         System.out.println("]");     } }  // This code is contributed by karthik.

Python3

# Python code implementation for the above program  # Function to check if a string is palindrome def is_palindrome(s, l):     for i in range(l):         if s[i] != s[l - i - 1]:             return False     return True  # Function to count the number of palindromic # substrings for each character of a string def find_count(s, l, k):     A = [0] * len(s)     for i in range(len(s)):         count = 0          # loop 1         for j in range(i - k + 1):             if j + k <= len(s):                 sub = s[j:j + k]                 if is_palindrome(sub, k):                     count += 1          # loop 2         for j in range(i + 1, l):             if j + k <= len(s):                 sub = s[j:j + k]                 if is_palindrome(sub, k):                     count += 1          A[i] = count      return A   s = "aababba"  # given string l = 7  # length of string k = 2  # length of substring  # Function call A = find_count(s, l, k)  print("Count of palindromic substrings for each character of string s is: ", A)  # This code is contributed by sankar.

// C# code implementation using System;  namespace PalindromicSubstrings {   class Program   {     // function to check if a string is palindrome     static bool IsPalin(string s, int l)     {       for (int i = 0; i < l; i++)       {         if (s[i] != s[l - i - 1])           return false;       }       return true;     }      // function to count the number of palindromic     // substrings for each character of a string     static void FindCount(string s, int l, int k, int[] A)     {       for (int i = 0; i < s.Length; i++)       {         int count = 0;          // loop 1         for (int j = 0; j <= i - k; j++)         {           if (j + k <= s.Length)           {             string sub = s.Substring(j, k);             if (IsPalin(sub, k))               count++;           }         }          // loop 2         for (int j = i + 1; j < l; j++)         {           if (j + k <= s.Length)           {             string sub = s.Substring(j, k);             if (IsPalin(sub, k))               count++;           }         }          A[i] = count;       }     }      static void Main(string[] args)     {       string s = "aababba"; // given string       int l = 7; // length of string       int k = 2; // length of substring        // array to store the count       int[] A = new int[s.Length];        // function call       FindCount(s, l, k, A);        Console.Write("Count of palindromic substrings for each character of string s is: [ ");       for (int i = 0; i < s.Length; i++)         Console.Write($"{A[i]} ");       Console.WriteLine("]");     }   } }

JavaScript

<script>     // JavaScript program for the above approach          // function to check if a string is palindrome     function isPalin(s, l) {       for (let i = 0; i < l; i++) {     if (s.charAt(i) !== s.charAt(l - i - 1)) {       return false;     }       }       return true;     }          // function to count the number of palindromic     // substrings for each character of a string     function findCount(s, l, k) {       const A = []; // array to store the count       for (let i = 0; i < s.length; i++) {     let count = 0;          // loop 1     for (let j = 0; j <= i - k; j++) {       if (j + k <= s.length) {         const sub = s.substring(j, j + k);         if (isPalin(sub, k)) {           count++;         }       }     }          // loop 2     for (let j = i + 1; j < l; j++) {       if (j + k <= s.length) {         const sub = s.substring(j, j + k);         if (isPalin(sub, k)) {           count++;         }       }     }          A[i] = count;       }       return A;     }          // main function     function main() {       const s = 'aababba'; // given string       const l = 7; // length of string       const k = 2; // length of substring            // function call       const A = findCount(s, l, k);            console.log(`Count of palindromic substrings for       each character of string s is: [ ${A.join(' ')} ]`);     }          // calling main function     main();          // This code is contributed by Susobhan Akhuli </script>

Output

Count of palindromic substrings for each character of string s is: [ 1 1 2 2 1 1 2 ]

Time Complexity: O(l²) where l is the length of the string
Auxiliary Space: O(l) where l is the length of the Resultant Array

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Count all palindromic Substrings for each character in a given String

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