Given a string, the task is to count all palindrome sub string in a given string. Length of palindrome sub string 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 similar problem below.
Find all distinct palindromic sub-strings of a given string
The above problem can be recursively defined.
Initial Values : i = 0, j = n-1;
Given string 'str'
CountPS(i, j)
// If length of string is 2 then we
// check both character are same or not
If (j == i+1)
return str[i] == str[j]
// this condition shows that in recursion if i crosses
// j then it will be a invalid substring or
// if i==j that means only one character is remaining
// and we require substring of length 2
//in both the conditions we need to return 0
Else if(i == j || i > j) return 0;
Else If str[i..j] is PALINDROME
// increment count by 1 and check for
// rest palindromic substring (i, j-1), (i+1, j)
// remove common palindrome substring (i+1, j-1)
return countPS(i+1, j) + countPS(i, j-1) + 1 -
countPS(i+1, j-1);
Else // if NOT PALINDROME
// We check for rest palindromic substrings (i, j-1)
// and (i+1, j)
// remove common palindrome substring (i+1 , j-1)
return countPS(i+1, j) + countPS(i, j-1) -
countPS(i+1 , j-1);
If we draw recursion tree of above recursive solution, we can observe overlapping Subproblems. Since the problem has overlapping sub-problems, we can solve it efficiently using Dynamic Programming.
Below is a Dynamic Programming based solution.
C++
// C++ program to find palindromic substrings of a string#include <bits/stdc++.h>using namespace std;// Returns total number of palindrome substring of// length greater than equal to 2int CountPS(char str[], int n){ //to store the count of palindrome subarrays int ans=0; // P[i][j] = true if substring str[i..j] is palindrome, // else false bool P[n][n]; memset(P, false, sizeof(P)); // palindrome of single length for (int i = 0; i < n; i++){ P[i][i] = true; } // Palindromes of length more than or equal to 2. We start with // a gap of length 1 and fill the DP table in a way that // gap between starting and ending indexes increases one // by one by outer loop. for (int gap = 2; gap <=n; gap++) { // Pick starting point for current gap for (int i = 0; i <= n-gap; i++) { // Set ending point int j = gap + i-1; // check If current string is palindrome of length 2 if(i==j-1){ P[i][j]=(str[i]==str[j]); }else { //check if the current substring is palindrome of length more than 2 //if characters at index i,j are equal check for string in between them //from i+1 to j-1 if it is palindrome or not P[i][j]=(str[i]==str[j] && P[i+1][j-1]); } //if substring from index i to j is palindrome increase the count if(P[i][j]){ ans++; } } } // return total palindromic substrings return ans;}// Driver codeint main(){ char str[] = "abaab"; int n = strlen(str); cout << CountPS(str, n) << endl; return 0;} |
Java
// Java program to find palindromic substrings of a stringpublic class GFG { // Returns total number of palindrome substring of // length greater than equal to 2 static int CountPS(char str[], int n) { // create empty 2-D matrix that counts all // palindrome substring. dp[i][j] stores counts of // palindromic substrings in st[i..j] int dp[][] = new int[n][n]; // P[i][j] = true if substring str[i..j] is // palindrome, else false boolean P[][] = new boolean[n][n]; // palindrome of single length for (int i = 0; i < n; i++) P[i][i] = true; // palindrome of length 2 for (int i = 0; i < n - 1; i++) { if (str[i] == str[i + 1]) { P[i][i + 1] = true; dp[i][i + 1] = 1; } } // Palindromes of length more than 2. This loop is // similar to Matrix Chain Multiplication. We start // with a gap of length 2 and fill the DP table in a // way that gap between starting and ending indexes // increases one by one by outer loop. for (int gap = 2; gap < n; gap++) { // Pick starting point for current gap for (int i = 0; i < n - gap; i++) { // Set ending point int j = gap + i; // If current string is palindrome if (str[i] == str[j] && P[i + 1][j - 1]) P[i][j] = true; // Add current palindrome substring ( + 1) // and rest palindrome substring (dp[i][j-1] // + dp[i+1][j]) remove common palindrome // substrings (- dp[i+1][j-1]) if (P[i][j] == true) dp[i][j] = dp[i][j - 1] + dp[i + 1][j] + 1 - dp[i + 1][j - 1]; else dp[i][j] = dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1]; } } // return total palindromic substrings return dp[0][n - 1]; } // Driver code public static void main(String[] args) { String str = "abaab"; System.out.println( CountPS(str.toCharArray(), str.length())); }} |
Python3
# Python3 program to find palindromic# substrings of a string# Returns total number of palindrome# substring of length greater than# equal to 2def CountPS(str, n): # create empty 2-D matrix that counts # all palindrome substring. dp[i][j] # stores counts of palindromic # substrings in st[i..j] dp = [[0 for x in range(n)] for y in range(n)] # P[i][j] = true if substring str[i..j] # is palindrome, else false P = [[False for x in range(n)] for y in range(n)] # palindrome of single length for i in range(n): P[i][i] = True # palindrome of length 2 for i in range(n - 1): if (str[i] == str[i + 1]): P[i][i + 1] = True dp[i][i + 1] = 1 # Palindromes of length more than 2. This # loop is similar to Matrix Chain Multiplication. # We start with a gap of length 2 and fill DP # table in a way that the gap between starting and # ending indexes increase one by one by # outer loop. for gap in range(2, n): # Pick a starting point for the current gap for i in range(n - gap): # Set ending point j = gap + i # If current string is palindrome if (str[i] == str[j] and P[i + 1][j - 1]): P[i][j] = True # Add current palindrome substring ( + 1) # and rest palindrome substring (dp[i][j-1] + # dp[i+1][j]) remove common palindrome # substrings (- dp[i+1][j-1]) if (P[i][j] == True): dp[i][j] = (dp[i][j - 1] + dp[i + 1][j] + 1 - dp[i + 1][j - 1]) else: dp[i][j] = (dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1]) # return total palindromic substrings return dp[0][n - 1]# Driver Codeif __name__ == "__main__": str = "abaab" n = len(str) print(CountPS(str, n))# This code is contributed by ita_c |
C#
// C# program to find palindromic// substrings of a stringusing System;class GFG { // Returns total number of // palindrome substring of // length greater than equal to 2 public static int CountPS(char[] str, int n) { // create empty 2-D matrix that counts // all palindrome substring. dp[i][j] // stores counts of palindromic // substrings in st[i..j] int[][] dp = RectangularArrays.ReturnRectangularIntArray( n, n); // P[i][j] = true if substring str[i..j] // is palindrome, else false bool[][] P = RectangularArrays.ReturnRectangularBoolArray( n, n); // palindrome of single length for (int i = 0; i < n; i++) { P[i][i] = true; } // palindrome of length 2 for (int i = 0; i < n - 1; i++) { if (str[i] == str[i + 1]) { P[i][i + 1] = true; dp[i][i + 1] = 1; } } // Palindromes of length more than 2. // This loop is similar to Matrix Chain // Multiplication. We start with a gap // of length 2 and fill DP table in a // way that gap between starting and // ending indexes increases one by one // by outer loop. for (int gap = 2; gap < n; gap++) { // Pick starting point for current gap for (int i = 0; i < n - gap; i++) { // Set ending point int j = gap + i; // If current string is palindrome if (str[i] == str[j] && P[i + 1][j - 1]) { P[i][j] = true; } // Add current palindrome substring // ( + 1) and rest palindrome substring // (dp[i][j-1] + dp[i+1][j]) remove common // palindrome substrings (- dp[i+1][j-1]) if (P[i][j] == true) { dp[i][j] = dp[i][j - 1] + dp[i + 1][j] + 1 - dp[i + 1][j - 1]; } else { dp[i][j] = dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1]; } } } // return total palindromic substrings return dp[0][n - 1]; } public static class RectangularArrays { public static int[][] ReturnRectangularIntArray( int size1, int size2) { int[][] newArray = new int[size1][]; for (int array1 = 0; array1 < size1; array1++) { newArray[array1] = new int[size2]; } return newArray; } public static bool[][] ReturnRectangularBoolArray( int size1, int size2) { bool[][] newArray = new bool[size1][]; for (int array1 = 0; array1 < size1; array1++) { newArray[array1] = new bool[size2]; } return newArray; } } // Driver Code public static void Main(string[] args) { string str = "abaab"; Console.WriteLine( CountPS(str.ToCharArray(), str.Length)); }}// This code is contributed by Shrikant13 |
PHP
<?php// PHP program to find palindromic substrings// of a string // Returns total number of palindrome // substring of length greater than equal to 2 function CountPS($str, $n) { // create empty 2-D matrix that counts // all palindrome substring. dp[i][j] // stores counts of palindromic // substrings in st[i..j] $dp = array(array()); for ($i = 0; $i < $n; $i++) for($j = 0; $j < $n; $j++) $dp[$i][$j] = 0; // P[i][j] = true if substring str[i..j] // is palindrome, else false $P = array(array()); for ($i = 0; $i < $n; $i++) for($j = 0; $j < $n; $j++) $P[$i][$j] = false; // palindrome of single length for ($i= 0; $i< $n; $i++) $P[$i][$i] = true; // palindrome of length 2 for ($i = 0; $i < $n - 1; $i++) { if ($str[$i] == $str[$i + 1]) { $P[$i][$i + 1] = true; $dp[$i][$i + 1] = 1; } } // Palindromes of length more than 2. This // loop is similar to Matrix Chain Multiplication. // We start with a gap of length 2 and fill DP // table in a way that gap between starting and // ending indexes increases one by one by // outer loop. for ($gap = 2; $gap < $n; $gap++) { // Pick starting point for current gap for ($i = 0; $i < $n - $gap; $i++) { // Set ending point $j = $gap + $i; // If current string is palindrome if ($str[$i] == $str[$j] && $P[$i + 1][$j - 1]) $P[$i][$j] = true; // Add current palindrome substring (+ 1) // and rest palindrome substring (dp[i][j-1] + // dp[i+1][j]) remove common palindrome // substrings (- dp[i+1][j-1]) if ($P[$i][$j] == true) $dp[$i][$j] = $dp[$i][$j - 1] + $dp[$i + 1][$j] + 1 - $dp[$i + 1][$j - 1]; else $dp[$i][$j] = $dp[$i][$j - 1] + $dp[$i + 1][$j] - $dp[$i + 1][$j - 1]; } } // return total palindromic substrings return $dp[0][$n - 1]; } // Driver Code $str = "abaab"; $n = strlen($str); echo CountPS($str, $n);// This code is contributed by Ryuga?> |
Javascript
<script>// Javascript program to find palindromic substrings of a string // Returns total number of palindrome substring of // length greater than equal to 2 function CountPS(str,n) { // create empty 2-D matrix that counts all // palindrome substring. dp[i][j] stores counts of // palindromic substrings in st[i..j] let dp=new Array(n); // P[i][j] = true if substring str[i..j] is // palindrome, else false let P=new Array(n); for(let i=0;i<n;i++) { dp[i]=new Array(n); P[i]=new Array(n); for(let j=0;j<n;j++) { dp[i][j]=0; P[i][j]=false; } } // palindrome of single length for (let i = 0; i < n; i++) P[i][i] = true; // palindrome of length 2 for (let i = 0; i < n - 1; i++) { if (str[i] == str[i + 1]) { P[i][i + 1] = true; dp[i][i + 1] = 1; } } // Palindromes of length more than 2. This loop is // similar to Matrix Chain Multiplication. We start // with a gap of length 2 and fill the DP table in a // way that gap between starting and ending indexes // increases one by one by outer loop. for (let gap = 2; gap < n; gap++) { // Pick starting point for current gap for (let i = 0; i < n - gap; i++) { // Set ending point let j = gap + i; // If current string is palindrome if (str[i] == str[j] && P[i + 1][j - 1]) P[i][j] = true; // Add current palindrome substring ( + 1) // and rest palindrome substring (dp[i][j-1] // + dp[i+1][j]) remove common palindrome // substrings (- dp[i+1][j-1]) if (P[i][j] == true) dp[i][j] = dp[i][j - 1] + dp[i + 1][j] + 1 - dp[i + 1][j - 1]; else dp[i][j] = dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1]; } } // return total palindromic substrings return dp[0][n - 1]; } // Driver code let str = "abaab"; document.write( CountPS(str.split(""), str.length)); // This code is contributed by avanitrachhadiya2155 </script> |
Output
3
Time Complexity: O(n2)
Auxiliary Space: O(n2)
Method 2: This approach uses Top Down DP i.e memoized version of recursion.
Recursive soln: 1. Here base condition comes out to be i>j if we hit this condition, return 1. 2. We check for each and every i and j, if the characters are equal, if that is not the case, return 0. 3. Call the is_palindrome function again with incremented i and decremented j. 4. Check this for all values of i and j by applying 2 for loops.
Implementation:
C++
#include <bits/stdc++.h>using namespace std;int dp[1001][1001]; // 2D matrixbool isPal(string s, int i, int j){ // Base condition if (i > j) return 1; // check if the recursive tree // for given i, j // has already been executed if (dp[i][j] != -1) return dp[i][j]; // If first and last characters of // substring are unequal if (s[i] != s[j]) return dp[i][j] = 0; // memoization return dp[i][j] = isPal(s, i + 1, j - 1); }int countSubstrings(string s){ memset(dp, -1, sizeof(dp)); int n = s.length(); int count = 0; // 2 for loops are required to check for // all the palindromes in the string. for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { // Increment count for every palindrome if (isPal(s, i, j)) count++; } } // return total palindromic substrings return count;}// Driver codeint main(){ string s = "abbaeae"; cout << countSubstrings(s); //"bb" , "abba" ,"aea", "eae" are // the 4 palindromic substrings. // This code is contributed by Bhavneet Singh return 0;} |
Java
import java.util.*;public class Main{ static int dp[][] = new int[1001][1001]; // 2D matrix public static int isPal(String s, int i, int j) { // Base condition if (i > j) return 1; // check if the recursive tree // for given i, j // has already been executed if (dp[i][j] != -1) return dp[i][j]; // If first and last characters of // substring are unequal if (s.charAt(i) != s.charAt(j)) return dp[i][j] = 0; // memoization return dp[i][j] = isPal(s, i + 1, j - 1); } public static int countSubstrings(String s) { for (int[] row: dp) { Arrays.fill(row, -1); } int n = s.length(); int count = 0; // 2 for loops are required to check for // all the palindromes in the string. for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { // Increment count for every palindrome if (isPal(s, i, j) != 0) count++; } } // return total palindromic substrings return count; } public static void main(String[] args) { String s = "abbaeae"; System.out.println(countSubstrings(s)); }}// This code is contributed by divyeshrabadiya07 |
Python3
# 2D matrixdp = [[-1 for i in range(1001)] for j in range(1001)]def isPal(s, i, j): # Base condition if (i > j): return 1 # Check if the recursive tree # for given i, j # has already been executed if (dp[i][j] != -1): return dp[i][j] # If first and last characters of # substring are unequal if (s[i] != s[j]): dp[i][j] = 0 return dp[i][j] # Memoization dp[i][j] = isPal(s, i + 1, j - 1) return dp[i][j]def countSubstrings(s): n = len(s) count = 0 # 2 for loops are required to check for # all the palindromes in the string. for i in range(n): for j in range(i + 1, n): # Increment count for every palindrome if (isPal(s, i, j)): count += 1 # Return total palindromic substrings return count# Driver codes = "abbaeae"print(countSubstrings(s))# This code is contributed by rag2127 |
C#
using System;class GFG{// 2D matrixstatic int[,] dp = new int[1001, 1001]; static int isPal(string s, int i, int j){ // Base condition if (i > j) return 1; // Check if the recursive tree // for given i, j // has already been executed if (dp[i, j] != -1) return dp[i, j]; // If first and last characters of // substring are unequal if (s[i] != s[j]) return dp[i, j] = 0; // Memoization return dp[i, j] = isPal(s, i + 1, j - 1); } static int countSubstrings(string s){ for(int i = 0; i < 1001; i++) { for(int j = 0; j < 1001; j++) { dp[i, j] = -1; } } int n = s.Length; int count = 0; // 2 for loops are required to check for // all the palindromes in the string. for(int i = 0; i < n; i++) { for(int j = i + 1; j < n; j++) { // Increment count for every palindrome if (isPal(s, i, j) != 0) count++; } } // Return total palindromic substrings return count;}// Driver Code static void Main(){ string s = "abbaeae"; Console.WriteLine(countSubstrings(s));}}// This code is contributed by divyesh072019 |
Javascript
<script> var dp = Array(1001).fill().map(()=>Array(1001).fill(-1)); // 2D matrix function isPal( s , i , j) { // Base condition if (i > j) return 1; // check if the recursive tree // for given i, j // has already been executed if (dp[i][j] != -1) return dp[i][j]; // If first and last characters of // substring are unequal if (s.charAt(i) != s.charAt(j)) return dp[i][j] = 0; // memoization return dp[i][j] = isPal(s, i + 1, j - 1); } function countSubstrings( s) { var n = s.length; var count = 0; // 2 for loops are required to check for // all the palindromes in the string. for (i = 0; i < n; i++) { for (j = i + 1; j < n; j++) { // Increment count for every palindrome if (isPal(s, i, j) != 0) count++; } } // return total palindromic substrings return count; } // Driver code var s = "abbaeae"; document.write(countSubstrings(s)); // This code is contributed by Rajput-Ji</script> |
Output
4
Time Complexity: O(n3)
Auxiliary Space: O(n2)
Count All Palindrome Sub-Strings in a String | Set 2
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