# Count of Palindromic substrings in an Index range

Given a string str of small alphabetic characters other than this we will be given many substrings of this string in form of index tuples. We need to find out the count of the palindromic substrings in given substring range.
Examples:

```Input : String str = "xyaabax"
Range1 = (3, 5)
Range2 = (2, 3)
Output : 4
3
For Range1,  substring is "aba"
Count of palindromic substring in "aba" is
four : "a", "b", "aba", "a"
For Range2,  substring is "aa"
Count of palindromic substring in "aa" is
3 : "a", "a", "aa"
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Prerequisite : Count All Palindrome Sub-Strings in a String

We can solve this problem using dynamic programming. First we will make a 2D array isPalin, isPalin[i][j] will be 1 if string(i..j) is a palindrome otherwise it will be 0. After constructing isPalin we will construct another 2D array dp, dp[i][j] will tell the count of palindromic substring in string(i..j)
Now we can write the relation among isPalin and dp values as shown below,

```
// isPalin[i][j] will be 1 if ith and jth characters
// are equal and mid substring str(i+1..j-1) is also
// a palindrome
isPalin[i][j] = (str[i] == str[j]) and
(isPalin[i + 1][j – 1])

// Similar to set theory we can write the relation among
// dp values as,
// dp[i][j] will be addition of number of palindromes from
// i to j-1 and i+1 to j  subtracting palindromes from i+1
// to j-1 because they are counted twice once in dp[i][j-1]
// and then in dp[i + 1][j] plus 1 if str(i..j) is also a
// palindrome
dp[i][j] = dp[i][j-1] + dp[i+1][j] - dp[i+1][j-1] +
isPalin[i][j];
```

Total time complexity of solution will be O(length ^ 2) for constructing dp array then O(1) per query.

## C/C++

 `// C++ program to query number of palindromic ` `// substrings of a string in a range ` `#include ` `using` `namespace` `std; ` `#define M 50 ` ` `  `// Utility method to construct the dp array ` `void` `constructDP(``int` `dp[M][M], string str) ` `{ ` `    ``int` `l = str.length(); ` ` `  `    ``// declare 2D array isPalin, isPalin[i][j] will ` `    ``// be 1 if str(i..j) is palindrome ` `    ``int` `isPalin[l + 1][l + 1]; ` ` `  `    ``// initialize dp and isPalin array by zeros ` `    ``for` `(``int` `i = 0; i <= l; i++) { ` `        ``for` `(``int` `j = 0; j <= l; j++) { ` `            ``isPalin[i][j] = dp[i][j] = 0; ` `        ``} ` `    ``} ` ` `  `    ``// loop for starting index of range ` `    ``for` `(``int` `i = l - 1; i >= 0; i--) { ` ` `  `        ``// initialize value for one character strings as 1 ` `        ``isPalin[i][i] = 1; ` `        ``dp[i][i] = 1; ` ` `  `        ``// loop for ending index of range ` `        ``for` `(``int` `j = i + 1; j < l; j++) { ` ` `  `            ``/* isPalin[i][j] will be 1 if ith and ` `               ``jth characters are equal and mid ` `               ``substring str(i+1..j-1) is also a ` `               ``palindrome             */` `            ``isPalin[i][j] = (str[i] == str[j] && (i + 1 > j - 1 || isPalin[i + 1][j - 1])); ` ` `  `            ``/* dp[i][j] will be addition of number ` `               ``of palindromes from i to j-1 and i+1 ` `               ``to j subtracting palindromes from i+1 ` `               ``to j-1 (as counted twice) plus 1 if ` `               ``str(i..j) is also a palindrome */` `            ``dp[i][j] = dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1] + isPalin[i][j]; ` `        ``} ` `    ``} ` `} ` ` `  `// method returns count of palindromic substring in range (l, r) ` `int` `countOfPalindromeInRange(``int` `dp[M][M], ``int` `l, ``int` `r) ` `{ ` `    ``return` `dp[l][r]; ` `} ` ` `  `// Driver code to test above methods ` `int` `main() ` `{ ` `    ``string str = ``"xyaabax"``; ` ` `  `    ``int` `dp[M][M]; ` `    ``constructDP(dp, str); ` ` `  `    ``int` `l = 3; ` `    ``int` `r = 5; ` ` `  `    ``cout << countOfPalindromeInRange(dp, l, r); ` `    ``return` `0; ` `} `

## Java

 `// Java program  to query number of palindromic ` `// substrings of a string in a range ` `import` `java.io.*; ` ` `  `class` `GFG { ` `    ``// Function to construct the dp array ` `    ``static` `void` `constructDp(``int` `dp[][], String str) ` `    ``{ ` `        ``int` `l = str.length(); ` ` `  `        ``// declare 2D array isPalin, isPalin[i][j] will ` `        ``// be 1 if str(i..j) is palindrome ` `        ``int``[][] isPalin = ``new` `int``[l + ``1``][l + ``1``]; ` ` `  `        ``// initialize dp and isPalin array by zeros ` `        ``for` `(``int` `i = ``0``; i <= l; i++) { ` `            ``for` `(``int` `j = ``0``; j <= l; j++) { ` `                ``isPalin[i][j] = dp[i][j] = ``0``; ` `            ``} ` `        ``} ` ` `  `        ``// loop for starting index of range ` `        ``for` `(``int` `i = l - ``1``; i >= ``0``; i--) { ` `            ``// initialize value for one character strings as 1 ` `            ``isPalin[i][i] = ``1``; ` `            ``dp[i][i] = ``1``; ` ` `  `            ``// loop for ending index of range ` `            ``for` `(``int` `j = i + ``1``; j < l; j++) { ` `                ``/* isPalin[i][j] will be 1 if ith and ` `                ``jth characters are equal and mid ` `                ``substring str(i+1..j-1) is also a ` `                ``palindrome             */` `                ``isPalin[i][j] = (str.charAt(i) == str.charAt(j) && (i + ``1` `> j - ``1` `|| (isPalin[i + ``1``][j - ``1``]) != ``0``)) ? ``1` `: ``0``; ` ` `  `                ``/* dp[i][j] will be addition of number ` `                ``of palindromes from i to j-1 and i+1 ` `                ``to j subtracting palindromes from i+1 ` `                ``to j-1 (as counted twice) plus 1 if ` `                ``str(i..j) is also a palindrome */` `                ``dp[i][j] = dp[i][j - ``1``] + dp[i + ``1``][j] - dp[i + ``1``][j - ``1``] + isPalin[i][j]; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// method returns count of palindromic substring in range (l, r) ` `    ``static` `int` `countOfPalindromeInRange(``int` `dp[][], ``int` `l, ``int` `r) ` `    ``{ ` `        ``return` `dp[l][r]; ` `    ``} ` ` `  `    ``// driver program ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `MAX = ``50``; ` `        ``String str = ``"xyaabax"``; ` `        ``int``[][] dp = ``new` `int``[MAX][MAX]; ` `        ``constructDp(dp, str); ` ` `  `        ``int` `l = ``3``; ` `        ``int` `r = ``5``; ` `        ``System.out.println(countOfPalindromeInRange(dp, l, r)); ` `    ``} ` `} ` ` `  `// Contributed by Pramod Kumar `

## Python3

 `# Python3 program to query the number of  ` `# palindromic substrings of a string in a range  ` `M ``=` `50` ` `  `# Utility method to construct the dp array  ` `def` `constructDP(dp, string):  ` ` `  `    ``l ``=` `len``(string)  ` ` `  `    ``# declare 2D array isPalin, isPalin[i][j]  ` `    ``# will be 1 if str(i..j) is palindrome ` `    ``# and initialize it with zero ` `    ``isPalin ``=` `[[``0` `for` `i ``in` `range``(l ``+` `1``)] ` `                  ``for` `j ``in` `range``(l ``+` `1``)]  ` ` `  `    ``# loop for starting index of range  ` `    ``for` `i ``in` `range``(l ``-` `1``, ``-``1``, ``-``1``):  ` ` `  `        ``# initialize value for one  ` `        ``# character strings as 1  ` `        ``isPalin[i][i], dp[i][i] ``=` `1``, ``1` ` `  `        ``# loop for ending index of range  ` `        ``for` `j ``in` `range``(i ``+` `1``, l):  ` ` `  `            ``# isPalin[i][j] will be 1 if ith and jth  ` `            ``# characters are equal and mid substring  ` `            ``# str(i+1..j-1) is also a palindrome ` `            ``isPalin[i][j] ``=` `(string[i] ``=``=` `string[j] ``and` `                            ``(i ``+` `1` `> j ``-` `1` `or` `isPalin[i ``+` `1``][j ``-` `1``]))  ` ` `  `            ``# dp[i][j] will be addition of number  ` `            ``# of palindromes from i to j-1 and i+1  ` `            ``# to j subtracting palindromes from i+1  ` `            ``# to j-1 (as counted twice) plus 1 if  ` `            ``# str(i..j) is also a palindrome  ` `            ``dp[i][j] ``=` `(dp[i][j ``-` `1``] ``+` `dp[i ``+` `1``][j] ``-`  `                        ``dp[i ``+` `1``][j ``-` `1``] ``+` `isPalin[i][j])  ` ` `  `# Method returns count of palindromic  ` `# substring in range (l, r)  ` `def` `countOfPalindromeInRange(dp, l, r): ` `    ``return` `dp[l][r]  ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``:  ` ` `  `    ``string ``=` `"xyaabax"` `    ``dp ``=` `[[``0` `for` `i ``in` `range``(M)]  ` `             ``for` `j ``in` `range``(M)] ` `     `  `    ``constructDP(dp, string)  ` ` `  `    ``l, r ``=` `3``, ``5` `    ``print``(countOfPalindromeInRange(dp, l, r)) ` ` `  `# This code is contributed by Rituraj Jain `

## C#

 `// C# program to query number of palindromic ` `// substrings of a string in a range ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// Function to construct the dp array ` `    ``static` `void` `constructDp(``int``[, ] dp, ``string` `str) ` `    ``{ ` `        ``int` `l = str.Length; ` ` `  `        ``// declare 2D array isPalin, isPalin[i][j] ` `        ``// will be 1 if str(i..j) is palindrome ` `        ``int``[, ] isPalin = ``new` `int``[l + 1, l + 1]; ` ` `  `        ``// initialize dp and isPalin array by zeros ` `        ``for` `(``int` `i = 0; i <= l; i++) { ` `            ``for` `(``int` `j = 0; j <= l; j++) { ` `                ``isPalin[i, j] = dp[i, j] = 0; ` `            ``} ` `        ``} ` ` `  `        ``// loop for starting index of range ` `        ``for` `(``int` `i = l - 1; i >= 0; i--) { ` `             `  `            ``// initialize value for one  ` `            ``// character strings as 1 ` `            ``isPalin[i, i] = 1; ` `            ``dp[i, i] = 1; ` ` `  `            ``// loop for ending index of range ` `            ``for` `(``int` `j = i + 1; j < l; j++) { ` `                 `  `                ``/* isPalin[i][j] will be 1 if ith and ` `                ``jth characters are equal and mid ` `                ``substring str(i+1..j-1) is also a ` `                ``palindrome*/` `                ``isPalin[i, j] = (str[i] == str[j] && (i + 1 > j - 1 ||  ` `                                ``(isPalin[i + 1, j - 1]) != 0)) ? 1 : 0; ` ` `  `                ``/* dp[i][j] will be addition of number ` `                ``of palindromes from i to j-1 and i+1 ` `                ``to j subtracting palindromes from i+1 ` `                ``to j-1 (as counted twice) plus 1 if ` `                ``str(i..j) is also a palindrome */` `                ``dp[i, j] = dp[i, j - 1] + dp[i + 1, j] -  ` `                           ``dp[i + 1, j - 1] + isPalin[i, j]; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// method returns count of palindromic ` `    ``// substring in range (l, r) ` `    ``static` `int` `countOfPalindromeInRange(``int``[, ] dp, ` `                                      ``int` `l, ``int` `r) ` `    ``{ ` `        ``return` `dp[l, r]; ` `    ``} ` ` `  `    ``// driver program ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `MAX = 50; ` `        ``string` `str = ``"xyaabax"``; ` `        ``int``[, ] dp = ``new` `int``[MAX, MAX]; ` `        ``constructDp(dp, str); ` ` `  `        ``int` `l = 3; ` `        ``int` `r = 5; ` `        ``Console.WriteLine(countOfPalindromeInRange(dp, l, r)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 `= 0; ``\$i``--)  ` `    ``{  ` ` `  `        ``// initialize value for one character ` `        ``// strings as 1  ` `        ``\$isPalin``[``\$i``][``\$i``] = 1;  ` `        ``\$dp``[``\$i``][``\$i``] = 1;  ` ` `  `        ``// loop for ending index of range  ` `        ``for` `(``\$j` `= ``\$i` `+ 1; ``\$j` `< ``\$l``; ``\$j``++)  ` `        ``{  ` ` `  `            ``/* isPalin[i][j] will be 1 if ith and  ` `            ``jth characters are equal and mid  ` `            ``substring str(i+1..j-1) is also a  ` `            ``palindrome         */` `            ``\$isPalin``[``\$i``][``\$j``] = (``\$str``[``\$i``] == ``\$str``[``\$j``] && ` `                               ``(``\$i` `+ 1 > ``\$j` `- 1 ||  ` `                                ``\$isPalin``[``\$i` `+ 1][``\$j` `- 1]));  ` ` `  `            ``/* dp[i][j] will be addition of number  ` `            ``of palindromes from i to j-1 and i+1  ` `            ``to j subtracting palindromes from i+1  ` `            ``to j-1 (as counted twice) plus 1 if  ` `            ``str(i..j) is also a palindrome */` `            ``\$dp``[``\$i``][``\$j``] = ``\$dp``[``\$i``][``\$j` `- 1] + ``\$dp``[``\$i` `+ 1][``\$j``] -  ` `                          ``\$dp``[``\$i` `+ 1][``\$j` `- 1] + ``\$isPalin``[``\$i``][``\$j``];  ` `        ``}  ` `    ``}  ` `    ``return` `\$dp` `; ` `}  ` ` `  `// method returns count of palindromic ` `// substring in range (l, r)  ` `function` `countOfPalindromeInRange(``\$dp``, ``\$l``, ``\$r``)  ` `{  ` `    ``return` `\$dp``[``\$l``][``\$r``];  ` `}  ` ` `  `// Driver code ` `\$str` `= ``"xyaabax"``;  ` ` `  `\$dp` `= ``array``(``array``()); ` ` `  `for``(``\$i` `= 0; ``\$i` `< ``\$GLOBALS``[``'M'``]; ``\$i``++ ) ` `    ``for``(``\$j` `= 0; ``\$j` `< ``\$GLOBALS``[``'M'``]; ``\$j``++) ` `        ``\$dp``[``\$i``][``\$j``] = 0; ` `         `  `\$dp` `= constructDP(``\$dp``, ``\$str``);  ` ` `  `\$l` `= 3;  ` `\$r` `= 5;  ` ` `  `echo` `countOfPalindromeInRange(``\$dp``, ``\$l``, ``\$r``); ` ` `  `// This code is contributed by Ryuga ` `?> `

Output:

```4
```

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