Count of Palindromic substrings in an Index range
Last Updated :
13 Jul, 2022
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"
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++ program to query number of palindromic
// substrings of a string in a range
#include <bits/stdc++.h>
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 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 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# 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
// PHP program to query number of palindromic
// substrings of a string in a range
$GLOBALS['M'] = 50;
// Utility method to construct the dp array
function constructDP($dp, $str)
{
$l = strlen($str);
// declare 2D array isPalin, isPalin[i][j]
// will be 1 if str(i..j) is palindrome
$isPalin = array(array());
// initialize dp and isPalin array by zeros
for ($i = 0; $i <= $l; $i++)
{
for ($j = 0; $j <= $l; $j++)
{
$isPalin[$i][$j] = $dp[$i][$j] = 0;
}
}
// loop for starting index of range
for ($i = $l - 1; $i >= 0; $i--)
{
// initialize value for one character
// strings as 1
$isPalin[$i][$i] = 1;
$dp[$i][$i] = 1;
// loop for ending index of range
for ($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
?>
<script>
// Javascript program to query number of palindromic
// substrings of a string in a range
// Function to construct the dp array
function constructDp(dp, str)
{
let l = str.length;
// declare 2D array isPalin, isPalin[i][j] will
// be 1 if str(i..j) is palindrome
let isPalin = new Array(l + 1);
// initialize dp and isPalin array by zeros
for (let i = 0; i <= l; i++) {
isPalin[i] = new Array(l + 1);
for (let j = 0; j <= l; j++) {
isPalin[i][j] = dp[i][j] = 0;
}
}
// loop for starting index of range
for (let 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 (let 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)
function countOfPalindromeInRange(dp, l, r)
{
return dp[l][r];
}
let MAX = 50;
let str = "xyaabax";
let dp = new Array(MAX);
for (let i = 0; i < MAX; i++) {
dp[i] = new Array(MAX);
for (let j = 0; j < MAX; j++) {
dp[i][j] = 0;
}
}
constructDp(dp, str);
let l = 3;
let r = 5;
document.write(countOfPalindromeInRange(dp, l, r));
</script>
Time complexity : O(l2)
Auxiliary Space : O(l2)
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