Answer to Question #256251 in Python for rasi
6. Find out the number of palindrome strings in a given string - Link
def count_palindromes[string]:
count = 0
string = string.lower[]
for i in range[len[string] - 1]:
for j in range[i + 2, len[string] + 1]:
substr = string[i:j]
if substr[:len[substr] // 2] == substr[-1:-[len[substr] // 2] - 1:-1]:
count += 1
return count
Learn more about our help with Assignments: Python
Palindrome - 3
This Program name is Palindrome - 2. Write a Python program to Palindrome - 3
The below link contains Palindrome - 3 question, explanation and test cases
//drive.google.com/file/d/1ZELb2KSvV36MM4kiF-dTFTC43CeUiFH9/view?usp=sharing
We need exact output when the code was run
s = input['Enter your words: ']
s = s.lower[]
s2 = s.split[]
s3 = ''.join[s2]
if s3 == s3[::-1]:
print["True"]
else:
print["False"]
Enter your words: No melon no lemon
True
Enter your words: Race Cars
FalseLearn more about our help with Assignments: Python
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++
#include
using namespace std;
int CountPS[char str[], int
n]
{
int ans=0;
bool P[n][n];
memset[P, false, sizeof[P]];
for [int i = 0; i < n; i++]{
P[i][i] = true;
}
for [int gap = 2; gap
Javascript
function
CountPS[str,n]
{
let dp=new Array[n];
let P=new Array[n];
for[let i=0;i j]
return 1;
if [dp[i][j] != -1]
return dp[i][j];
if [s[i] != s[j]]
return dp[i][j] = 0;
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;
for [int i = 0; i < n; i++]
{
for [int j = i + 1; j < n; j++]
{
if [isPal[s, i, j]]
count++;
}
}
return count;
}
int main[]
{
string s = "abbaeae";
cout j]
return 1;
if [dp[i][j] != -1]
return dp[i][j];
if [s.charAt[i] != s.charAt[j]]
return
dp[i][j] = 0;
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;
for [int i = 0; i < n; i++]
{
for
[int j = i + 1; j < n; j++]
{
if [isPal[s, i, j] != 0]
count++;
}
}
return count;
}
public static void main[String[] args] {
String s = "abbaeae";
System.out.println[countSubstrings[s]];
}
}
Python3
dp = [[-1 for
i in range[1001]]
for j in range[1001]]
def isPal[s, i, j]:
if [i > j]:
return 1
if [dp[i][j] != -1]:
return dp[i][j]
if [s[i] != s[j]]:
dp[i][j] = 0
return dp[i][j]
dp[i][j] = isPal[s, i +
1, j - 1]
return dp[i][j]
def countSubstrings[s]:
n = len[s]
count = 0
for i in range[n]:
for j in range[i + 1, n]:
if [isPal[s, i, j]]:
count += 1
return count
s = "abbaeae"
print[countSubstrings[s]]
C#
using System;
class GFG{
static int[,] dp = new int[1001, 1001];
static int isPal[string s, int i, int j]
{
if [i > j]
return 1;
if [dp[i, j] != -1]
return dp[i, j];
if [s[i] != s[j]]
return dp[i, j] = 0;
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;
for[int i = 0; i < n; i++]
{
for[int j = i + 1; j < n; j++]
{
if [isPal[s, i, j] != 0]
count++;
}
}
return count;
}
static void Main[]
{
string s = "abbaeae";
Console.WriteLine[countSubstrings[s]];
}
}
Javascript
var dp = Array[1001].fill[].map[[]=>Array[1001].fill[-1]];
function isPal[ s , i , j]
{
if [i > j]
return 1;
if
[dp[i][j] != -1]
return dp[i][j];
if [s.charAt[i] != s.charAt[j]]
return dp[i][j] = 0;
return dp[i][j] = isPal[s, i + 1, j - 1];
}
function countSubstrings[ s] {
var n = s.length;
var count = 0;
for [i = 0; i < n; i++] {
for [j = i + 1; j < n; j++] {
if [isPal[s, i, j] != 0]
count++;
}
}
return count;
}
var
s = "abbaeae";
document.write[countSubstrings[s]];
Count All Palindrome Sub-Strings in a String | Set 2
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