Count of ways to split given string into two non-empty palindromes
Last Updated :
25 Jul, 2023
Given a string S, the task is to find the number of ways to split the given string S into two non-empty palindromic strings.
Examples:
Input: S = "aaaaa"
Output: 4
Explanation:
Possible Splits: {"a", "aaaa"}, {"aa", "aaa"}, {"aaa", "aa"}, {"aaaa", "a"}
Input: S = "abacc"
Output: 1
Explanation:
Only possible split is "aba", "cc".
Naive Approach: The naive approach is to split the string at each possible index and check if both the substrings are palindromic or not. If yes then increment the count for that index. Print the final count.
Below is the implementation of the above approach:
C++
// C++ Program to implement
// the above approach
#include<bits/stdc++.h>
using namespace std;
// Function to check whether the
// substring from l to r is
// palindrome or not
bool isPalindrome(int l, int r,
string& s)
{
while (l <= r) {
// If characters at l and
// r differ
if (s[l] != s[r])
// Not a palindrome
return false;
l++;
r--;
}
// If the string is
// a palindrome
return true;
}
// Function to count and return
// the number of possible splits
int numWays(string& s)
{
int n = s.length();
// Stores the count
// of splits
int ans = 0;
for (int i = 0;
i < n - 1; i++) {
// Check if the two substrings
// after the split are
// palindromic or not
if (isPalindrome(0, i, s)
&& isPalindrome(i + 1,
n - 1, s)) {
// If both are palindromes
ans++;
}
}
// Print the final count
return ans;
}
// Driver Code
int main()
{
string S = "aaaaa";
cout << numWays(S);
return 0;
}
// Java program to implement
// the above approach
class GFG{
// Function to check whether the
// substring from l to r is
// palindrome or not
public static boolean isPalindrome(int l, int r,
String s)
{
while (l <= r)
{
// If characters at l and
// r differ
if (s.charAt(l) != s.charAt(r))
// Not a palindrome
return false;
l++;
r--;
}
// If the string is
// a palindrome
return true;
}
// Function to count and return
// the number of possible splits
public static int numWays(String s)
{
int n = s.length();
// Stores the count
// of splits
int ans = 0;
for(int i = 0; i < n - 1; i++)
{
// Check if the two substrings
// after the split are
// palindromic or not
if (isPalindrome(0, i, s) &&
isPalindrome(i + 1, n - 1, s))
{
// If both are palindromes
ans++;
}
}
// Print the final count
return ans;
}
// Driver Code
public static void main(String args[])
{
String S = "aaaaa";
System.out.println(numWays(S));
}
}
// This code is contributed by SoumikMondal
# Python3 program to implement
# the above approach
# Function to check whether the
# substring from l to r is
# palindrome or not
def isPalindrome(l, r, s):
while (l <= r):
# If characters at l and
# r differ
if (s[l] != s[r]):
# Not a palindrome
return bool(False)
l += 1
r -= 1
# If the string is
# a palindrome
return bool(True)
# Function to count and return
# the number of possible splits
def numWays(s):
n = len(s)
# Stores the count
# of splits
ans = 0
for i in range(n - 1):
# Check if the two substrings
# after the split are
# palindromic or not
if (isPalindrome(0, i, s) and
isPalindrome(i + 1, n - 1, s)):
# If both are palindromes
ans += 1
# Print the final count
return ans
# Driver Code
S = "aaaaa"
print(numWays(S))
# This code is contributed by divyeshrabadiya07
// C# program to implement
// the above approach
using System;
class GFG{
// Function to check whether the
// substring from l to r is
// palindrome or not
public static bool isPalindrome(int l, int r,
string s)
{
while (l <= r)
{
// If characters at l and
// r differ
if (s[l] != s[r])
// Not a palindrome
return false;
l++;
r--;
}
// If the string is
// a palindrome
return true;
}
// Function to count and return
// the number of possible splits
public static int numWays(string s)
{
int n = s.Length;
// Stores the count
// of splits
int ans = 0;
for(int i = 0; i < n - 1; i++)
{
// Check if the two substrings
// after the split are
// palindromic or not
if (isPalindrome(0, i, s) &&
isPalindrome(i + 1, n - 1, s))
{
// If both are palindromes
ans++;
}
}
// Print the final count
return ans;
}
// Driver Code
public static void Main(string []args)
{
string S = "aaaaa";
Console.Write(numWays(S));
}
}
// This code is contributed by Rutvik
<script>
// Javascript program to implement
// the above approach
// Function to check whether the
// substring from l to r is
// palindrome or not
function isPalindrome(l, r, s)
{
while (l <= r)
{
// If characters at l and
// r differ
if (s[l] != s[r])
// Not a palindrome
return false;
l++;
r--;
}
// If the string is
// a palindrome
return true;
}
// Function to count and return
// the number of possible splits
function numWays(s)
{
let n = s.length;
// Stores the count
// of splits
let ans = 0;
for(let i = 0; i < n - 1; i++)
{
// Check if the two substrings
// after the split are
// palindromic or not
if (isPalindrome(0, i, s) &&
isPalindrome(i + 1, n - 1, s))
{
// If both are palindromes
ans++;
}
}
// Print the final count
return ans;
}
let S = "aaaaa";
document.write(numWays(S));
</script>
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized using the Hashing and Rabin-Karp Algorithm to store Prefix and Suffix Hashes of the string. Follow the steps below to solve the problem:
- Compute prefix and suffix hash of the given string.
- For every index i in the range [1, N - 1], check if the two substrings [0, i - 1] and [i, N - 1] are palindrome or not.
- To check if a substring [l, r] is a palindrome or not, simply check:
PrefixHash[l - r] = SuffixHash[l - r]
- For every index i for which two substrings are found to be palindromic, increase the count.
- Print the final value of count.
Below is the implementation of the above approach:
C++
// C++ Program to implement
// the above approach
#include
using namespace std;
// Modulo for rolling hash
const int MOD = 1e9 + 9;
// Small prime for rolling hash
const int P = 37;
// Maximum length of string
const int MAXN = 1e5 + 5;
// Stores prefix hash
vector prefixHash(MAXN);
// Stores suffix hash
vector suffixHash(MAXN);
// Stores inverse modulo
// of P for prefix
vector inversePrefix(MAXN);
// Stores inverse modulo
// of P for suffix
vector inverseSuffix(MAXN);
int n;
int power(int x, int y, int mod)
{
// Function to compute
// power under modulo
if (x == 0)
return 0;
int ans = 1;
while (y > 0) {
if (y & 1)
ans = (1LL * ans * x)
% MOD;
x = (1LL * x * x) % MOD;
y >>= 1;
}
return ans;
}
// Precompute hashes for the
// given string
void preCompute(string& s)
{
int x = 1;
for (int i = 0; i 0)
prefixHash[i]
= (prefixHash[i]
+ prefixHash[i - 1])
% MOD;
// Compute inverse modulo
// of P ^ i for division
// using Fermat Little theorem
inversePrefix[i] = power(x, MOD - 2,
MOD);
x = (1LL * x * P) % MOD;
}
x = 1;
// Calculate suffix hash
for (int i = n - 1; i >= 0; i--) {
// Calculate and store hash
suffixHash[i]
= (1LL * int(s[i]
- 'a' + 1)
* x)
% MOD;
if (i 0
? prefixHash[l - 1]
: 0);
h = (h + MOD) % MOD;
h = (1LL * h * inversePrefix[l])
% MOD;
return h;
}
// Function to return Suffix
// Hash of substring
int getSuffixHash(int l, int r)
{
// Calculate suffix hash
// from l to r
int h = suffixHash[l]
- (r < n - 1
? suffixHash[r + 1]
: 0);
h = (h + MOD) % MOD;
h = (1LL * h * inverseSuffix[r])
% MOD;
return h;
}
int numWays(string& s)
{
n = s.length();
// Compute prefix and
// suffix hashes
preCompute(s);
// Stores the number of
// possible splits
int ans = 0;
for (int i = 0;
i < n - 1; i++) {
int preHash = getPrefixHash(0, i);
int sufHash = getSuffixHash(0, i);
// If the substring s[0]...s[i]
// is not palindromic
if (preHash != sufHash)
continue;
preHash = getPrefixHash(i + 1,
n - 1);
sufHash = getSuffixHash(i + 1,
n - 1);
// If the substring (i + 1, n - 1)
// is not palindromic
if (preHash != sufHash)
continue;
// If both are palindromic
ans++;
}
return ans;
}
// Driver Code
int main()
{
string s = "aaaaa";
int ans = numWays(s);
cout << ans << endl;
return 0;
}
// Java Program to implement
// the above approach
import java.util.*;
public class Main {
// Modulo for rolling hash
static final int MOD = 1_000_000_007;
// Small prime for rolling hash
static final int P = 37;
// Maximum length of string
static final int MAXN = 100_005;
// Stores prefix hash
static int[] prefixHash = new int[MAXN];
// Stores suffix hash
static int[] suffixHash = new int[MAXN];
// Stores inverse modulo
// of P for prefix
static int[] inversePrefix = new int[MAXN];
// Stores inverse modulo
// of P for suffix
static int[] inverseSuffix = new int[MAXN];
static int n;
static int power(int x, int y, int mod) {
// Function to compute
// power under modulo
if (x == 0)
return 0;
int ans = 1;
while (y > 0) {
if ((y & 1) == 1)
ans = (int)(((long)ans * x) % mod);
x = (int)(((long)x * x) % mod);
y >>= 1;
}
return ans;
}
// Precompute hashes for the
// given string
static void preCompute(String s) {
int x = 1;
for (int i = 0; i < n; i++) {
prefixHash[i] = (int)(((long)(s.charAt(i) - 'a' + 1) * x) % MOD);
// Compute inverse modulo
// of P ^ i for division
// using Fermat Little theorem
if (i > 0)
prefixHash[i] = (prefixHash[i] + prefixHash[i - 1]) % MOD;
inversePrefix[i] = power(x, MOD - 2, MOD);
x = (int)(((long)x * P) % MOD);
}
x = 1;
// Calculate suffix hash
for (int i = n - 1; i >= 0; i--) {
// Calculate and store hash
suffixHash[i] = (int)(((long)(s.charAt(i) - 'a' + 1) * x) % MOD);
if (i < n - 1)
suffixHash[i] = (suffixHash[i] + suffixHash[i + 1]) % MOD;
inverseSuffix[i] = power(x, MOD - 2, MOD);
x = (int)(((long)x * P) % MOD);
}
}
static int getPrefixHash(int l, int r) {
int h = prefixHash[r];
if (l > 0)
h = (h - prefixHash[l - 1] + MOD) % MOD;
h = (int)(((long)h * inversePrefix[l]) % MOD);
return h;
}
// Function to return Suffix
// Hash of substring
static int getSuffixHash(int l, int r) {
int h = suffixHash[l];
if (r < n - 1)
h = (h - suffixHash[r + 1] + MOD) % MOD;
h = (int)(((long)h * inverseSuffix[r]) % MOD);
return h;
}
static int numWays(String s) {
n = s.length();
// Calculate suffix hash
// from l to r
preCompute(s);
// Stores the number of
// possible splits
int ans = 0;
for (int i = 0; i < n - 1; i++) {
int preHash = getPrefixHash(0, i);
int sufHash = getSuffixHash(0, i);
// If the substring s[0]...s[i]
// is not palindromic
if (preHash != sufHash)
continue;
preHash = getPrefixHash(i + 1, n - 1);
sufHash = getSuffixHash(i + 1, n - 1);
// If the substring (i + 1, n - 1)
// is not palindromic
if (preHash != sufHash)
continue;
// If both are palindromic
ans++;
}
return ans;
}
// Driver Code
public static void main(String[] args) {
String s = "aaaaa";
int ans = numWays(s);
System.out.println(ans);
}
}
// Contributed by adityasha4x71
# Python Program to implement
# the above approach
# Modulo for rolling hash
MOD = 10**9 + 9
# Small prime for rolling hash
P = 37
# Maximum length of string
MAXN = 10**5 + 5
# Stores prefix hash
prefixHash = [0] * MAXN
# Stores suffix hash
suffixHash = [0] * MAXN
# Stores inverse modulo
# of P for prefix
inversePrefix = [0] * MAXN
# Stores inverse modulo
# of P for suffix
inverseSuffix = [0] * MAXN
def power(x, y, mod):
# Function to compute
# power under modulo
if x == 0:
return 0
ans = 1
while y > 0:
if y & 1:
ans = (ans * x) % mod
x = (x * x) % mod
y >>= 1
return ans
# Precompute hashes for the
# given string
def preCompute(s):
global prefixHash, suffixHash, inversePrefix, inverseSuffix, P, MOD
x = 1
for i in range(len(s)):
# Calculate and store hash
prefixHash[i] = ((ord(s[i]) - ord('a') + 1) * x) % MOD
# Calculate prefix sum
if i > 0:
prefixHash[i] = (prefixHash[i] + prefixHash[i - 1]) % MOD
# Compute inverse modulo
# of P ^ i for division
# using Fermat Little theorem
inversePrefix[i] = power(x, MOD - 2, MOD)
x = (x * P) % MOD
x = 1
# Calculate suffix hash
for i in range(len(s) - 1, -1, -1):
# Calculate and store hash
suffixHash[i] = ((ord(s[i]) - ord('a') + 1) * x) % MOD
if i < len(s) - 1:
suffixHash[i] = (suffixHash[i] + suffixHash[i + 1]) % MOD
# Compute inverse modulo
# of P ^ i for division
# using Fermat Little theorem
inverseSuffix[i] = power(x, MOD - 2, MOD)
x = (x * P) % MOD
# Function to return Prefix
# Hash of substring
def getPrefixHash(l, r):
global prefixHash, inversePrefix, P, MOD
# Calculate prefix hash
# from l to r
h = prefixHash[r] - (prefixHash[l - 1] if l > 0 else 0)
h = (h + MOD) % MOD
h = (h * inversePrefix[l]) % MOD
return h
# Function to return Suffix
# Hash of substring
def getSuffixHash(l, r):
global suffixHash, inverseSuffix, P, MOD
# Calculate suffix hash
# from l to r
h = suffixHash[l] - (suffixHash[r + 1] if r < len(suffixHash) - 1 else 0)
h = (h + MOD) % MOD
h = (h * inverseSuffix[r]) % MOD
return h
def numWays(s):
global n, preHash, sufHash
n = len(s)
# Compute prefix and
# suffix hashes
preCompute(s)
# Stores the number of
# possible splits
ans = 0
for i in range(n - 1):
preHash = getPrefixHash(0, i)
sufHash = getSuffixHash(0, i)
# If the substring s[0]...s[i]
# is not palindromic
if (preHash != sufHash):
continue
preHash = getPrefixHash(i + 1,
n - 1)
sufHash = getSuffixHash(i + 1,
n - 1)
# If the substring (i + 1, n - 1)
# is not palindromic
if (preHash != sufHash):
continue
# If both are palindromic
ans += 1
return ans
# Driver Code
s = "aaaaa"
ans = numWays(s)
print(ans)
// C# program for the above approach
using System;
public class GFG
{
// Modulo for rolling hash
static readonly int MOD = 1_000_000_007;
// Small prime for rolling hash
static readonly int P = 37;
// Maximum length of string
static readonly int MAXN = 100_005;
// Stores prefix hash
static int[] prefixHash = new int[MAXN];
// Stores suffix hash
static int[] suffixHash = new int[MAXN];
// Stores inverse modulo
// of P for prefix
static int[] inversePrefix = new int[MAXN];
// Stores inverse modulo
// of P for suffix
static int[] inverseSuffix = new int[MAXN];
static int n;
// Function to compute
// power under modulo
static int power(int x, int y, int mod)
{
if (x == 0)
return 0;
int ans = 1;
while (y > 0)
{
if ((y & 1) == 1)
ans = (int)(((long)ans * x) % mod);
x = (int)(((long)x * x) % mod);
y >>= 1;
}
return ans;
}
// Precompute hashes for the
// given string
static void preCompute(string s)
{
int x = 1;
for (int i = 0; i < n; i++)
{
prefixHash[i] = (int)(((long)(s[i] - 'a' + 1) * x) % MOD);
// Compute inverse modulo
// of P ^ i for division
// using Fermat Little theorem
if (i > 0)
prefixHash[i] = (prefixHash[i] + prefixHash[i - 1]) % MOD;
inversePrefix[i] = power(x, MOD - 2, MOD);
x = (int)(((long)x * P) % MOD);
}
x = 1;
// Calculate suffix hash
for (int i = n - 1; i >= 0; i--)
{
// Calculate and store hash
suffixHash[i] = (int)(((long)(s[i] - 'a' + 1) * x) % MOD);
if (i < n - 1)
suffixHash[i] = (suffixHash[i] + suffixHash[i + 1]) % MOD;
inverseSuffix[i] = power(x, MOD - 2, MOD);
x = (int)(((long)x * P) % MOD);
}
}
// Function to return Prefix
// Hash of substring
static int getPrefixHash(int l, int r)
{
int h = prefixHash[r];
if (l > 0)
h = (h - prefixHash[l - 1] + MOD) % MOD;
h = (int)(((long)h * inversePrefix[l]) % MOD);
return h;
}
// Function to return Suffix
// Hash of the substring
static int getSuffixHash(int l, int r)
{
int h = suffixHash[l];
if (r < n - 1)
h = (h - suffixHash[r + 1] + MOD) % MOD;
h = (int)(((long)h * inverseSuffix[r]) % MOD);
return h;
}
static int numWays(string s)
{
n = s.Length;
// Calculate suffix hash
// from l to r
preCompute(s);
// Stores the number of
// possible splits
int ans = 0;
for (int i = 0; i < n - 1; i++)
{
int preHash = getPrefixHash(0, i);
int sufHash = getSuffixHash(0, i);
// If the substring s[0]...s[i]
// is not palindromic
if (preHash != sufHash)
continue;
preHash = getPrefixHash(i + 1, n - 1);
sufHash = getSuffixHash(i + 1, n - 1);
// If the substring (i + 1, n - 1)
// is not palindromic
if (preHash != sufHash)
continue;
// If both are palindromic
ans++;
}
return ans;
}
// Driver Code
public static void Main(string[] args)
{
string s = "aaaaa";
int ans = numWays(s);
Console.WriteLine(ans);
}
}
// This code is contributed by princekumaras
// Modulo for rolling hash
const MOD = BigInt(10 ** 9 + 9);
// Small prime for rolling hash
const P = BigInt(37);
// Maximum length of string
const MAXN = 10 ** 5 + 5;
// Stores prefix hash
const prefixHash = new Array(MAXN).fill(0n);
// Stores suffix hash
const suffixHash = new Array(MAXN).fill(0n);
// Stores inverse modulo
// of P for prefix
const inversePrefix = new Array(MAXN).fill(0n);
// Stores inverse modulo
// of P for suffix
const inverseSuffix = new Array(MAXN).fill(0n);
function power(x, y, mod) {
// Function to compute
// power under modulo
if (x == 0n) {
return 0n;
}
let ans = 1n;
while (y > 0) {
if (y & 1n) {
ans = (ans * x) % mod;
}
x = (x * x) % mod;
y >>= 1n;
}
return ans;
}
// Precompute hashes for the
// given string
function preCompute(s) {
let x = 1n;
for (let i = 0; i < s.length; i++) {
// Calculate and store hash
prefixHash[i] = ((BigInt(s.charCodeAt(i) - "a".charCodeAt(0) + 1) * x) % MOD);
// Calculate prefix sum
if (i > 0) {
prefixHash[i] = (prefixHash[i] + prefixHash[i - 1]) % MOD;
}
// Compute inverse modulo
// of P ^ i for division
// using Fermat Little theorem
inversePrefix[i] = power(x, MOD - 2n, MOD);
x = (x * P) % MOD;
}
x = 1n;
// Calculate suffix hash
for (let i = s.length - 1; i >= 0; i--) {
// Calculate and store hash
suffixHash[i] = ((BigInt(s.charCodeAt(i) - "a".charCodeAt(0) + 1) * x) % MOD);
if (i < s.length - 1) {
suffixHash[i] = (suffixHash[i] + suffixHash[i + 1]) % MOD;
}
// Compute inverse modulo
// of P ^ i for division
// using Fermat Little theorem
inverseSuffix[i] = power(x, MOD - 2n, MOD);
x = (x * P) % MOD;
}
}
// Function to return Prefix
// Hash of substring
function getPrefixHash(l, r) {
// Calculate prefix hash
// from l to r
let h = prefixHash[r] - (prefixHash[l - 1n] || 0n);
h = (h + MOD) % MOD;
h = (h * inversePrefix[l]) % MOD;
return h;
}
// Function to return Suffix
// Hash of substring
function getSuffixHash(l, r) {
// Calculate suffix hash
// from l to r
let h = suffixHash[l] - (suffixHash[r + 1] || 0n);
h = (h + MOD) % MOD;
h = (h * inverseSuffix[r]) % MOD;
return h;
}
function numW
Time Complexity: O(N * log(109))
Auxiliary Space: O(N)
Approach Name: Split String into Palindromes
Steps:
- Define a function named count_palindrome_splits that takes a string S as input.
- Initialize a variable count to 0.
- Loop through each possible index i to split the string from 1 to len(S)-1.
- Check if both the substrings formed by the split are palindromes.
- If yes, increment count.
- Return the count.
C++
#include <iostream>
#include <string>
using namespace std;
bool is_palindrome(string s)
{
return s == string(s.rbegin(), s.rend());
}
int count_palindrome_splits(string S)
{
int count = 0;
for (int i = 1; i < S.length(); i++) {
string left_substring = S.substr(0, i);
string right_substring = S.substr(i);
if (is_palindrome(left_substring)
&& is_palindrome(right_substring)) {
count++;
}
}
return count;
}
int main()
{
string S = "aaaaa";
cout << count_palindrome_splits(S) << endl; // Output: 4
return 0;
}
import java.util.*;
public class Main {
public static boolean isPalindrome(String s)
{
return s.equals(
new StringBuilder(s).reverse().toString());
}
public static int countPalindromeSplits(String S)
{
int count = 0;
for (int i = 1; i < S.length(); i++) {
String leftSubstring = S.substring(0, i);
String rightSubstring = S.substring(i);
if (isPalindrome(leftSubstring)
&& isPalindrome(rightSubstring)) {
count++;
}
}
return count;
}
public static void main(String[] args)
{
String S = "aaaaa";
System.out.println(
countPalindromeSplits(S)); // Output: 4
}
}
def count_palindrome_splits(S):
count = 0
for i in range(1, len(S)):
left_substring = S[:i]
right_substring = S[i:]
if is_palindrome(left_substring) and is_palindrome(right_substring):
count += 1
return count
def is_palindrome(s):
return s == s[::-1]
# Example usage
S = "aaaaa"
print(count_palindrome_splits(S)) # Output: 4
using System;
public class Program {
public static int CountPalindromeSplits(string s)
{
int count = 0;
for (int i = 1; i < s.Length; i++) {
string leftSubstring = s.Substring(0, i);
string rightSubstring = s.Substring(i);
if (IsPalindrome(leftSubstring)
&& IsPalindrome(rightSubstring)) {
count++;
}
}
return count;
}
public static bool IsPalindrome(string s)
{
char[] charArray = s.ToCharArray();
Array.Reverse(charArray);
string reversedString = new string(charArray);
return s == reversedString;
}
public static void Main()
{
string s = "aaaaa";
Console.WriteLine(
CountPalindromeSplits(s)); // Output: 4
}
}
//Funtion to check palindrome
function isPalindrome(s) {
return s === s.split('').reverse().join('');
}
function countPalindromeSplits(S) {
let count = 0;
for (let i = 1; i < S.length; i++) {
// finding the left and right substring
let leftSubstring = S.substring(0, i);
let rightSubstring = S.substring(i);
if (isPalindrome(leftSubstring) && isPalindrome(rightSubstring)) {
count++;
}
}
return count;
}
// Main function
function main() {
let S = "aaaaa";
// Function call
console.log(countPalindromeSplits(S)); // Output: 4
}
main();
Time Complexity: O(n^2) where n is the length of the input string S.
Auxiliary Space: O(n) where n is the length of the input string S.
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