Count substrings of a given string whose anagram is a palindrome
Last Updated :
15 Jul, 2025
Given a string S of length N containing only lowercase alphabets, the task is to print the count of substrings of the given string whose anagram is palindromic.
Examples:
Input: S = "aaaa"
Output: 10
Explanation:
Possible substrings are {"a", "a", "a", "a", "aa", "aa", "aa", "aaa", "aaa", "aaaa"}. Since all substrings are have palindromic anagrams, the required answer is 10.
Input: S = "abc"
Output: 3
Naive Approach: The idea is to generate all substrings of the given string and for each substring, check whether its anagram is a palindrome or not. Keep increasing the count of substrings for which the above condition is found to be true. Finally, print the count of all such substrings.
Time Complexity: O(N3)
Auxiliary Space: O(N)
Bit Masking Approach: The idea is to generate the masks of 26-bit numbers as there are 26 characters. Also, observe that if the anagram of some string is a palindrome, then the frequencies of its characters except at most one character must be even.
Follow the steps below to solve the problem:
- Traverse the string and initialize a variable X = 0 at each index.
- From every ithe index, traverse the string over a range of indices [i, N - 1], and for each character S[j], calculate Bitwise XOR of X and 2(S[j] - 'a'), where 0th bit represents if the frequency of a is odd, 1st bit represents if the frequency of b is odd, and so on.
- Then, check if X & (X - 1) is equal to 0 or not. If found to be true, then increment the count by 1.
Illustration:
Suppose, X = (0001000)2
=> (X - 1) will be represented as (0000111)2.
Therefore, X & (X - 1) = 0
- Once the above steps are exhausted, print the count obtained.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print count of substrings
// whose anagrams are palindromic
void countSubstring(string s)
{
// Stores the answer
int res = 0;
// Iterate over the string
for (int i = 0;
i < s.length(); i++) {
int x = 0;
for (int j = i;
j < s.length(); j++) {
// Set the current character
int temp = 1 << s[j] - 'a';
// Parity update
x ^= temp;
if ((x & (x - 1)) == 0)
res++;
}
}
// Print the final count
cout << res;
}
// Driver Code
int main()
{
string str = "aaa";
// Function Call
countSubstring(str);
return 0;
}
// Java program for
// the above approach
class GFG{
// Function to print count of subStrings
// whose anagrams are palindromic
static void countSubString(String s)
{
// Stores the answer
int res = 0;
// Iterate over the String
for (int i = 0; i < s.length(); i++)
{
int x = 0;
for (int j = i; j < s.length(); j++)
{
// Set the current character
int temp = 1 << s.charAt(j) - 'a';
// Parity update
x ^= temp;
if ((x & (x - 1)) == 0)
res++;
}
}
// Print the final count
System.out.print(res);
}
// Driver Code
public static void main(String[] args)
{
String str = "aaa";
// Function Call
countSubString(str);
}
}
// This code is contributed by shikhasingrajput
# Python3 program for
# the above approach
# Function to print count of subStrings
# whose anagrams are palindromic
def countSubString(s):
# Stores the answer
res = 0;
# Iterate over the String
for i in range(len(s)):
x = 0;
for j in range(i, len(s)):
# Set the current character
temp = 1 << ord(s[j]) - ord('a');
# Parity update
x ^= temp;
if ((x & (x - 1)) == 0):
res += 1;
# Print final count
print(res);
# Driver Code
if __name__ == '__main__':
str = "aaa";
# Function Call
countSubString(str);
# This code is contributed by 29AjayKumar
// C# program for
// the above approach
using System;
class GFG{
// Function to print count of subStrings
// whose anagrams are palindromic
static void countSubString(String s)
{
// Stores the answer
int res = 0;
// Iterate over the String
for (int i = 0; i < s.Length; i++)
{
int x = 0;
for (int j = i; j < s.Length; j++)
{
// Set the current character
int temp = 1 << s[j] - 'a';
// Parity update
x ^= temp;
if ((x & (x - 1)) == 0)
res++;
}
}
// Print the readonly count
Console.Write(res);
}
// Driver Code
public static void Main(String[] args)
{
String str = "aaa";
// Function Call
countSubString(str);
}
}
// This code is contributed by shikhasingrajput
<script>
// JavaScript program for
//the above approach
// Function to print count of subStrings
// whose anagrams are palindromic
function countSubString(s)
{
// Stores the answer
let res = 0;
// Iterate over the String
for (let i = 0; i < s.length; i++)
{
let x = 0;
for (let j = i; j < s.length; j++)
{
// Set the current character
let temp = 1 << s[j] - 'a';
// Parity update
x ^= temp;
if ((x & (x - 1)) == 0)
res++;
}
}
// Print the final count
document.write(res);
}
// Driver Code
let str = "aaa";
// Function Call
countSubString(str);
// This code is contributed by souravghosh0416.
</script>
Time Complexity: O(N2)
Auxiliary Space: O(N)
Efficient Approach: To optimize the above Bit Masking technique, the idea is to use a Map. Follow the steps below to solve the problem:
- Initialize a map to store the frequencies of the masks. Initialize a variable X = 0.
- Traverse the string and for every ith index, calculate Bitwise XOR of X and 2(S[j] - 'a') and update the answer by adding the frequency of the current value of X in the Map because if any substring from 0 to j has the same mask as that of X, which is a mask for substring from 0 to i, then substring S[j + 1, i] will have even frequencies, where j < i.
- Also add the frequencies of masks X xor 2k, where, 0 ? k < 26. After that, increment the frequency of X by 1.
- Repeat the above steps for each index of the given string.
- After traversing the given string, print the required answer.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to get the count of substrings
// whose anagrams are palindromic
void countSubstring(string s)
{
// Store the answer
int answer = 0;
// Map to store the freq of masks
unordered_map<int, int> m;
// Set frequency for mask
// 00...00 to 1
m[0] = 1;
// Store mask in x from 0 to i
int x = 0;
for (int j = 0; j < s.length(); j++) {
x ^= 1 << (s[j] - 'a');
// Update answer
answer += m[x];
for (int i = 0; i < 26; ++i) {
answer += m[x ^ (1 << i)];
}
// Update frequency
m[x] += 1;
}
// Print the answer
cout << answer;
}
// Driver Code
int main()
{
string str = "abab";
// Function Call
countSubstring(str);
return 0;
}
// Java program for
// the above approach
import java.util.*;
class GFG {
// Function to get the count of subStrings
// whose anagrams are palindromic
static void countSubString(String s)
{
// Store the answer
int answer = 0;
// Map to store the freq of masks
HashMap<Integer,
Integer> m = new HashMap<Integer,
Integer>();
// Set frequency for mask
// 00...00 to 1
m.put(0, 1);
// Store mask in x from 0 to i
int x = 0;
for (int j = 0; j < s.length(); j++)
{
x ^= 1 << (s.charAt(j) - 'a');
// Update answer
answer += m.containsKey(x) ? m.get(x) : 0;
for (int i = 0; i < 26; ++i)
{
answer += m.containsKey(x ^ (1 << i)) ?
m.get(x ^ (1 << i)) : 0;
}
// Update frequency
if (m.containsKey(x))
m.put(x, m.get(x) + 1);
else
m.put(x, 1);
}
// Print the answer
System.out.print(answer);
}
// Driver Code
public static void main(String[] args)
{
String str = "abab";
// Function Call
countSubString(str);
}
}
// This code is contributed by shikhasingrajput
# Python3 program for the above approach
from collections import defaultdict
# Function to get the count of substrings
# whose anagrams are palindromic
def countSubstring(s):
# Store the answer
answer = 0
# Map to store the freq of masks
m = defaultdict(lambda : 0)
# Set frequency for mask
# 00...00 to 1
m[0] = 1
# Store mask in x from 0 to i
x = 0
for j in range(len(s)):
x ^= 1 << (ord(s[j]) - ord('a'))
# Update answer
answer += m[x]
for i in range(26):
answer += m[x ^ (1 << i)]
# Update frequency
m[x] += 1
# Print the answer
print(answer)
# Driver Code
str = "abab"
# Function call
countSubstring(str)
# This code is contributed by Shivam Singh
// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to get the count of
// subStrings whose anagrams
// are palindromic
static void countSubString(String s)
{
// Store the answer
int answer = 0;
// Map to store the freq of masks
Dictionary<int,
int> m = new Dictionary<int,
int>();
// Set frequency for mask
// 00...00 to 1
m.Add(0, 1);
// Store mask in x from 0 to i
int x = 0;
for (int j = 0; j < s.Length; j++)
{
x ^= 1 << (s[j] - 'a');
// Update answer
answer += m.ContainsKey(x) ? m[x] : 0;
for (int i = 0; i < 26; ++i)
{
answer += m.ContainsKey(x ^ (1 << i)) ?
m[x ^ (1 << i)] : 0;
}
// Update frequency
if (m.ContainsKey(x))
m[x] = m[x] + 1;
else
m.Add(x, 1);
}
// Print the answer
Console.Write(answer);
}
// Driver Code
public static void Main(String[] args)
{
String str = "abab";
// Function Call
countSubString(str);
}
}
// This code is contributed by shikhasingrajput
<script>
// JavaScript program for the above approach
// Function to get the count of substrings
// whose anagrams are palindromic
function countSubstring(s)
{
// Store the answer
var answer = 0;
// Map to store the freq of masks
var m = new Map();
// Set frequency for mask
// 00...00 to 1
m.set(0, 1);
// Store mask in x from 0 to i
var x = 0;
for (var j = 0; j < s.length; j++) {
x ^= 1 << (s[j].charCodeAt(0) - 'a'.charCodeAt(0));
// Update answer
answer += m.has(x)? m.get(x):0;
for (var i = 0; i < 26; ++i) {
answer += m.has(x ^ (1 << i))?m.get(x ^ (1 << i)):0;
}
// Update frequency
if(m.has(x))
m.set(x, m.get(x)+1)
else
m.set(x, 1)
}
// Print the answer
document.write( answer);
}
// Driver Code
var str = "abab";
// Function Call
countSubstring(str);
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
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