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Number of substrings with each character occurring even times
• Difficulty Level : Medium
• Last Updated : 24 Jun, 2021

Given a string S consisting of N lowercase alphabets, the task is to count the number of substrings whose frequency of each character is even.

Examples:

Input: S = “abbaa”
Output: 4
Explanation:
The substrings having frequency of each character is even are {“abba”, “aa”, “bb”, “bbaa”}.
Therefore, the count is 4.

Input: S = “geeksforgeeks”
Output: 2

Naive Approach: The simplest approach to solve the given problem is to generate all possible substrings of the given string and count those substrings having even frequency of every character. After checking for all substrings, print the total count obtained.

Below is the implementation of the above approach:

## Java

 `// Java program for the above approach` `import` `java.io.*;` `import` `java.lang.*;` `import` `java.util.*;` `public` `class` `GFG ` `{`   `    ``// Function to count substrings having` `    ``// even frequency of each character` `    ``static` `int` `subString(String s, ``int` `n)` `    ``{`   `        ``// Stores the total` `        ``// count of substrings` `        ``int` `count = ``0``;`   `        ``// Traverse the range [0, N]:` `        ``for` `(``int` `i = ``0``; i < n; i++) {`   `            ``// Traverse the range [i + 1, N]` `            ``for` `(``int` `len = i + ``1``; len <= n; len++) {`   `                ``// Stores the substring over` `                ``// the range of indices [i, len]` `                ``String test_str = s.substring(i, len);`   `                ``// Stores the frequency of characters` `                ``HashMap res` `                    ``= ``new` `HashMap<>();`   `                ``// Count frequency of each character` `                ``for` `(``char` `keys : test_str.toCharArray()) {` `                    ``res.put(keys,` `                            ``res.getOrDefault(keys, ``0``) + ``1``);` `                ``}`   `                ``int` `flag = ``0``;`   `                ``// Traverse the dictionary` `                ``for` `(``char` `keys : res.keySet()) {`   `                    ``// If any of the keys` `                    ``// have odd count` `                    ``if` `(res.get(keys) % ``2` `!= ``0``) {`   `                        ``flag = ``1``;` `                        ``break``;` `                    ``}` `                ``}` `              `  `                ``// Otherwise` `                ``if` `(flag == ``0``)` `                    ``count += ``1``;` `            ``}` `        ``}`   `        ``// Return count` `        ``return` `count;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``String S = ``"abbaa"``;` `        ``int` `N = S.length();` `        ``System.out.println(subString(S, N));` `    ``}` `}`   `// This code is contributed by Kingash.`

## Python3

 `# Python program for the above approach`   `# Function to count substrings having` `# even frequency of each character` `def` `subString(s, n):`   `    ``# Stores the total` `    ``# count of substrings` `    ``count ``=` `0`   `    ``# Traverse the range [0, N]:` `    ``for` `i ``in` `range``(n):` `      `  `        ``# Traverse the range [i + 1, N]` `        ``for` `len` `in` `range``(i ``+` `1``, n ``+` `1``):` `          `  `            ``# Stores the substring over` `            ``# the range of indices [i, len]` `            ``test_str ``=` `(s[i: ``len``])`   `            ``# Stores the frequency of characters` `            ``res ``=` `{}`   `            ``# Count frequency of each character` `            ``for` `keys ``in` `test_str:` `                ``res[keys] ``=` `res.get(keys, ``0``) ``+` `1` `                `  `            ``flag ``=` `0` `            `  `            ``# Traverse the dictionary` `            ``for` `keys ``in` `res:` `              `  `                ``# If any of the keys` `                ``# have odd count` `                ``if` `res[keys] ``%` `2` `!``=` `0``:`   `                    ``flag ``=` `1` `                    ``break` `                    `  `            ``# Otherwise` `            ``if` `flag ``=``=` `0``:` `                ``count ``+``=` `1`   `    ``# Return count` `    ``return` `count`     `# Driver Code`   `S ``=` `"abbaa"` `N ``=` `len``(S)` `print``(subString(S, N))`

## Javascript

 ``

Output:

`4`

Time Complexity: O(N2 * 26)
Auxiliary Space: O(N)

Efficient Approach: The above approach can be optimized by using the concept of Bitmasking and dictionary. Follow the below steps to solve the problem:

• Initialize a dictionary, say hash to store the count of a character.
• Initialize two variables, say count as 0 and pre as 0 to store the total count of substrings having an even count of each character and to store the mask of characters included in the substring.
• Traverse the given string and perform the following steps:
• Flip the (S[i] – ‘a’)th bit in the variable pre.
• Increment the count by hash[pre] and the count of pre in the hash.
• After completing the above steps, print the value of count as the result.

Below is the implementation of the above approach:

## C++

 `// C ++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to count substrings having` `// even frequency of each character` `int` `subString(string s, ``int` `n)` `{`   `    ``// Stores the count of a character` `    ``map<``int``, ``int``> hash;` `    ``hash = 1;`   `    ``// Stores bitmask` `    ``int` `pre = 0;`   `    ``// Stores the count of substrings` `    ``// with even count of each character` `    ``int` `count = 0;`   `    ``// Traverse the string S` `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``// Flip the ord(i)-97 bits in pre` `        ``pre ^= (1 << ``int``(s[i]) - 97);`   `        ``// Increment the count by hash[pre]` `        ``count += hash[pre];`   `        ``// Increment count of pre in hash` `        ``hash[pre] = hash[pre] + 1;` `    ``}`   `    ``// Return the total count obtained` `    ``return` `count;` `}`   `// Driver Code` `int` `main()` `{` `    ``string S = ``"abbaa"``;` `    ``int` `N = S.length();` `    ``cout << (subString(S, N));` `}`   `// THIS CODE IS CONTRIBUTED BY UKASP.`

## Java

 `// Java program for the above approach` `import` `java.util.*;` `import` `java.lang.*;`   `class` `GFG{` `    `  `// Function to count substrings having` `// even frequency of each character` `static` `int` `subString(String s, ``int` `n)` `{` `    `  `    ``// Stores the count of a character` `    ``Map hash = ``new` `HashMap<>();` `    ``hash.put(``0``, ``1``);`   `    ``// Stores bitmask` `    ``int` `pre = ``0``;`   `    ``// Stores the count of substrings` `    ``// with even count of each character` `    ``int` `count = ``0``;`   `    ``// Traverse the string S` `    ``for``(``int` `i = ``0``; i < n; i++) ` `    ``{` `        `  `        ``// Flip the ord(i)-97 bits in pre` `        ``pre ^= (``1` `<< (``int``)(s.charAt(i) - ``97``));`   `        ``// Increment the count by hash[pre]` `        ``count += hash.getOrDefault(pre, ``0``);`   `        ``// Increment count of pre in hash` `        ``hash.put(pre, hash.getOrDefault(pre, ``0``) + ``1``);` `    ``}`   `    ``// Return the total count obtained` `    ``return` `count;` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``String S = ``"abbaa"``;` `    ``int` `N = S.length();` `    `  `    ``System.out.print(subString(S, N));` `}` `}`   `// This code is contributed by offbeat`

## Python3

 `# Python program for the above approach`   `# Function to count substrings having` `# even frequency of each character` `def` `subString(s, n):`   `    ``# Stores the count of a character` `    ``hash` `=` `{``0``: ``1``}`   `    ``# Stores bitmask` `    ``pre ``=` `0`   `    ``# Stores the count of substrings` `    ``# with even count of each character` `    ``count ``=` `0`   `    ``# Traverse the string S` `    ``for` `i ``in` `s:` `      `  `        ``# Flip the ord(i)-97 bits in pre` `        ``pre ^``=` `(``1` `<< ``ord``(i) ``-` `97``)`   `        ``# Increment the count by hash[pre]` `        ``count ``+``=` `hash``.get(pre, ``0``)`   `        ``# Increment count of pre in hash` `        ``hash``[pre] ``=` `hash``.get(pre, ``0``) ``+` `1` `        `  `    ``# Return the total count obtained` `    ``return` `count`     `# Driver Code`   `S ``=` `"abbaa"` `N ``=` `len``(S)` `print``(subString(S, N))`

## C#

 `// C# program for the above approach` `using` `System.IO;` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG{`   `// Function to count substrings having` `// even frequency of each character` `static` `int` `subString(``string` `s, ``int` `n)` `{` `    `  `    ``// Stores the count of a character` `    ``Dictionary<``int``, ` `               ``int``> hash = ``new` `Dictionary<``int``, ` `                                          ``int``>();`   `    ``hash = 1;`   `    ``// Stores bitmask` `    ``int` `pre = 0;`   `    ``// Stores the count of substrings` `    ``// with even count of each character` `    ``int` `count = 0;`   `    ``// Traverse the string S` `    ``for``(``int` `i = 0; i < n; i++)` `    ``{` `        `  `        ``// Flip the ord(i)-97 bits in pre` `        ``pre ^= (1 << (``int``)(s[i]) - 97);`   `        ``// Increment the count by hash[pre]` `        ``if` `(hash.ContainsKey(pre))` `            ``count += hash[pre];` `        ``else` `            ``count += 0;`   `        ``// Increment count of pre in hash` `        ``if` `(hash.ContainsKey(pre))` `            ``hash[pre] = hash[pre] + 1;` `        ``else` `            ``hash.Add(pre, 1);` `    ``}`   `    ``// Return the total count obtained` `    ``return` `count;` `}`   `// Driver code` `static` `void` `Main()` `{` `    ``String S = ``"abbaa"``;` `    ``int` `N = S.Length;` `    `  `    ``Console.WriteLine(subString(S, N));` `}` `}`   `// This code is contributed by sk944795`

## Javascript

 ``

Output:

`4`

Time Complexity: O(N)
Auxiliary Space: O(N)

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