# Hackerrank: Sherlock and anagram (optimal time complexity)

Problem statement

Given a string $S$, find the number of "unordered anagrammatic pairs" of substrings.

Input Format

First line contains $T$, the number of testcases. Each testcase consists of string $S$ in one line.

Constraints

$1 \leq T \leq 10$

$2 \leq length(S) \leq 100$

String $S$ contains only the lowercase letters of the English alphabet.

Output Format

For each testcase, print the required answer in one line.

Sample Input#00

2

abba

abcd


Sample Output#00

4

0


Sample Input#01

5

ifailuhkqq

hucpoltgty

ovarjsnrbf

pvmupwjjjf

iwwhrlkpek


Sample Output#01

3

2

2

6

3


Explanation

Sample00

Let's say $S[i,j]$ denotes the substring $S_i, S_i+1,\ldots ,S_j$ .

testcase 1:

For $S=abba$, anagrammatic pairs are: $\lbrace S[1,1],S[4,4]\rbrace ,\lbrace S[1,2],S[3,4]\rbrace ,\lbrace S[2,2],S[3,3]\rbrace, \lbrace S[1,3],S[2,4]\rbrace$.

testcase 2:

No anagrammatic pairs.

My introduction of algorithm:

Algorithm with time complexity O(26*N*N), N is the string's length

To solve the problem to beat time complexity $O(N^3)$ implementation, I chose to use an idea similar to a more famous problem: Leetcode 238 Product of array except itself, use less time by doing dynamic programming technique. Related to sherlock and anagrams, the time complexity can be lowered from $O(N^3)$ to $O(N^2)$.

Here is more detail. We take advantage of alphabetic size is limited and constant, $26$ chars, and then work with substring (denote $S_i$) starting from $0$ to $i$, $0$ < $i$ < $n$ ($n$ is string's length) to calculate frequency table, in other words, calculate all $O(N^2)$ substrings' frequency table use only $N$ substrings' preprocessed frequency table. Any substring starting from $i$ and ending $j$'s frequency table can be calculated using the difference value between $S_j$'s frequency table and $S_i$'s frequency table for those $26$ alphabetic numbers.

So, the preprocessed frequency table size is $O(N)$ ( $N$ is the length of string) instead of $O(N^2)$ based on each of substrings. For any of substrings, there are only $26$ calculation to compute for each alphabet number, so the time complexity goes to $O(26N^2)$ = $O(N^2)$.

We also need to discuss the time complexity to build a dictionary with key and value, where key is associated with frequency table hashed key, value is how many substring with the hashed key. Since the hashed key cannot be more than substring number in total, the time to build the dictionary (refer to function ConstructHashedAnagramsDictionary's variable hashedAnagramsDictionary). So the time complexity is $O(N^2)$.

I'd like to ask for help on the time complexity analysis, make sure that the time complexity overall is $O(N^2)$, not $O(N^3)$. The code passes all the test cases on HackerRank.

using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.IO;
using System.Linq;
using System.Text;

class Solution
{
public class HashedAnagramString
{
/*
* Make sure that two anagram strings will have some hashed code
*
* @frequencyTable - Dictionary<char, int>
* The frequency table has to be sorted first and then construct
* a string with each char in alphabetic numbers concatenated by
* its occurrences.
*
*/
public static string GetHashedAnagram(int[] frequencyTable)
{
StringBuilder key = new StringBuilder();

for (int i = 0; i < 26; i++)
{
int value = frequencyTable[i];
if (value > 0)
{
char c = (char)(i + 'a');
key.Append(c.ToString() + value);
}
}

return key.ToString();
}
}

static void Main(String[] args)
{
ProcessInput();
//RunSampleTestcase();
}

public static void RunSampleTestcase()
{
var hashedAnagramsDictionary = ConstructHashedAnagramsDictionary("abba");

var pairs = CalculatePairs(hashedAnagramsDictionary);

Debug.Assert(pairs == 4);
}

public static void ProcessInput()
{
var queries = int.Parse(Console.ReadLine());

while (queries-- > 0)
{
var input = Console.ReadLine();

var hashedAnagramsDictionary = ConstructHashedAnagramsDictionary(input);

Console.WriteLine(CalculatePairs(hashedAnagramsDictionary));
}
}

/*
* Prepare fequence table for N substring which starts from index 0
* Do not need include all other substrings as we know there are total O(N*N) substrings.
*
* Work on one small test case:
* "abcd"
* so the frequency table for substrings, all starts from index = 0:
* a
* ab
* abc
* abcd
* Only for those 4 substrings, not all of them.
*
* Time complexity:  O(N * N)
* Space complexity: O(26*N), N is the string's length
*/
public static int[][] PrepareFequencyTableForOnlyNSubstring(string input)
{
if (input == null || input.Length == 0)
{
return null;
}

int length = input.Length;
int[][] frequencyTables = new int[length][];

for (int i = 0; i < length; i++)
{
frequencyTables[i] = new int[26];
}

for (int start = 0; start < length; start++) // go over the string once from the beginning
{
char current = input[start];
int charIndex = current - 'a';

for (int index = start; index < length; index++)
{
frequencyTables[index][charIndex]++;
}
}

return frequencyTables;
}

/*
* What should be taken cared of in the design?
* Time complexity: O(26 * N * N), N is the string length.
*
*
* I think that it is same idea as Leetcode 238 Product of array except itself. We
* take advantage of alphabetic size is limited and constant, 26 chars, and then work
* with substring (denote Si) starting from 0 to i, 0 < i < n ( n is string's length)
* to calculate frequency table, and any substring starting from i and ending j can be
* Sj's frequency table - Si's frequency table for those 26 alphabetic numbers.
* So, the preprocessed frequency table size is O(N) ( N is the length of string)
* instead of O(N^2) based on each of substrings. For any of substrings, there are only 26
* calculation to compute for each alphabet number, so the time complexity goes
* to O(26N^2) = O(n^2)
*
* Update hashed anagram counting dictionary - a statistics, basically
* tell the fact like this:
* For example, test case string abba,
* substring ab -> hashed key a1b1, value is 2, because there are
* two substrings: "ab","ba" having hashed key: "a1b1"
* Here are all possible hashed keys:
* a1   - a, a,
* b1   - b, b
* a1b1 - ab, ba
* b2   - bb
* a1b2 - abb, bba
* a2b2 - abba
*
* Time complexity is O(N^2), not O(N^3).
*/
public static Dictionary<string, int> ConstructHashedAnagramsDictionary(string input)
{
var hashedAnagramsDictionary = new Dictionary<string, int>();

var length = input.Length;

// frequency table memo is using time O(N * N)
var fequencyTableMemo = PrepareFequencyTableForOnlyNSubstring(input);

for (int start = 0; start < length; start++)
{
for (int substringLength = 1; start + substringLength <= length; substringLength++)
{
var frequencyData = CalculateSubstringFequencyDiff(fequencyTableMemo, start, substringLength);

var key = HashedAnagramString.GetHashedAnagram(frequencyData);

// At most there are O(N*N) entry in the dictionary, go over once
if (hashedAnagramsDictionary.ContainsKey(key))
{
hashedAnagramsDictionary[key]++;
}
else
{
}
}
}

return hashedAnagramsDictionary;
}

/*
*
* @start - substring' start position in the original string
* @length - substring's length
* Just a simple minus calculation for each alphabet number.
*/
public static int[] CalculateSubstringFequencyDiff(int[][] fequencyTableMemo, int start, int length)
{
const int size = 26;
int[] difference = new int[size];

for (int i = 0; i < size; i++)
{
difference[i] = fequencyTableMemo[start + length - 1][i];
if (start > 0)
{
difference[i] -= fequencyTableMemo[start - 1][i];
}
}

return difference;
}

/*
* The formula to calculate pairs
* For example, test case string abba,
* substring ab -> hashed key a1b1, value is 2, because there are two substrings: "ab","ba" having hashed key: "a1b1"
* Here are all possible hashed keys:
* a1   - a, a,
* b1   - b, b
* a1b1 - ab, ba
* b2   - bb
* a1b2 - abb, bba
* a2b2 - abba
* So, how many pairs?
* should be 4, from 4 hashed keys: a1, b1, a1b1 and a2b2
*/
public static int CalculatePairs(Dictionary<string, int> hashedAnagrams)
{
// get pairs
int anagrammaticPairs = 0;

foreach (var count in hashedAnagrams)
{
anagrammaticPairs += count.Value * (count.Value - 1) / 2;
}

return anagrammaticPairs;
}
}