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Count Occurrences of Anagrams

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  • Difficulty Level : Medium
  • Last Updated : 25 Oct, 2022
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Given a word and a text, return the count of the occurrences of anagrams of the word in the text(For eg: anagrams of word for are for, ofr, rof etc.))

Examples: 

Input : forxxorfxdofr
            for
Output : 3
Explanation : Anagrams of the word for – for, orf, ofr appear in the text and hence the count is 3.

Input : aabaabaa
            aaba
Output : 4
Explanation : Anagrams of the word aaba – aaba, abaa each appear twice in the text and hence the count is 4.

A simple approach is to traverse from start of the string considering substrings of length equal to the length of the given word and then check if this substring has all the characters of word.

Implementation:

C++




// A Simple C++ program to count anagrams of a
// pattern in a text.
#include<bits/stdc++.h>
using namespace std;
 
// Function to find if two strings are equal
bool areAnagram(string s1, string s2)
{
    map<char, int> m;
    for(int i = 0; i < s1.length(); i++)
        m[s1[i]]++;
         
    for(int i = 0; i < s2.length(); i++)
        m[s2[i]]--;
         
    for(auto it = m.begin(); it != m.end(); it++)
        if(it -> second != 0)
            return false;
             
        return true;
}
 
int countAnagrams(string text, string word)
{
     
    // Initialize result
    int res = 0;
    for(int i = 0;
            i < text.length() - word.length() + 1;
            i++)
    {
         
        // Check if the word and substring are
        // anagram of each other.
        if (areAnagram(text.substr(i, word.length()),
                                      word))
            res++;
    }
    return res;
}
 
// Driver Code
int main()
{
    string text = "forxxorfxdofr";
    string word = "for";
     
    cout << countAnagrams(text, word);
     
    return 0;
}
 
// This code is contributed by probinsah


Java




// A Simple Java program to count anagrams of a
// pattern in a text.
import java.io.*;
import java.util.*;
 
public class GFG {
 
    // Function to find if two strings are equal
    static boolean araAnagram(String s1,
                              String s2)
    {
        // converting strings to char arrays
        char[] ch1 = s1.toCharArray();
        char[] ch2 = s2.toCharArray();
 
        // sorting both char arrays
        Arrays.sort(ch1);
        Arrays.sort(ch2);
 
        // Check for equality of strings
        if (Arrays.equals(ch1, ch2))
            return true;
        else
            return false;
    }
 
    static int countAnagrams(String text, String word)
    {
        int N = text.length();
        int n = word.length();
 
        // Initialize result
        int res = 0;
 
        for (int i = 0; i <= N - n; i++) {
 
            String s = text.substring(i, i + n);
 
            // Check if the word and substring are
            // anagram of each other.
            if (araAnagram(word, s))
                res++;
        }
     
        return res;
    }
 
    // Driver code
    public static void main(String args[])
    {
        String text = "forxxorfxdofr";
        String word = "for";
        System.out.print(countAnagrams(text, word));
    }
}


Python3




# A Simple Python program to count anagrams of a
# pattern in a text.
 
# Function to find if two strings are equal
def areAnagram(s1, s2):
    m = {}
    for i in range(len(s1)):
        if s1[i] not in m:
            m[s1[i]] = 1
        else:
            cnt = m[s1[i]]
            m.pop(s1[i])
            m[s1[i]] = cnt + 1
 
    for j in range(len(s2)):
        if((s2[j] in m) == False):
            return False
        else:
            cnt = m[s2[j]]
            m.pop(s2[j])
            m[s2[j]] = cnt - 1
         
    for it in m.values():
       if(it != 0):
          return False
    return True
 
def countAnagrams(text, word):
     
    # Initialize result
    res = 0
    for i in range(len(text)- len(word)+1):
         
        # Check if the word and substring are
        # anagram of each other.
 
        if (areAnagram(text[i:i+len(word)], word)):
            res += 1
    return res
 
# Driver Code
 
text = "forxxorfxdofr"
word = "for"
     
print(countAnagrams(text, word))
 
# This code is contributed by shinjanpatra


Javascript




<script>// A Simple JavaScript program to count anagrams of a
// pattern in a text.
 
// Function to find if two strings are equal
function areAnagram(s1, s2)
{
    let m = new Map();
    for(let i = 0; i < s1.length ; i++)
    {
        if(m.has(s1[i])===false){
            m.set(s1[i],1)
        }
        else{
            let cnt = m.get(s1[i]);
            m.delete(s1[i]);
            m.set(s1[i],cnt+1);
        }
    }
 
    for(let j = 0; j < s1.length; j++)
    {
        if(m.has(s2[j])===false){
            return false;
        }
        else{
            let cnt = m.get(s2[j]);
            m.delete(s2[j]);
            m.set(s2[j],cnt-1);
        }
    }
         
    for(const it in m.values()){
       if(it !== 0)return false
    }
    return true;
}
 
function countAnagrams(text, word)
{
     
    // Initialize result
    let res = 0;
    for(let i = 0;i < text.length - word.length + 1;i++)
    {
         
        // Check if the word and substring are
        // anagram of each other.
 
        if (areAnagram(text.substring(i, i+word.length), word)) res++;
    }
    return res;
}
 
// Driver Code
 
let text = "forxxorfxdofr";
let word = "for";
     
document.write(countAnagrams(text, word));
 
// This code is contributed by shinjanpatra
</script>


Output

3

Complexity Analysis:

  • Time Complexity: O(l1logl1 + l2logl2)
  • Auxiliary space: O(l1 + l2). 

An Efficient Solution is to use a count array to check for anagrams, we can construct the current count window from the previous window in O(1) time using the sliding window concept. 

Implementation:

C++




#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
    static int findAnagrams(const std::string& text,
                            const std::string& word)
    {
        int text_length = text.length();
        int word_length = word.length();
        if (text_length < 0 || word_length < 0
            || text_length < word_length)
            return 0;
 
        constexpr int CHARACTERS = 256;
        int count = 0;
        int index = 0;
        std::array<char, CHARACTERS> wordArr;
        wordArr.fill(0);
        std::array<char, CHARACTERS> textArr;
        textArr.fill(0);
 
        // till window size
        for (; index < word_length; index++) {
            wordArr[CHARACTERS - word[index]]++;
            textArr[CHARACTERS - text[index]]++;
        }
        if (wordArr == textArr)
            count += 1;
        // next window
        for (; index < text_length; index++) {
            textArr[CHARACTERS - text[index]]++;
            textArr[CHARACTERS
                    - text[index - word_length]]--;
 
            if (wordArr == textArr)
                count += 1;
        }
        return count;
    }
};
 
int main()
{
    const std::string& text = "forxxorfxdofr";
    const std::string& word = "for";
 
    cout << Solution::findAnagrams(text, word);
    return 0;
}


Java




// An efficient Java program to count anagrams of a
// pattern in a text.
import java.io.*;
import java.util.*;
 
class Solution {
    public static int countAnagrams(String s, String p)
    {
        // change CHARACTERS to support range of supported
        // characters
        int CHARACTERS = 256;
        int sn = s.length();
        int pn = p.length();
        int count = 0;
        if (sn < 0 || pn < 0 || sn < pn)
            return 0;
 
        char[] pArr = new char[CHARACTERS];
        char[] sArr = new char[CHARACTERS];
        int i = 0;
        // till window size
        for (; i < pn; i++) {
            sArr[CHARACTERS - s.charAt(i)]++;
            pArr[CHARACTERS - p.charAt(i)]++;
        }
        if (Arrays.equals(pArr, sArr))
            count += 1;
        // next window
        for (; i < sn; i++) {
            sArr[CHARACTERS - s.charAt(i)]++;
            sArr[CHARACTERS - s.charAt(i - pn)]--;
 
            if (Arrays.equals(pArr, sArr))
                count += 1;
        }
        return count;
    }
    // Driver code
    public static void main(String args[])
    {
        String text = "forxxorfxdofr";
        String word = "for";
        System.out.print(countAnagrams(text, word));
    }
}


Python3




# A Simple Python program to count anagrams of a
# pattern in a text with the help of sliding window problem
string = "forxxorfxdofr"
ptr = "for"
n = len(string)
k = len(ptr)
temp = []
d = {}
for i in ptr:
    if i in d:
        d[i] += 1
    else:
        d[i] = 1
i = 0
j = 0
count = len(d)
 
ans = 0
 
 
while j < n:
    if string[j] in d:
        d[string[j]] -= 1
        if d[string[j]] == 0:
            count -= 1
    if (j-i+1) < k:
        j += 1
    elif (j-i+1) == k:
        if count == 0:
            ans += 1
 
        if string[i] in d:
            d[string[i]] += 1
            if d[string[i]] == 1:
                count += 1
 
        i += 1
        j += 1
print(ans)


Javascript




<script>
 
// A Simple JavaScript program to count anagrams of a
// pattern in a text with the help of sliding window problem
 
let string = "forxxorfxdofr"
let ptr = "for"
let n = string.length
let k = ptr.length
let temp = []
let d = new Map();
for(let i = 0; i < ptr.length; i++)
{
    if(d.has(ptr[i]))
        d.set(ptr[i], d.get(ptr[i])+1)
    else
        d.set(ptr[i], 1)
}
 
let i = 0
let j = 0
let count = d.size
 
let ans = 0
 
while(j < n)
{
    if(d.has(string[j]))
    {
        d.set(string[j], d.get(string[j]) - 1)
        if(d.get(string[j]) == 0)
            count--
    }
    if(j - i + 1 < k)
        j++
    else if(j - i + 1 == k){
        if(count == 0)
            ans++
 
        if(d.has(string[i]))
        {
            d.set(string[i],d.get(string[i])+1)
            if(d.get(string[i]) == 1)
                count++
        }
        i++
        j++
    }
}
 
document.write(ans)
 
// This code is contributed by shinjanpatra
 
</script>


Output

3

Complexity Analysis:

  • Time Complexity: O(l1 + l2)
  • Auxiliary space: O(l1 + l2). 

Please suggest if someone has a better solution that is more efficient in terms of space and time.
This article is contributed by Aarti_Rathi.


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