Count subsequences in first string which are anagrams of the second string

Given two strings str1 and str2 of length n1 and n2 respectively. The problem is to count all the subsequences of str1 which are anagrams of str2.

Examples:

Input : str1 = "abacd", str2 = "abc"
Output : 2
Index of characters in the 2 subsequences are:
{0, 1, 3} = {a, b, c} = abc and
{1, 2, 3} = {b, a, c} = bac
The above two subsequences of str1
are anagrams of str2.

Input : str1 = "geeksforgeeks", str2 = "geeks"
Output : 48

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Create two arrays freq1[] and freq2[] each of size ’26’ implemented as hash tables to store the frequencies of each character of str1 and str2 respectively. Let n1 and n2 be the lengths of str1 and str2 respectively. Now implement the following algorithm:

countSubsequences(str1, str2)
for i = 0 to n1-1
freq1[str1[i] - 'a']++
for i = 0 n2-1
freq2[str2[i] - 'a']++

Initialize count = 1
for i = 0 to 25
if freq2[i] != 0 then
if freq2[i] <= freq1[i] then
count = count * binomialCoeff(freq1[i], freq2[i])
else
return 0
return count

Let freq1[i] is represented as n and freq2[i] as r. Now, binomialCoeff(n, r) mentioned in the algorithm above is nothing but binomial coefficient nCr .Refer this post for its implementation.

Explanation:
Let frequency of some character say ch in ‘str2’ is ‘x’ and in ‘str1’ is ‘y’. If y < x, then no subsequence of 'str1' exists which could be an anagram of 'str2'. Otherwise yCx gives the number of occurrences of ch selected which will contribute to the total count of required subsequences.

C++

 // C++ implementation to // count subsequences in  // first string which are   // anagrams of the second // string #include using namespace std;    #define SIZE 26    // Returns value of Binomial // Coefficient C(n, k) int binomialCoeff(int n, int k) {     int res = 1;        // Since C(n, k) = C(n, n-k)     if (k > n - k)         k = n - k;        // Calculate value of     // [n * (n-1) *---* (n-k+1)] /     // [k * (k-1) *----* 1]     for (int i = 0; i < k; ++i) {         res *= (n - i);         res /= (i + 1);     }        return res; }    // function to count subsequences  // in first string which are  // anagrams of the second string int countSubsequences(string str1, string str2) {     // hash tables to store frequencies      // of each character     int freq1[SIZE], freq2[SIZE];        int n1 = str1.size();     int n2 = str2.size();        // Initialize     memset(freq1, 0, sizeof(freq1));     memset(freq2, 0, sizeof(freq2));        // store frequency of each     // character of 'str1'     for (int i = 0; i < n1; i++)         freq1[str1[i] - 'a']++;        // store frequency of each      // character of 'str2'     for (int i = 0; i < n2; i++)         freq2[str2[i] - 'a']++;        // to store the total count     // of subsequences     int count = 1;        for (int i = 0; i < SIZE; i++)            // if character (i + 'a')          // exists in 'str2'         if (freq2[i] != 0) {                    // if this character's frequency                  // in 'str2' in less than or                 // equal to its frequency in                  // 'str1' then accumulate its                  // contribution to the count                 // of subsequences. If its                  // frequency in 'str1' is 'n'                  // and in 'str2' is 'r', then                  // its contribution wil be nCr,                 //  where C is the binomial                  // coefficient.             if (freq2[i] <= freq1[i])                 count = count * binomialCoeff(freq1[i], freq2[i]);                // else return 0 as there could             // be no subsequence which is an             // anagram of 'str2'             else                 return 0;         }        // required count of subsequences     return count; }    // Driver program to test above int main() {     string str1 = "abacd";     string str2 = "abc";     cout << "Count = "         << countSubsequences(str1, str2);     return 0; }

Java

 // Java implementation to // count subsequences in  // first string which are  // anagrams of the second // string import java.util.*; import java.lang.*;    public class GfG{            public final static int SIZE = 26;            // Returns value of Binomial     // Coefficient C(n, k)     public static int binomialCoeff(int n,                                      int k)     {         int res = 1;            // Since C(n, k) = C(n, n-k)         if (k > n - k)             k = n - k;            // Calculate value of          // [n * (n-1) *---* (n-k+1)] /         // [k * (k-1) *----* 1]         for (int i = 0; i < k; ++i)          {             res *= (n - i);             res /= (i + 1);         }            return res;     }        // function to count subsequences      // in first string which are      // anagrams of the second string     public static int countSubsequences(String str,                                          String str3)     {         // hash tables to store frequencies          // of each character         int[] freq1 = new int [SIZE];          int[] freq2 = new int [SIZE];            char[] str1 = str.toCharArray();         char[] str2 = str3.toCharArray();                    int n1 = str.length();         int n2 = str3.length();            // store frequency of each         // character of 'str1'         for (int i = 0; i < n1; i++)             freq1[str1[i] - 'a']++;            // store frequency of each          // character of 'str2'         for (int i = 0; i < n2; i++)             freq2[str2[i] - 'a']++;            // to store the total count         // of subsequences         int count = 1;            for (int i = 0; i < SIZE; i++)                // if character (i + 'a')             // exists in 'str2'             if (freq2[i] != 0) {                    // if this character's frequency                  // in 'str2' in less than or                 // equal to its frequency in                  // 'str1' then accumulate its                  // contribution to the count                 // of subsequences. If its                  // frequency in 'str1' is 'n'                  // and in 'str2' is 'r', then                  // its contribution wil be nCr,                 // where C is the binomial                  // coefficient.                 if (freq2[i] <= freq1[i])                     count = count * binomialCoeff(freq1[i], freq2[i]);                    // else return 0 as there could                 // be no subsequence which is an                 // anagram of 'str2'                 else                     return 0;             }            // required count of subsequences         return count;     }            // Driver function     public static void main(String argc[])     {         String str1 = "abacd";         String str2 = "abc";                    System.out.println("Count = " +         countSubsequences(str1, str2));     } }    /* This code is contributed by Sagar Shukla */

Python

 # Python3 implementation to count # subsequences in first which are  # anagrams of the second    # import library import numpy as np    SIZE = 26    # Returns value of Binomial # Coefficient C(n, k) def binomialCoeff(n, k):            res = 1        # Since C(n, k) = C(n, n-k)     if (k > n - k):         k = n - k        # Calculate value of     # [n * (n-1) *---* (n-k+1)] /     # [k * (k-1) *----* 1]     for i in range(0, k):         res = res * (n - i)         res = int(res / (i + 1))            return res       # Function to count subsequences  # in first which are anagrams # of the second  def countSubsequences(str1, str2):            # Hash tables to store frequencies      # of each character     freq1 = np.zeros(26, dtype = np.int)      freq2 = np.zeros(26, dtype = np.int)         n1 = len(str1)     n2 = len(str2)        # Store frequency of each     # character of 'str1'     for i in range(0, n1):         freq1[ord(str1[i]) - ord('a') ] += 1        # Store frequency of each      # character of 'str2'     for i in range(0, n2):                    freq2[ord(str2[i]) - ord('a')] += 1        # To store the total count     # of subsequences     count = 1        for i in range(0, SIZE):                    # if character (i + 'a')          # exists in 'str2'         if (freq2[i] != 0):             # if this character's frequency              # in 'str2' in less than or             # equal to its frequency in              # 'str1' then accumulate its              # contribution to the count             # of subsequences. If its              # frequency in 'str1' is 'n'              # and in 'str2' is 'r', then              # its contribution wil be nCr,             # where C is the binomial              # coefficient.             if (freq2[i] <= freq1[i]):                 count = count * binomialCoeff(freq1[i], freq2[i])                # else return 0 as there could             # be no subsequence which is an             # anagram of 'str2'             else:                 return 0                # required count of subsequences     return count    # Driver code str1 = "abacd" str2 = "abc" ans = countSubsequences(str1, str2) print ("Count = ", ans)       # This code contributed by saloni1297

C#

 // C# implementation to // count subsequences in  // first string which are  // anagrams of the second // string using System;    class GfG {            public static int SIZE = 26;            // Returns value of Binomial     // Coefficient C(n, k)     public static int binomialCoeff(int n,                                      int k)     {         int res = 1;            // Since C(n, k) = C(n, n-k)         if (k > n - k)             k = n - k;            // Calculate value of          // [n * (n-1) *---* (n-k+1)] /         // [k * (k-1) *----* 1]         for (int i = 0; i < k; ++i)          {             res *= (n - i);             res /= (i + 1);         }            return res;     }        // function to count subsequences      // in first string which are      // anagrams of the second string     public static int countSubsequences(String str,                                          String str3)     {         // hash tables to store frequencies          // of each character         int[] freq1 = new int [SIZE];          int[] freq2 = new int [SIZE];            char[] str1 = str.ToCharArray();         char[] str2 = str3.ToCharArray();                    int n1 = str.Length;         int n2 = str3.Length;            // store frequency of each         // character of 'str1'         for (int i = 0; i < n1; i++)             freq1[str1[i] - 'a']++;            // store frequency of each          // character of 'str2'         for (int i = 0; i < n2; i++)             freq2[str2[i] - 'a']++;            // to store the total count         // of subsequences         int count = 1;            for (int i = 0; i < SIZE; i++)                // if character (i + 'a')             // exists in 'str2'             if (freq2[i] != 0) {                    // if this character's frequency                  // in 'str2' in less than or                 // equal to its frequency in                  // 'str1' then accumulate its                  // contribution to the count                 // of subsequences. If its                  // frequency in 'str1' is 'n'                  // and in 'str2' is 'r', then                  // its contribution wil be nCr,                 // where C is the binomial                  // coefficient.                 if (freq2[i] <= freq1[i])                     count = count * binomialCoeff(freq1[i],                                                    freq2[i]);                    // else return 0 as there could                 // be no subsequence which is an                 // anagram of 'str2'                 else                     return 0;             }            // required count of subsequences         return count;     }            // Driver code     public static void Main(String[] argc)     {         String str1 = "abacd";         String str2 = "abc";                    Console.Write("Count = " +         countSubsequences(str1, str2));     } }    // This code is contributed by parashar

PHP

 \$n - \$k)         \$k = \$n - \$k;        // Calculate value of     // [n * (n-1) *---* (n-k+1)] /     // [k * (k-1) *----* 1]     for (\$i = 0; \$i < \$k; ++\$i)     {         \$res *= (\$n - \$i);         \$res /= (\$i + 1);     }        return \$res; }    // function to count  // subsequences in  // first string which  // are anagrams of the // second string function countSubsequences(\$str1,                            \$str2) {     global \$SIZE;            // hash tables to     // store frequencies      // of each character     \$freq1 = array();     \$freq2 = array();        // Initialize     for (\$i = 0;           \$i < \$SIZE; \$i++)     {         \$freq1[\$i] = 0;         \$freq2[\$i] = 0;     }      \$n1 = strlen(\$str1);     \$n2 = strlen(\$str2);        // store frequency of each     // character of 'str1'     for (\$i = 0; \$i < \$n1; \$i++)         \$freq1[ord(\$str1[\$i]) -                 ord('a')]++;        // store frequency of each      // character of 'str2'     for (\$i = 0; \$i < \$n2; \$i++)         \$freq2[ord(\$str2[\$i]) -                 ord('a')]++;        // to store the total count     // of subsequences     \$count = 1;        for (\$i = 0; \$i < \$SIZE; \$i++)            // if character (i + 'a')          // exists in 'str2'         if (\$freq2[\$i] != 0)         {                // if this character's frequency              // in 'str2' in less than or             // equal to its frequency in              // 'str1' then accumulate its              // contribution to the count             // of subsequences. If its              // frequency in 'str1' is 'n'              // and in 'str2' is 'r', then              // its contribution wil be nCr,             // where C is the binomial              // coefficient.             if (\$freq2[\$i] <= \$freq1[\$i])                 \$count = \$count *                           binomialCoeff(\$freq1[\$i],                                         \$freq2[\$i]);                // else return 0 as there              // could be no subsequence              // which is an anagram of             // 'str2'             else                 return 0;         }        // required count      // of subsequences     return \$count; }    // Driver Code \$str1 = "abacd"; \$str2 = "abc"; echo ("Count = ".        countSubsequences(\$str1,                          \$str2));    // This code is contributed by  // Manish Shaw(manishshaw1) ?>

Output :

Count = 2

Time Complexity: O(n1 + n2) + O(max), where max is the maximum frequency.

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