Number of subsequences as “ab” in a string repeated K times

Given a String S, consider a new string formed by repeating the S exactly K times. We need find the number of subsequences as “ab” in the newly formed string.

Examples :

Input : S = "abcb"
        K = 2
Output : 6
Here, Given string is repeated 2 times and
we get a new string "abcbabcb"
Below are 6 occurrences of "ab"
abcbabcb
abcbabcb
abcbabcb
abcbabcb
abcbabcb
abcbabcb

Input : S = "aacbd"
        K = 1
Output : 2

Naive Approach: Finding no.of subsequences of “ab” is in fact finding a pair s[i], s[j] (i < j) where s[i] = ‘a’, s[j] = ‘b’.
We can do this by using two nested for loops and count the no. of pairs.
We can improve this approach in a single traversal of the string. Let us consider an index j, s[j] =’b’, if we find no.of index i‘s such that s[i] = ‘a’ and i < j, then it is same as no.of subsequences of "ab" till j. This can be done by maintaining count of a’s by traversing the array and add the count to our answer at position where s[i] =’b .
Time Complexity:O(|S|*K)



Efficient Approach:
Let T be the newly formed string
T = s1 + s2 + s3 + ….. + sk;
where si is the ith occurrence of the string s.
Here, occurrence of “ab” in T are as follows:
1)”ab” lies completely in the some of occurrence of string S, so we can simply find occurrences of “ab” in Si.Let it be C. So, total no.of occurrences of “ab” in T will be C*K.

2) Otherwise, ”a” lies strictly inside some string Si and “b” lies inside some other string Sj, (i < j). In this way finding no.of occurrences of "ab" will be choosing two occurrences of string S from K occurrences(KC2) and multiplying it with no. of occurrences of “a” in Si and no.of occurrences of “b” in Sj.
As, Si = Sj = S.
Time Complexity: O(|S|), for counting no.of “a”s and no.of “b”s.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP code to find number of subsequences of
// "ab" in the string S which is repeated K times.
#include <bits/stdc++.h>
using namespace std;
  
int countOccurrences(string s, int K)
{
    int n = s.length();
    int C, c1 = 0, c2 = 0;
    for (int i = 0; i < n; i++) {
        if (s[i] == 'a')
            c1++; // Count of 'a's
        if (s[i] == 'b') {
            c2++; // Count of 'b's
            C += c1; // occurrence of "ab"s in string S
        }
    }
  
    // Add following two :
    // 1) K * (Occurrences of "ab" in single string)
    // 2) a is from one string and b is from other.
    return C * K + (K * (K - 1) / 2) * c1 * c2;
}
  
// Driver code
int main()
{
    string S = "abcb";
    int k = 2;
    cout << countOccurrences(S, k) << endl;
    return 0;
}

chevron_right






                        filter_none









Java














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// Java code to find number of subsequences of 


// "ab" in the string S which is repeated K times. 


  


import java.io.*; 


  


class GFG { 


  


    static int countOccurrences(String s, int K) 


    { 


        int n = s.length(); 


        int C = 0, c1 = 0, c2 = 0; 


        for (int i = 0; i < n; i++) { 


            if (s.charAt(i) == 'a') 


                c1++; // Count of 'a's 


            if (s.charAt(i) == 'b') { 


                c2++; // Count of 'b's 


  


                // occurrence of "ab"s 


                // in string S 


                C += c1; 


            } 


        } 


  


        // Add following two : 


        // 1) K * (Occurrences of "ab" in single string) 


        // 2) a is from one string and b is from other. 


        return C * K + (K * (K - 1) / 2) * c1 * c2; 


    } 


  


    // Driver code 


    public static void main(String[] args) 


    { 


        String S = "abcb"; 


        int k = 2; 


  


        System.out.println(countOccurrences(S, k)); 


    } 


} 


  


// This code is contributed by vt_m. 



















                        chevron_right







                        filter_none









Python3














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# Python3 code to find number of  


# subsequences of "ab" in the  


# string S which is repeated K times. 


  


def countOccurrences (s, K): 


    n = len(s) 


    c1 = 0


    c2 = 0


    C = 0


    for i in range(n): 


        if s[i] == 'a': 


            c1+= 1 # Count of 'a's 


        if s[i] == 'b': 


            c2+= 1 # Count of 'b's 


              


            # occurrence of "ab"s in string S 


            C += c1  


              


    # Add following two : 


    # 1) K * (Occurrences of "ab" in single string) 


    # 2) a is from one string and b is from other. 


    return C * K + (K * (K - 1) / 2) * c1 * c2 


      


      


# Driver code 


S = "abcb"


k = 2


print (countOccurrences(S, k)) 


  


# This code is contributed by "Abhishek Sharma 44" 



















                        chevron_right







                        filter_none









C#














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// C# code to find number of subsequences 


// of "ab" in the string S which is 


// repeated K times. 


using System; 


  


class GFG { 


  


    static int countOccurrences(string s, int K) 


    { 


  


        int n = s.Length; 


        int C = 0, c1 = 0, c2 = 0; 


  


        for (int i = 0; i < n; i++) { 


  


            if (s[i] == 'a') 


  


                // Count of 'a's 


                c1++; 


            if (s[i] == 'b') { 


  


                // Count of 'b's 


                c2++; 


  


                // occurrence of "ab"s 


                // in string S 


                C += c1; 


            } 


        } 


  


        // Add following two : 


        // 1) K * (Occurrences of "ab" in 


        // single string) 


        // 2) a is from one string and b 


        // is from other. 


        return C * K + (K * (K - 1) / 2) * c1 * c2; 


    } 


  


    // Driver code 


    public static void Main() 


    { 


  


        string S = "abcb"; 


        int k = 2; 


  


        Console.WriteLine( 


            countOccurrences(S, k)); 


    } 


} 


  


// This code is contributed by vt_m. 



















                        chevron_right







                        filter_none









PHP














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















<?php 


// PHP code to find number of  


// subsequences of "ab" in the  


// string S which is repeated K times. 


  


function countOccurrences($s, $K) 


{ 


    $n = strlen($s); 


    $C = 0; $c1 = 0; $c2 = 0; 


    for ($i = 0; $i < $n; $i++)  


    { 


        if ($s[$i] == 'a') 


            // Count of 'a's 


            $c1++;  


        if ($s[$i] == 'b')  


        { 


            // Count of 'b's 


            $c2++;  


              


            // occurrence of "ab"s 


            // in string S 


            $C = $C+ $c1;  


        } 


    } 


  


    // Add following two : 


    // 1) K * (Occurrences of "ab" 


    //    in single string) 


    // 2) a is from one string and 


    //    b is from other. 


    return $C * $K + ($K * ($K - 1) / 2) *  


                                $c1 * $c2; 


} 


  


// Driver code 


$S = "abcb"; 


$k = 2; 


echo countOccurrences($S, $k), "\n"; 


  


// This code is contributed by ajit. 


?> 



















                        chevron_right







                        filter_none










Output :


 6 



Time complexity:O(|S|).



          
          
          
          

            

    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:
Harsha_Mogali
Check out this Author's contributed articles.
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on  the "Improve Article" button below.


Improved By :  vt_m, jit_t, Shishir_Kumar