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

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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 . 

Implementation:-

C++

// 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)
{
      //to store answer
      int ans=0;
   
      //Count of a
      int c=0;
   
      string temp="";
   
      //repeating s K times
      for(int i=0;i<K;i++)
    {
          temp+=s;
    }
   
      //iterating over temp
      for(int i=0;i<temp.length();i++)
    {
          if(temp[i]=='a')c++;
       
          //if found b then ab can be formed c times for that b
          //because c times a is present before b
          else if(temp[i]=='b')ans+=c;
    }
      return ans;
       
}
 
// Driver code
int main()
{
    string S = "abcb";
    int k = 2;
    cout << countOccurrences(S, k) << endl;
    return 0;
}
//Code contributed by shubhamrajput6156

                    

Java

// Java code to find number of subsequences of
// "ab" in the string S which is repeated K times.
 
import java.io.*;
 
class GFG {
     
    public static int countOccurrences(String s, int K)
    {
          //to store answer
          int ans = 0;
 
          //Count of a
          int c = 0;
 
          String temp = "";
 
          //repeating s K times
          for (int i = 0;i < K;i++)
          {
              temp += s;
          }
 
          //iterating over temp
          for (int i = 0;i < temp.length();i++)
          {
              if (temp.charAt(i) == 'a')
              {
                  c++;
              }
 
              //if found b then ab can be formed c times for that b
              //because c times a is present before b
              else if (temp.charAt(i) == 'b')
              {
                  ans += c;
              }
          }
          return ans;
 
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String S = "abcb";
        int k = 2;
        System.out.print(countOccurrences(S, k));
        System.out.print("\n");
    }
}
 
//Code contributed by bhardwajji

                    

Python3

# Python code to find number of subsequences of
# "ab" in the string S which is repeated K times.
 
def countOccurrences(s, K):
   
      # to store answer
    ans = 0
     
    # Count of a
    c = 0
    temp = ""
     
    # repeating s K times
    for i in range(K):
        temp += s
         
    # iterating over temp
    for i in range(len(temp)):
        if temp[i] == 'a':
            c += 1
             
        # if found b then ab can be formed c times for that b
        # because c times a is present before b
        elif temp[i] == 'b':
            ans += c
    return ans
 
# driver code
S = "abcb"
k = 2
print(countOccurrences(S, k))

                    

C#

// C# code to find number of subsequences of
// "ab" in the string S which is repeated K times.
using System;
 
class MainClass {
   
  static int countOccurrences(string s, int K) {
       
    // to store answer
    int ans = 0;
     
    // Count of a
    int c = 0;
    string temp = "";
     
    // repeating s K times
    for (int i = 0; i < K; i++) {
        temp += s;
    }
     
    // iterating over temp
    for (int i = 0; i < temp.Length; i++) {
        if (temp[i] == 'a') {
            c += 1;
        }
         
        // if found b then ab can be formed c times for that b
        // because c times a is present before b
        else if (temp[i] == 'b') {
            ans += c;
        }
    }
    return ans;
  }
 
  public static void Main (string[] args) {
    // driver code
    string S = "abcb";
    int k = 2;
    Console.WriteLine(countOccurrences(S, k));
  }
}
 
//This code is contributed by shivamsharma215

                    

Javascript

// Javascript code to find number of subsequences of
// "ab" in the string S which is repeated K times.
function countOccurrences(s, K) {
  //to store answer
  let ans = 0;
   
  //Count of a
  let c = 0;
  let temp = "";
   
  //repeating s K times
  for (let i = 0; i < K; i++) {
    temp += s;
  }
   
  //iterating over temp
  for (let i = 0; i < temp.length; i++) {
    if (temp[i] == 'a') {
      c++;
    }
     //if found b then ab can be formed c times for that b
    //because c times a is present before b
    else if (temp[i] == 'b') {
      ans += c;
    }
  }
  return ans;
}
 
// Driver code
const S = "abcb";
const k = 2;
console.log(countOccurrences(S, k));

                    

Output
6

Time Complexity:O(|S|*K)
Auxiliary Space :- 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++

// 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;
}

                    

Java

// 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.

                    

Python3

# 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"

                    

C#

// 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.

                    

PHP

<?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.
?>

                    

Javascript

<script>
 
// JavaScript code to find number of subsequences of
// "ab" in the string S which is repeated K times.
function countOccurrences(s, K)
{
    let n = s.length;
    let C = 0, c1 = 0, c2 = 0;
     
    for(let i = 0; i < n; i++)
    {
        if (s[i] == 'a')
            c1++; // Count of 'a's
        if (s[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
let S = "abcb";
let k = 2;
 
document.write(countOccurrences(S, k));
 
// This code is contributed by susmitakundugoaldanga
 
</script>

                    

Output
6

Time complexity:O(|S|).
Auxiliary Space: O(|S|)), where n is the length of the string. This is because when string is passed to any function it is passed by value and creates a copy of itself in stack.  



Last Updated : 23 Mar, 2023
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