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Count subsequences 01 in string generated by concatenation of given numeric string K times

Given a string S and a positive integer K, the task is to find the number of subsequences “01” in the string generated by concatenation of the given numeric string S K times.

Examples:

Input: S = “0171”, K = 2
Output: 6
Explanation:
The string formed by concatenation of S, K number of times is “01710171”. There are total 6 possible subsequences which are marked as bold = {“01710171″, “01710171″, “01710171″, “01710171“, “01710171″, “01710171“}.

Input: S = “230013110087”, K = 2
Output: 24   

Naive Approach: The simplest approach to solve the given problem is to generate the resultant string by concatenating S, K number of times and then find all possible pairs (i, j) from the string such that (i < j) and S[i] = 0 and S[j] = 1.

Time Complexity: O((N*K)2)
Auxiliary Space: O(N*K)

Efficient Approach: The task can also be optimized by observing the following 2 Cases:

  • Case 1: Substring “01” strictly inside each occurrence of S in P. Let suppose C be the count of occurrences of “01” in S, then in P it would be C*K.
  • Case 2: When ‘0‘ lies inside at ith occurrence of S and ‘1‘ lies inside some jth occurrence to form a subsequence “01” such that i < j, then finding the number of occurrences of “01” will be the same as choosing the two strings or occurrence of strings in P given by ((K)*(K – 1))/2. Let that value be Si and Sj and multiplying it by the number of occurrences of ‘0’ in Si(denoted by cnt0) and a number of occurrences of ‘1’ in Sj(denoted by cnt1) gives the number of subsequences of “01”.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate the number of
// subsequences of "01"
int countSubsequence(string S, int N,
                     int K)
{
    // Store count of 0's and 1's
    int C = 0, C1 = 0, C0 = 0;
 
    for (int i = 0; i < N; i++) {
        if (S[i] == '1')
            C1++;
        else if (S[i] == '0')
            C0++;
    }
 
    // Count of subsequences without
    // concatenation
    int B1 = 0;
    for (int i = 0; i < N; i++) {
        if (S[i] == '1')
            B1++;
        else if (S[i] == '0')
            C = C + (C1 - B1);
    }
 
    // Case 1
    int ans = C * K;
 
    // Case 2
    ans += (C1 * C0 * (((K) * (K - 1)) / 2));
 
    // Return the total count
    return ans;
}
 
// Driver Code
int main()
{
    string S = "230013110087";
    int K = 2;
    int N = S.length();
 
    cout << countSubsequence(S, N, K);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG {
 
    // Function to calculate the number of
    // subsequences of "01"
    static int countSubsequence(String S, int N, int K)
    {
        // Store count of 0's and 1's
        int C = 0, C1 = 0, C0 = 0;
 
        for (int i = 0; i < N; i++) {
            if (S.charAt(i) == '1')
                C1++;
            else if (S.charAt(i) == '0')
                C0++;
        }
 
        // Count of subsequences without
        // concatenation
        int B1 = 0;
        for (int i = 0; i < N; i++) {
            if (S.charAt(i) == '1')
                B1++;
            else if (S.charAt(i) == '0')
                C = C + (C1 - B1);
        }
 
        // Case 1
        int ans = C * K;
 
        // Case 2
        ans += (C1 * C0 * (((K) * (K - 1)) / 2));
 
        // Return the total count
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String S = "230013110087";
        int K = 2;
        int N = S.length();
 
        System.out.println(countSubsequence(S, N, K));
    }
}
 
// This code  is contributed by Potta Lokesh

Python3

# python program for the above approach
 
 
# Function to calculate the number of
# subsequences of "01"
def countSubsequence(S, N, K):
 
        # Store count of 0's and 1's
    C = 0
    C1 = 0
    C0 = 0
 
    for i in range(0, N):
 
        if (S[i] == '1'):
            C1 += 1
        elif (S[i] == '0'):
            C0 += 1
 
        # Count of subsequences without
        # concatenation
    B1 = 0
 
    for i in range(0, N):
        if (S[i] == '1'):
            B1 += 1
        elif (S[i] == '0'):
            C = C + (C1 - B1)
 
        # Case 1
    ans = C * K
 
    # Case 2
 
    ans += (C1 * C0 * (((K) * (K - 1)) // 2))
 
    # Return the total count
    return ans
 
 
# Driver Code
if __name__ == "__main__":
 
    S = "230013110087"
    K = 2
    N = len(S)
 
    print(countSubsequence(S, N, K))
 
    # This code is contributed by rakeshsahni

C#

// C# implementation for the above approach
using System;
class GFG
{
 
    // Function to calculate the number of
    // subsequences of "01"
    static int countSubsequence(string S, int N, int K)
    {
       
        // Store count of 0's and 1's
        int C = 0, C1 = 0, C0 = 0;
 
        for (int i = 0; i < N; i++) {
            if (S[i] == '1')
                C1++;
            else if (S[i] == '0')
                C0++;
        }
 
        // Count of subsequences without
        // concatenation
        int B1 = 0;
        for (int i = 0; i < N; i++) {
            if (S[i] == '1')
                B1++;
            else if (S[i] == '0')
                C = C + (C1 - B1);
        }
 
        // Case 1
        int ans = C * K;
 
        // Case 2
        ans += (C1 * C0 * (((K) * (K - 1)) / 2));
 
        // Return the total count
        return ans;
    }
 
    // Driver Code
    public static void Main()
    {
        string S = "230013110087";
        int K = 2;
        int N = S.Length;
 
        Console.Write(countSubsequence(S, N, K));
    }
}
 
// This code is contributed by sanjoy_62.

Javascript

<script>
// Javascript program for the above approach
 
// Function to calculate the number of
// subsequences of "01"
function countSubsequence(S, N, K) {
  // Store count of 0's and 1's
  let C = 0,
    C1 = 0,
    C0 = 0;
 
  for (let i = 0; i < N; i++) {
    if (S[i] == "1") C1++;
    else if (S[i] == "0") C0++;
  }
 
  // Count of subsequences without
  // concatenation
  let B1 = 0;
  for (let i = 0; i < N; i++) {
    if (S[i] == "1") B1++;
    else if (S[i] == "0") C = C + (C1 - B1);
  }
 
  // Case 1
  let ans = C * K;
 
  // Case 2
  ans += C1 * C0 * ((K * (K - 1)) / 2);
 
  // Return the total count
  return ans;
}
 
// Driver Code
 
let S = "230013110087";
let K = 2;
let N = S.length;
 
document.write(countSubsequence(S, N, K));
 
// This code is contributed by gfgking.
</script>

Output: 

24

 

Time Complexity: O(N)
Auxiliary Space: O(1)


 


Reffered: https://www.geeksforgeeks.org

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Type:
 Geek
Category:
 Coding
Sub Category:
 Tutorial
Uploaded by:
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