# 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 ` `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

 ``

Output:

`24`

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

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