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

• Difficulty Level : Expert
• Last Updated : 18 Nov, 2021

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:

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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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