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Count of Binary Strings possible as per given conditions
• Last Updated : 17 Mar, 2021

Given two integers N and M, where N denotes the count of ‘0’ and M denotes the count of ‘1’, and an integer K, the task is to find the maximum number of binary strings that can be generated of the following two types:

• A string can consist of K0‘s and a single ‘1‘.
• A string can consist of K1‘s and a single ‘0‘.

Examples:

Input: N = 4, M = 4, K = 2
Output:
Explanation:
Count of ‘0‘s = 4
Count of ‘1‘s = 4
Possible ways to combine 0’s and 1’s under given constraints are {“001”, “001”} or {“001”, “110”} or {“110”, “110”}
Therefore, at most 2 combinations exists in a selection.
Each combination can be arranged in K + 1 ways, i.e. “001” can be rearranged to form “010, “100” as well.
Therefore, the maximum possible strings that can be generated is 3 * 2 = 6

Input: N = 101, M = 231, K = 15
Output: 320

Approach:
Follow the steps below to solve the problem:

• Consider the following three conditions to generate maximum possible combinations of binary strings:
• Number of combinations cannot exceed N.
• Number of combinations cannot exceed M.
• Number of combinations cannot exceed (A+B)/(K + 1).
• Therefore, the maximum possible combinations are min(A, B, (A + B)/ (K + 1)).
• Therefore, the maximum possible strings that can be generated are (K + 1) * min(A, B, (A + B)/ (K + 1)).

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement``// the above approach``#include ``using` `namespace` `std;` `// Function to generate maximum``// possible strings that can be generated``long` `long` `countStrings(``long` `long` `A,``                    ``long` `long` `B,``                    ``long` `long` `K)``{` `    ``long` `long` `X = (A + B) / (K + 1);` `    ``// Maximum possible strings``    ``return` `(min(A, min(B, X)) * (K + 1));``}``int` `main()``{` `    ``long` `long` `N = 101, M = 231, K = 15;``    ``cout << countStrings(N, M, K);``    ``return` `0;``}`

## Java

 `// Java program to implement``// the above approach``import` `java.io.*;``import` `java.util.*;` `class` `GFG{` `// Function to generate maximum``// possible strings that can be generated``static` `long` `countStrings(``long` `A, ``long` `B,``                        ``long` `K)``{``    ``long` `X = (A + B) / (K + ``1``);` `    ``// Maximum possible strings``    ``return` `(Math.min(A, Math.min(B, X)) *``                                ``(K + ``1``));``}` `// Driver Code``public` `static` `void` `main (String[] args)``{``    ``long` `N = ``101``, M = ``231``, K = ``15``;``    ` `    ``System.out.print(countStrings(N, M, K));``}``}` `// This code is contributed by offbeat`

## Python3

 `# Python3 program to implement ``# the above approach `` ` `# Function to generate maximum ``# possible strings that can be``# generated ``def` `countStrings(A, B, K): ``  ` `    ``X ``=` `(A ``+` `B) ``/``/` `(K ``+` `1``) ``  ` `    ``# Maximum possible strings ``    ``return` `(``min``(A, ``min``(B, X)) ``*` `(K ``+` `1``)) ` `# Driver code``N, M, K ``=` `101``, ``231``, ``15` `print``(countStrings(N, M, K))` `# This code is contributed divyeshrabadiya07`

## C#

 `// C# program to implement``// the above approach``using` `System;` `class` `GFG{` `// Function to generate maximum``// possible strings that can be generated``static` `long` `countStrings(``long` `A, ``long` `B,``                        ``long` `K)``{``    ``long` `X = (A + B) / (K + 1);` `    ``// Maximum possible strings``    ``return` `(Math.Min(A, Math.Min(B, X)) *``                                ``(K + 1));``}` `// Driver Code``public` `static` `void` `Main (``string``[] args)``{``    ``long` `N = 101, M = 231, K = 15;``    ` `    ``Console.Write(countStrings(N, M, K));``}``}` `// This code is contributed by rock_cool`

## Javascript

 ``

Output:

`320`

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

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