# Count of Binary Strings possible as per given conditions

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: 6
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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 for 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  `

## C#

 `// C# program for 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 `

Output:

```320
```

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

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Improved By : offbeat, rock_cool