# 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
**K**‘**0**‘s and a single ‘**1**‘. - A string can consist of
**K**‘**1**‘s and a single ‘**0**‘.

**Examples:**

Input:N = 4, M = 4, K = 2Output:6Explanation:

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 inK + 1ways, 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 = 15Output: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)**.

- Number of combinations cannot exceed
- 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 <bits/stdc++.h>` `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

`<script>` `// JavaScript Program to implement` `// the above approach` `// Function to generate maximum` `// possible strings that can be generated` `function` `countStrings(A, B, K)` `{` ` ` `let X = Math.floor((A + B) / (K + 1));` ` ` `// Maximum possible strings` ` ` `return` `(Math.min(A, Math.min(B, X)) * (K + 1));` `}` ` ` `let N = 101, M = 231, K = 15;` ` ` `document.write(countStrings(N, M, K));` `// This code is contributed by Surbhi Tyagi.` `</script>` |

**Output:**

320

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