# Minimum count of Full Binary Trees such that the count of leaves is N

Given an integer **N** and an infinite number of Full Binary Trees of different depths, the task is to choose minimum number of trees such that the sum of the count of leaf nodes in each of the tree is **N**.

**Example:**

Input:N = 7

Output:3

Trees with depths 2, 1 and 0 can be picked

with the number of leaf nodes as 4, 2 and 1 respectively.

(4 + 2 + 1) = 7

Input:N = 1

Output:1

**Approach:** Since the number of leaf nodes in a full binary tree is always a power of two. So, the problem now gets reduced to finding the powers of 2 which give N when added together such that the total number of terms in the summation is minimum which is the required answer.

Since every power of 2 contains only one ‘1’ in its binary representation, so **N** will contain the same number of ‘1’s as the number of terms in summation (assuming we take the minimum number of terms). So, the problem further gets reduced to finding the number of set bits in **N** which can be easily calculated using the approach used in this post.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach ` `#include <iostream> ` `using` `namespace` `std; ` ` ` `// Function to return the minimum ` `// count of trees required ` `int` `minTrees(` `int` `n) ` `{ ` ` ` ` ` `// To store the count of ` ` ` `// set bits in n ` ` ` `int` `count = 0; ` ` ` ` ` `while` `(n) { ` ` ` `n &= (n - 1); ` ` ` `count++; ` ` ` `} ` ` ` `return` `count; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `n = 7; ` ` ` ` ` `cout << minTrees(n); ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the approach ` `class` `GFG ` `{ ` ` ` `// Function to return the minimum ` `// count of trees required ` `static` `int` `minTrees(` `int` `n) ` `{ ` ` ` ` ` `// To store the count of ` ` ` `// set bits in n ` ` ` `int` `count = ` `0` `; ` ` ` ` ` `while` `(n > ` `0` `) ` ` ` `{ ` ` ` `n &= (n - ` `1` `); ` ` ` `count++; ` ` ` `} ` ` ` `return` `count; ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `int` `n = ` `7` `; ` ` ` ` ` `System.out.print(minTrees(n)); ` `} ` `} ` ` ` `// This code is contributed by 29AjayKumar ` |

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## Python3

`# Python3 implementation of the approach ` ` ` `# Function to return the minimum ` `# count of trees required ` `def` `minTrees(n): ` ` ` ` ` `# To store the count of ` ` ` `# set bits in n ` ` ` `count ` `=` `0` `; ` ` ` ` ` `while` `(n): ` ` ` `n &` `=` `(n ` `-` `1` `); ` ` ` `count ` `+` `=` `1` `; ` ` ` `return` `count; ` ` ` `# Driver code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` `n ` `=` `7` `; ` ` ` `print` `(minTrees(n)); ` ` ` `# This code is contributed by 29AjayKumar ` |

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## C#

`// C# implementation of the approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` ` ` `// Function to return the minimum ` ` ` `// count of trees required ` ` ` `static` `int` `minTrees(` `int` `n) ` ` ` `{ ` ` ` ` ` `// To store the count of ` ` ` `// set bits in n ` ` ` `int` `count = 0; ` ` ` ` ` `while` `(n > 0) ` ` ` `{ ` ` ` `n &= (n - 1); ` ` ` `count++; ` ` ` `} ` ` ` `return` `count; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `n = 7; ` ` ` ` ` `Console.Write(minTrees(n)); ` ` ` `} ` `} ` ` ` `// This code is contributed by AnkitRai01 ` |

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**Output:**

3

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