# Minimize number of cuts required to break N length stick into N unit length sticks

Given an integer **N** denoting the length of a given stick, the task is to minimize the time required to split the stick into pieces of unit length, given that, a single cut is possible for any portion of stick at any instance of time.

**Examples:**

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend **live classes **with experts, please refer **DSA Live Classes for Working Professionals **and **Competitive Programming Live for Students**.

Input:N = 100Output:7Explanation:

(100 units) —> (2 portions of 50 units) —> (4 portions of 25 units) —> (4 portions of 12 units, 4 portions of 13 units) —> (12 portions of 6 units, 4 portions of 7 units) —> (28 portions of 3 units, 4 portions of 4 units) —> (28 portions of 1 unit, 36 portions of 2 units) —> (100 portions of 1 unit)

Input:N = 65Output:7Explanation:

(65 units) —> (1 portion of 32 units, 1 portion of 33 units) —> (3 portions of 16 units, 1 portion of 17 units) —> (7 portions of 8 units, 1 portions of 9 units) —> (15 portions of 4 units, 1 portion of 5 units) —> (31 portions of 2 units, 1 portions of 3 units) —> (63 portions of 1 unit, 1 portion of 2 units) —> (65 portions of 1 unit)

**Approach:**

Since we can cut any portion of the stick once at a particular instance of time, we need to maximize the portions after every cut. So, we will cut the stick into two parts of the longest possible length with the first cut. In the next instance, further cut both the obtained parts into two longest parts each in the next cut. Repeat this until **N** unit pieces are obtained.

Illustration:

N = 100

1st Cut: (50) + (50)

2nd Cut: (25) + (25) + (25) + (25)

3rd Cut: (12) + (13) + (12) + (13) + (12) + (13) + (12) + (13)

4th Cut: (6) + (6) + (6) + (7) + (6) + (6) + (6) + (7) + (6) + (6) + (6) + (7) + (6) + (6) + (6) + (7)

5th Cut: (3) + (3) + (3) + (3) + (3) + (3) + (3) + (4) + (3) + (3) + (3) + (3) + (3) + (3) + (3) + (4) + (3) + (3) + (3) + (3) + (3) + (3) + (3) + (4) + (3) + (3) + (3) + (3) + (3) + (3) + (3) + (4)

6th Cut: 28 portions of 1 unit, 36 portions of 2 units

7th Cut: 100 portions of 1 unit

Hence, the minimum time required to split an N length stick into 1 unit pieces is **ceil(log _{2}N)**.

Below is the implementation of the above approach:

## C++

`// C++ program to find minimum` `// time required to split a` `// stick of N length into` `// unit pieces` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to return the` `// minimum time required` `// to split stick of N into` `// length into unit pieces` `int` `min_time_to_cut(` `int` `N)` `{` ` ` `if` `(N == 0)` ` ` `return` `0;` ` ` `// Return the minimum` ` ` `// unit of time required` ` ` `return` `ceil` `(log2(N));` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 100;` ` ` `cout << min_time_to_cut(N);` ` ` `return` `0;` `}` |

## Java

`// Java program to find minimum` `// time required to split a` `// stick of N length into` `// unit pieces` `import` `java.lang.*;` `class` `GFG{` ` ` `// Function to return the` `// minimum time required` `// to split stick of N into` `// length into unit pieces` `static` `int` `min_time_to_cut(` `int` `N)` `{` ` ` `if` `(N == ` `0` `)` ` ` `return` `0` `;` ` ` ` ` `// Return the minimum` ` ` `// unit of time required` ` ` `return` `(` `int` `)Math.ceil(Math.log(N) /` ` ` `Math.log(` `2` `));` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `N = ` `100` `;` ` ` `System.out.print(min_time_to_cut(N));` `}` `}` `// This code is contributed by rock_cool` |

## Python3

`# Python3 program to find minimum` `# time required to split a stick` `# of N length into unit pieces` `import` `math` `# Function to return the` `# minimum time required` `# to split stick of N into` `# length into unit pieces` `def` `min_time_to_cut(N):` ` ` ` ` `if` `(N ` `=` `=` `0` `):` ` ` `return` `0` ` ` ` ` `# Return the minimum` ` ` `# unit of time required` ` ` `return` `int` `(math.log2(N)) ` `+` `1` `# Driver Code` `N ` `=` `100` `print` `(min_time_to_cut(N))` `# This code is contributed by Vishal Maurya` |

## C#

`// C# program to find minimum` `// time required to split a` `// stick of N length into` `// unit pieces` `using` `System;` `class` `GFG{` ` ` `// Function to return the` `// minimum time required` `// to split stick of N into` `// length into unit pieces` `static` `int` `min_time_to_cut(` `int` `N)` `{` ` ` `if` `(N == 0)` ` ` `return` `0;` ` ` ` ` `// Return the minimum` ` ` `// unit of time required` ` ` `return` `(` `int` `)Math.Ceiling(Math.Log(N) /` ` ` `Math.Log(2));` `}` `// Driver Code` `public` `static` `void` `Main()` `{` ` ` `int` `N = 100;` ` ` `Console.Write(min_time_to_cut(N));` `}` `}` `// This code is contributed by Code_Mech` |

## Javascript

`<script>` `// Javascript program to find minimum` `// time required to split a stick of` `// N length into unit pieces` `// Function to return the` `// minimum time required` `// to split stick of N into` `// length into unit pieces` `function` `min_time_to_cut(N)` `{` ` ` `if` `(N == 0)` ` ` `return` `0;` ` ` ` ` `// Return the minimum` ` ` `// unit of time required` ` ` `return` `Math.ceil(Math.log(N) /` ` ` `Math.log(2));` `}` `// Driver code` `let N = 100;` `document.write(min_time_to_cut(N));` `// This code is contributed by divyesh072019` `</script>` |

**Output:**

7

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