# Minimum number of cuts required to pay salary from N length Gold Bar

Given a gold bar of length **N** with **N** equal markings, the task is to find the minimum number of cuts required to pay the salary in **N days** such that on any **i ^{th}** day the worker has

**i**parts of the gold bar.

**Examples:**

Input:N = 5Output:2Explanation:

Divide the 5 length Gold bar into 3 part of length 1, 2, 2 by making 2 cuts. Then salary for each day will be as:

For Day 1 = 1 (Give 1 length bar)

For Day 2 = 2 – 1 = 1 (Give 2 length bar and take back 1 length bar)

For Day 3 = 1 (Give 1 length bar)

For Day 4 = 2 – 1 = 1 (Give 2 length bar and take back 1 length bar)

For Day 5 = 1 (Give 1 length bar)

Input:N = 15Output:3Explanation:

Divide the 15 length Gold bar into 4 part of length 1, 2, 4, 8 by making 3 cuts. Then salary for each day will be as:

For Day 1 = 1 (Give 1 length bar)

For Day 2 = 2 – 1 = 1 (Give 2 length bar and take back 1 length bar)

For Day 3 = 1 (Give 1 length bar)

For Day 4 = 4 – 2 – 1 = 1 (Give 4 length bar and take back 1 and 2 length bar)

For Day 5 = 1 (Give 1 length bar)

For Day 6 = 2 – 1 = 1 (Give 2 length bar and take back 1 length bar)

For Day 7 = 1 (Give 1 length bar)

For Day 8 = 8 – 4 – 2 – 1 = 1 (Give 8 length bar and take back 1, 2 and 4 length bar)

For Day 9 = 1 (Give 1 length bar)

For Day 10 = 2 – 1 = 1 (Give 2 length bar and take back 1 length bar)

For Day 11 = 1 (Give 1 length bar)

For Day 12 = 4 – 2 – 1 = 1 (Give 4 length bar and take back 1 and 2 length bar)

For Day 13 = 1 (Give 1 length bar)

For Day 14 = 2 – 1 = 1 (Give 2 length bar and take back 1 length bar)

For Day 15 = 1 (Give 1 length bar)

**Approach:**

As we know any number can be represented with the help of the numbers in the form of powers of 2. So If we cut the length of Gold Bar into the nearest integer to log_{2} (N) then we can represent any number up to N. **For example:** When N = 15, we can divide the number into the parts 1, 2, 4, 8 and using these numbers we can represent any number from 1 to 15, as shown below:

For 1 - 1 For 2 - 2 For 3 - 2 + 1 For 4 - 4 For 5 - 4 + 1 For 6 - 4 + 2 For 7 - 4 + 2 + 1 For 8 - 8 For 9 - 8 + 1 For 10 - 8 + 2 For 11 - 8 + 2 + 1 For 12 - 8 + 4 For 13 - 8 + 4 + 1 For 14 - 8 + 4 + 2 For 15 - 8 + 4 + 2 + 1

Below is the implementation of the above approach.

## C++

`// CPP Implementation to find` `// the minimum number of cuts to` `// pay the worker.` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the minimum` `// number of cuts to pay the worker.` `int` `pay(` `int` `n)` `{` ` ` `// Nearest Integer to the Log` ` ` `// value of the number N` ` ` `int` `cuts = ` `int` `(` `log` `(n)/` `log` `(2));` ` ` `return` `cuts;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `n = 5;` ` ` `int` `cuts = pay(n);` ` ` `cout << cuts << endl;` ` ` ` ` `// Cuts Required in the` ` ` `// Length of 15` ` ` `n = 15;` ` ` `cuts = pay(n);` ` ` `cout<<(cuts);` ` ` ` ` `return` `0;` `}` `// This code is contributed by mohit kumar 29` |

## Java

`// JAVA Implementation to find` `// the minimum number of cuts to` `// pay the worker.` `class` `GFG` `{` `// Function to find the minimum` `// number of cuts to pay the worker.` `static` `int` `pay(` `int` `n)` `{` ` ` `// Nearest Integer to the Log` ` ` `// value of the number N` ` ` `int` `cuts = (` `int` `) (Math.log(n)/Math.log(` `2` `));` ` ` `return` `cuts;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `n = ` `5` `;` ` ` `int` `cuts = pay(n);` ` ` `System.out.print(cuts +` `"\n"` `);` ` ` ` ` `// Cuts Required in the` ` ` `// Length of 15` ` ` `n = ` `15` `;` ` ` `cuts = pay(n);` ` ` `System.out.print(cuts);` `}` `}` `// This code is contributed by 29AjayKumar` |

## Python

`# Python Implementation to find` `# the minimum number of cuts to` `# pay the worker.` `import` `math` `# Function to find the minimum` `# number of cuts to pay the worker.` `def` `pay(n):` ` ` ` ` `# Nearest Integer to the Log` ` ` `# value of the number N` ` ` `cuts ` `=` `int` `(math.log(n, ` `2` `))` ` ` ` ` `return` `cuts` ` ` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `n ` `=` `5` ` ` `cuts ` `=` `pay(n)` ` ` `print` `(cuts)` ` ` ` ` `# Cuts Required in the` ` ` `# Length of 15` ` ` `n ` `=` `15` ` ` `cuts ` `=` `pay(n)` ` ` `print` `(cuts)` |

## C#

`// C# Implementation to find` `// the minimum number of cuts to` `// pay the worker.` `using` `System;` `class` `GFG` `{` `// Function to find the minimum` `// number of cuts to pay the worker.` `static` `int` `pay(` `int` `n)` `{` ` ` `// Nearest int to the Log` ` ` `// value of the number N` ` ` `int` `cuts = (` `int` `) (Math.Log(n)/Math.Log(2));` ` ` `return` `cuts;` `}` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `n = 5;` ` ` `int` `cuts = pay(n);` ` ` `Console.Write(cuts +` `"\n"` `);` ` ` ` ` `// Cuts Required in the` ` ` `// Length of 15` ` ` `n = 15;` ` ` `cuts = pay(n);` ` ` `Console.Write(cuts);` `}` `}` `// This code is contributed by 29AjayKumar` |

## Javascript

`<script>` `// JavaScript Implementation to find` `// the minimum number of cuts to` `// pay the worker.` `// Function to find the minimum` `// number of cuts to pay the worker.` `function` `pay(n)` `{` ` ` `// Nearest Integer to the Log` ` ` `// value of the number N` ` ` `let cuts = parseInt(Math.log(n)/Math.log(2));` ` ` `return` `cuts;` `}` `// Driver code` ` ` `let n = 5;` ` ` `let cuts = pay(n);` ` ` `document.write(cuts + ` `"<br>"` `);` ` ` ` ` `// Cuts Required in the` ` ` `// Length of 15` ` ` `n = 15;` ` ` `cuts = pay(n);` ` ` `document.write(cuts + ` `"<br>"` `);` `</script>` |

**Output:**

2 3

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