Master theorem is used to determine the Big – O upper bound on functions which possess recurrence, i.e which can be broken into sub problems. **Master Theorem For Subtract and Conquer Recurrences**:

Let T(n) be a function defined on positive n as shown below:

for some constants c, a>0, b>0, k>=0 and function f(n). If f(n) is O(n^{k}), then

1. If a<1 then T(n) = O(n^{k})

2. If a=1 then T(n) = O(n^{k+1})

3. if a>1 then T(n) = O(n^{k}a^{n/b})**Proof of above theorem( By substitution method )**:

From above function, we have:

T(n) = aT(n-b) + f(n)

T(n-b) = aT(n-2b) + f(n-b)

T(n-2b) = aT(n-3b) + f(n-2b)

Now,

T(n-b) = a^{2}T(n-3b) + af(n-2b) + f(n-b)

T(n) = a^{3}T(n-3b) + a^{2}f(n-2b) + af(n-b) + f(n)

T(n) = Σ^{i=0 to n }a^{i} f(n-ib) + constant, where f(n-ib) is O(n-ib)

T(n) = O(n^{k} Σ^{i=0 to n/b }a^{i} )

Where,

If a<1 then Σ^{i=0 to n/b }a^{i} = O(1), T(n) = O(n^{k})

If a=1 then Σ^{i=0 to n/b }a^{i} = O(n), T(n) = O(n^{k+1})

If a>1 then Σ^{i=0 to n/b }a^{i} = O(a^{n/b}), T(n) = O(n^{k}a^{n/b})**Consider the following program for nth fibonacci number**:

## C++

`#include<stdio.h>` `int` `fib(` `int` `n)` `{` ` ` `if` `(n <= 1)` ` ` `return` `n;` ` ` `return` `fib(n-1) + fib(n-2);` `}` ` ` `int` `main ()` `{` ` ` `int` `n = 9;` ` ` `printf` `(` `"%d"` `, fib(n));` ` ` `getchar` `();` ` ` `return` `0;` `}` |

## Python3

`# Python3 code for the above approach` `def` `fib(n):` ` ` `if` `(n <` `=` `1` `):` ` ` `return` `n` ` ` `return` `fib(n ` `-` `1` `) ` `+` `fib(n ` `-` `2` `)` `# Driver code` `n ` `=` `9` `print` `(fib(n))` `# This code is contributed` `# by sahishelangia` |

## Java

`//Java code for above the approach.` `class` `clg` `{` ` ` `static` `int` `fib(` `int` `n)` `{` `if` `(n <= ` `1` `)` ` ` `return` `n;` `return` `fib(n-` `1` `) + fib(n-` `2` `);` `}` `// Driver Code` `public` `static` `void` `main (String[] args)` `{` `int` `n = ` `9` `;` `System.out.println( fib(n));` `}` `}` `// This code is contributed by Mukul Singh.` |

## C#

`// C# code for above the approach.` `using` `System;` ` ` `class` `GFG` `{` ` ` `static` `int` `fib(` `int` `n)` ` ` `{` ` ` `if` `(n <= 1)` ` ` `return` `n;` ` ` `return` `fib(n - 1) + fib(n - 2);` ` ` `}` ` ` ` ` `// Driver Code` ` ` `public` `static` `void` `Main(String[] args)` ` ` `{` ` ` `int` `n = 9;` ` ` `Console.WriteLine(fib(n));` ` ` `}` `}` `// This code has been contributed` `// by Rajput-Ji` |

## PHP

`<?php` `// PHP code for the above approach` `function` `fib(` `$n` `)` `{` ` ` `if` `(` `$n` `<= 1)` ` ` `return` `$n` `;` ` ` `return` `fib(` `$n` `- 1) +` ` ` `fib(` `$n` `- 2);` `}` `// Driver Code` `$n` `= 9;` `echo` `fib(` `$n` `);` `// This code is contributed` `// by Akanksha Rai` `?>` |

## Javascript

`<script>` ` ` `// Javascript code for above the approach.` ` ` ` ` `function` `fib(n)` ` ` `{` ` ` `if` `(n <= 1)` ` ` `return` `n;` ` ` `return` `fib(n - 1) + fib(n - 2);` ` ` `}` ` ` ` ` `let n = 9;` ` ` `document.write(fib(n));` ` ` `</script>` |

**Output**

34

**Time complexity Analysis:**

The recursive function can be defined as, T(n) = T(n-1) + T(n-2)

- For Worst Case, Let T(n-1) ≈ T(n-2)

T(n) = 2T(n-1) + c

where,f(n) = O(1)

∴ k=0, a=2, b=1;

T(n) = O(n^{0}2^{n/1})

= O(2^{n})

- For Best Case, Let T(n-2) ≈ T(n-1)

T(n) = 2T(n-2) + c

where,f(n) = O(1)

∴ k=0, a=2, b=2;

T(n) = O(n^{0}2^{n/2})

= O(2^{n/2})

**More Examples**:

**Example-1**:

T(n) = 3T(n-1), n>0

= c, n<=0

Sol:a=3, b=1, f(n)=0 so k=0;

Since a>0, T(n) = O(n^{k}a^{n/b})

T(n)= O(n^{0}3^{n/1})

T(n)= 3^{n}

**Example-2**:

T(n) = T(n-1) + n(n-1), if n>=2

= 1, if n=1

Sol:a=1, b=1, f(n)=n(n-1) so k=2;

Since a=1, T(n) = O(n^{k+1})

T(n)= O(n^{2+1})

T(n)= O(n^{3})

**Example-3**:

T(n) = 2T(n-1) – 1, if n>0

= 1, if n<=0

Sol: This recurrence can’t be solved using above method

since function is not of form**T(n) = aT(n-b) + f(n)**

This article is contributed by **Yash Singla**. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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