Related Articles

# Ways to represent a number as a sum of 1’s and 2’s

• Difficulty Level : Medium
• Last Updated : 04 May, 2021

Given a positive integer N. The task is to find the number of ways of representing N as a sum of 1s and 2s.
Examples:

```Input : N = 3
Output : 3
3 can be represented as (1+1+1), (2+1), (1+2).

Input : N = 5
Output : 8```

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.

For N = 1, answer is 1.
For N = 2. (1 + 1), (2), answer is 2.
For N = 3. (1 + 1 + 1), (2 + 1), (1 + 2), answer is 3.
For N = 4. (1 + 1 + 1 + 1), (2 + 1 + 1), (1 + 2 + 1), (1 + 1 + 2), (2 + 2) answer is 5.
And so on.
It can be observed that it form Fibonacci Series. So, the number of ways of representing N as a sum of 1s and 2s is (N + 1)th Fibonacci number.
How?
We can easily see that the recursive function is exactly the same as Fibonacci Numbers. To obtain the sum of N, we can add 1 to N – 1. Also, we can add 2 to N – 2. And only 1 and 2 are allowed to make the sum N. So, to obtain sum N using 1s and 2s, total ways are: number of ways to obtain (N – 1) + number of ways to obtain (N – 2).
We can find N’th Fibonacci Number in O(Log n) time. Please refer to method 5 of this post.
Below is the implementation of this approach:

## C++

 `// C++ program to find number of ways to representing``// a number as a sum of 1's and 2's``#include ``using` `namespace` `std;` `// Function to multiply matrix.``void` `multiply(``int` `F, ``int` `M)``{``    ``int` `x =  F*M + F*M;``    ``int` `y =  F*M + F*M;``    ``int` `z =  F*M + F*M;``    ``int` `w =  F*M + F*M;` `    ``F = x;``    ``F = y;``    ``F = z;``    ``F = w;``}` `// Power function in log n``void` `power(``int` `F, ``int` `n)``{``    ``if``( n == 0 || n == 1)``        ``return``;``    ``int` `M = {{1,1},{1,0}};` `    ``power(F, n/2);``    ``multiply(F, F);` `    ``if` `(n%2 != 0)``        ``multiply(F, M);``}` `/* function that returns (n+1)th Fibonacci number``   ``Or number of ways to represent n as sum of 1's``   ``2's */``int` `countWays(``int` `n)``{``    ``int` `F = {{1,1},{1,0}};``    ``if` `(n == 0)``        ``return` `0;``    ``power(F, n);``    ``return` `F;``}` `// Driver program``int` `main()``{``    ``int` `n = 5;``    ``cout << countWays(n) << endl;``    ``return` `0;``}`

## Java

 `// Java program to find number of``// ways to representing a number``// as a sum of 1's and 2's``class` `GFG``{` `// Function to multiply matrix.``    ``static` `void` `multiply(``int` `F[][], ``int` `M[][])``    ``{``        ``int` `x = F[``0``][``0``] * M[``0``][``0``] + F[``0``][``1``] * M[``1``][``0``];``        ``int` `y = F[``0``][``0``] * M[``0``][``1``] + F[``0``][``1``] * M[``1``][``1``];``        ``int` `z = F[``1``][``0``] * M[``0``][``0``] + F[``1``][``1``] * M[``1``][``0``];``        ``int` `w = F[``1``][``0``] * M[``0``][``1``] + F[``1``][``1``] * M[``1``][``1``];` `        ``F[``0``][``0``] = x;``        ``F[``0``][``1``] = y;``        ``F[``1``][``0``] = z;``        ``F[``1``][``1``] = w;``    ``}` `    ``// Power function in log n``    ``static` `void` `power(``int` `F[][], ``int` `n)``    ``{``        ``if` `(n == ``0` `|| n == ``1``)``        ``{``            ``return``;``        ``}``        ``int` `M[][] = {{``1``, ``1``}, {``1``, ``0``}};` `        ``power(F, n / ``2``);``        ``multiply(F, F);` `        ``if` `(n % ``2` `!= ``0``)``        ``{``            ``multiply(F, M);``        ``}``    ``}` `    ``/* function that returns (n+1)th Fibonacci number``    ``Or number of ways to represent n as sum of 1's``    ``2's */``    ``static` `int` `countWays(``int` `n)``    ``{``        ``int` `F[][] = {{``1``, ``1``}, {``1``, ``0``}};``        ``if` `(n == ``0``)``        ``{``            ``return` `0``;``        ``}``        ``power(F, n);``        ``return` `F[``0``][``0``];``    ``}` `    ``// Driver program``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `n = ``5``;``        ``System.out.println(countWays(n));``    ``}``}` `// This code contributed by Rajput-Ji`

## Python3

 `# Python3 program to find number of ways to``# representing a number as a sum of 1's and 2's` `# Function to multiply matrix.``def` `multiply(F, M):` `    ``x ``=` `F[``0``][``0``] ``*` `M[``0``][``0``] ``+` `F[``0``][``1``] ``*` `M[``1``][``0``]``    ``y ``=` `F[``0``][``0``] ``*` `M[``0``][``1``] ``+` `F[``0``][``1``] ``*` `M[``1``][``1``]``    ``z ``=` `F[``1``][``0``] ``*` `M[``0``][``0``] ``+` `F[``1``][``1``] ``*` `M[``1``][``0``]``    ``w ``=` `F[``1``][``0``] ``*` `M[``0``][``1``] ``+` `F[``1``][``1``] ``*` `M[``1``][``1``]` `    ``F[``0``][``0``] ``=` `x``    ``F[``0``][``1``] ``=` `y``    ``F[``1``][``0``] ``=` `z``    ``F[``1``][``1``] ``=` `w` `# Power function in log n``def` `power(F, n):` `    ``if``( n ``=``=` `0` `or` `n ``=``=` `1``):``        ``return``    ``M ``=` `[[``1``, ``1``],[``1``, ``0``]]` `    ``power(F, n ``/``/` `2``)``    ``multiply(F, F)` `    ``if` `(n ``%` `2` `!``=` `0``):``        ``multiply(F, M)` `#/* function that returns (n+1)th Fibonacci number``# Or number of ways to represent n as sum of 1's``# 2's */``def` `countWays(n):``    ``F ``=` `[[``1``, ``1``], [``1``, ``0``]]``    ``if` `(n ``=``=` `0``):``        ``return` `0``    ``power(F, n)` `    ``return` `F[``0``][``0``]` `# Driver Code``n ``=` `5``print``(countWays(n))` `# This code is contributed by mohit kumar`

## C#

 `// C# program to find number of``// ways to representing a number``// as a sum of 1's and 2's``class` `GFG``{` `    ``// Function to multiply matrix.``    ``static` `void` `multiply(``int` `[,]F, ``int` `[,]M)``    ``{``        ``int` `x = F[0,0] * M[0,0] + F[0,1] * M[1,0];``        ``int` `y = F[0,0] * M[0,1] + F[0,1] * M[1,1];``        ``int` `z = F[1,0] * M[0,0] + F[1,1] * M[1,0];``        ``int` `w = F[1,0] * M[0,1] + F[1,1] * M[1,1];` `        ``F[0,0] = x;``        ``F[0,1] = y;``        ``F[1,0] = z;``        ``F[1,1] = w;``    ``}` `    ``// Power function in log n``    ``static` `void` `power(``int` `[,]F, ``int` `n)``    ``{``        ``if` `(n == 0 || n == 1)``        ``{``            ``return``;``        ``}``        ``int` `[,]M = {{1, 1}, {1, 0}};` `        ``power(F, n / 2);``        ``multiply(F, F);` `        ``if` `(n % 2 != 0)``        ``{``            ``multiply(F, M);``        ``}``    ``}` `    ``/* function that returns (n+1)th Fibonacci number``    ``Or number of ways to represent n as sum of 1's``    ``2's */``    ``static` `int` `countWays(``int` `n)``    ``{``        ``int` `[,]F = {{1, 1}, {1, 0}};``        ``if` `(n == 0)``        ``{``            ``return` `0;``        ``}``        ``power(F, n);``        ``return` `F[0,0];``    ``}` `    ``// Driver program``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 5;``        ``System.Console.WriteLine(countWays(n));``    ``}``}` `// This code contributed by mits`

## Javascript

 ``

Output:

`8`

Time Complexity: O(logn).
This article is contributed by Anuj Chauhan. 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.