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

08 Mar, 2022
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
 
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++
Java
Python3
C#
Javascript
// C++ program to find number of ways to representing// a number as a sum of 1's and 2's#include <bits/stdc++.h>using namespace std; // Function to multiply matrix.void multiply(int F[2][2], int M[2][2]){    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 nvoid power(int F[2][2], int n){    if( n == 0 || n == 1)        return;    int M[2][2] = {{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[2][2] = {{1,1},{1,0}};    if (n == 0)        return 0;    power(F, n);    return F[0][0];} // Driver programint main(){    int n = 5;    cout << countWays(n) << endl;    return 0;}
// Java program to find number of// ways to representing a number// as a sum of 1's and 2'sclass 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 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 ndef 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 Coden = 5print(countWays(n)) # This code is contributed by mohit kumar
// C# program to find number of// ways to representing a number// as a sum of 1's and 2'sclass 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
<script> // Javascript program to find number of// ways to representing a number// as a sum of 1's and 2's // Function to multiply matrix.function multiply(F , M){    var x = F[0][0] * M[0][0] + F[0][1] * M[1][0];    var y = F[0][0] * M[0][1] + F[0][1] * M[1][1];    var z = F[1][0] * M[0][0] + F[1][1] * M[1][0];    var 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 nfunction power(F , n){    if (n == 0 || n == 1)    {        return;    }    var M = [[1, 1], [1, 0]];     power(F, parseInt(n / 2));    multiply(F, F);     if (n % 2 != 0)    {        multiply(F, M);    }} /* function that returns (n+1)th Fibonacci numberOr number of ways to represent n as sum of 1's2's */function countWays(n){    var F = [[1, 1], [1, 0]];    if (n == 0)    {        return 0;    }    power(F, n);    return F[0][0];} // Driver programvar n = 5;document.write(countWays(n)); // This code is contributed by 29AjayKumar </script>
Output:  
8
Time Complexity: O(logn).
Auxiliary Space: O(1).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 review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
Rajput-Ji
mohit kumar 29
Mithun Kumar
29AjayKumar
rohitsingh07052
Fibonacci
Bit Magic
Combinatorial
Bit Magic
Fibonacci
Combinatorial
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