# Count ways to express ‘n’ as sum of odd integers

Given an positive integer n. Count total number of ways to express ‘n’ as sum of odd positive integers.

```Input: 4
Output: 3

Explanation
There are only three ways to write 4
as sum of odd integers:
1. 1 + 3
2. 3 + 1
3. 1 + 1 + 1 + 1

Input: 5
Output: 5
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Simple approach is to find recursive nature of problem. The number ‘n’ can be written as sum of odd integers from either (n-1)th number or (n-2)th number. Let the total number of ways to write ‘n’ be ways(n). The value of ‘ways(n)’ can be written by recursive formula as follows:

```ways(n) = ways(n-1) + ways(n-2)
```

The above expression is actually the expression for Fibonacci numbers. Therefore problem is reduced to find the nth fibonnaci number.

```ways(1) = fib(1) = 1
ways(2) = fib(2) = 1
ways(3) = fib(2) = 2
ways(4) = fib(4) = 3
```

## C++

 `// C++ program to count ways to write ` `// number as sum of odd integers ` `#include ` `using` `namespace` `std; ` ` `  `// Function to calculate n'th Fibonacci number ` `int` `fib(``int` `n) ` `{ ` `  ``/* Declare an array to store Fibonacci numbers. */` `  ``int` `f[n+1]; ` `  ``int` `i; ` ` `  `  ``/* 0th and 1st number of the series are 0 and 1*/` `  ``f = 0; ` `  ``f = 1; ` ` `  `  ``for` `(i = 2; i <= n; i++) ` `  ``{ ` `      ``/* Add the previous 2 numbers in the series ` `         ``and store it */` `      ``f[i] = f[i-1] + f[i-2]; ` `  ``} ` ` `  `  ``return` `f[n]; ` `} ` ` `  `// Return number of ways to write 'n' ` `// as sum of odd integers ` `int` `countOddWays(``int` `n) ` `{ ` `    ``return` `fib(n); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 4; ` `    ``cout << countOddWays(n) << ``"\n"``; ` ` `  `    ``n = 5; ` `    ``cout << countOddWays(n); ` `   ``return` `0; ` `} `

## Java

 `// Java program to count ways to write ` `// number as sum of odd integers ` `import` `java.util.*; ` ` `  `class` `GFG { ` `     `  `// Function to calculate n'th Fibonacci number ` `static` `int` `fib(``int` `n) { ` `     `  `    ``/* Declare an array to store Fibonacci numbers. */` `    ``int` `f[] = ``new` `int``[n + ``1``]; ` `    ``int` `i; ` ` `  `    ``/* 0th and 1st number of the series are 0 and 1*/` `    ``f[``0``] = ``0``; ` `    ``f[``1``] = ``1``; ` ` `  `    ``for` `(i = ``2``; i <= n; i++) { ` `         `  `    ``/* Add the previous 2 numbers in the series ` `        ``and store it */` `    ``f[i] = f[i - ``1``] + f[i - ``2``]; ` `    ``} ` ` `  `    ``return` `f[n]; ` `} ` ` `  `// Return number of ways to write 'n' ` `// as sum of odd integers ` `static` `int` `countOddWays(``int` `n) ` `{ ` `    ``return` `fib(n); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) { ` `     `  `    ``int` `n = ``4``; ` `    ``System.out.print(countOddWays(n) + ``"\n"``); ` ` `  `    ``n = ``5``; ` `    ``System.out.print(countOddWays(n)); ` `} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python3

 `# Python code to count ways to write ` `# number as sum of odd integers ` ` `  `# Function to calculate n'th  ` `# Fibonacci number ` `def` `fib( n ): ` ` `  `    ``# Declare a list to store  ` `    ``# Fibonacci numbers. ` `    ``f``=``list``() ` `     `  `    ``# 0th and 1st number of the  ` `    ``# series are 0 and 1 ` `    ``f.append(``0``) ` `    ``f.append(``1``) ` `     `  `    ``i ``=` `2` `    ``while` `i

## C#

 `// C# program to count ways to write ` `// number as sum of odd integers ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// Function to calculate n'th ` `    ``// Fibonacci number ` `    ``static` `int` `fib(``int` `n) { ` `         `  `        ``/* Declare an array to store ` `        ``Fibonacci numbers. */` `        ``int` `[]f = ``new` `int``[n + 1]; ` `        ``int` `i; ` `     `  `        ``/* 0th and 1st number of the ` `        ``series are 0 and 1*/` `        ``f = 0; ` `        ``f = 1; ` `     `  `        ``for` `(i = 2; i <= n; i++) ` `        ``{ ` `             `  `            ``/* Add the previous 2 numbers ` `            ``in the series and store it */` `            ``f[i] = f[i - 1] + f[i - 2]; ` `        ``} ` `     `  `        ``return` `f[n]; ` `    ``} ` `     `  `    ``// Return number of ways to write 'n' ` `    ``// as sum of odd integers ` `    ``static` `int` `countOddWays(``int` `n) ` `    ``{ ` `        ``return` `fib(n); ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 4; ` `        ``Console.WriteLine(countOddWays(n)); ` `     `  `        ``n = 5; ` `        ``Console.WriteLine(countOddWays(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output:

```3
5
```

Note: The time complexity of the above implementation is O(n). It can be further optimized up-to O(Logn) time using Fibonacci function optimization by Matrix Exponential.

This article is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.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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Improved By : vt_m