Count of different ways to express N as the sum of 1, 3 and 4 - GeeksforGeeks

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Count of different ways to express N as the sum of 1, 3 and 4
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Difficulty Level :
Easy
Given N, count the number of ways to express N as sum of 1, 3 and 4.
Examples: 
Input :  N = 4
Output : 4 
Explanation: 1+1+1+1 
             1+3
             3+1 
             4 

Input : N = 5 
Output : 6
Explanation: 1 + 1 + 1 + 1 + 1
             1 + 4
             4 + 1
             1 + 1 + 3
             1 + 3 + 1
             3 + 1 + 1
Approach : Divide the problem into sub-problems for solving it. Let DP[n] be the be the number of ways to write N as the sum of 1, 3, and 4. Consider one possible solution with n = x1 + x2 + x3 + ... xn. If the last number is 1, then sum of the remaining numbers is n-1. So the number that ends with 1 is equal to DP[n-1]. Taking other cases into account where the last number is 3 and 4. The final recurrence would be: 
DPn = DPn-1 + DPn-3 + DPn-4
Base case :
DP[0] = DP[1] = DP[2] = 1 and DP[3] = 2
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// C++ program to illustrate the number of// ways to represent N as sum of 1, 3 and 4.#include <bits/stdc++.h>using namespace std; // function to count the number of// ways to represent n as sum of 1, 3 and 4int countWays(int n){    int DP[n + 1];     // base cases    DP[0] = DP[1] = DP[2] = 1;    DP[3] = 2;     // iterate for all values from 4 to n    for (int i = 4; i <= n; i++)        DP[i] = DP[i - 1] + DP[i - 3] + DP[i - 4];         return DP[n];} // driver codeint main(){    int n = 10;    cout << countWays(n);    return 0;}
// Java program to illustrate// the number of ways to represent// N as sum of 1, 3 and 4. class GFG {     // Function to count the    // number of ways to represent    // n as sum of 1, 3 and 4    static int countWays(int n)    {        int DP[] = new int[n + 1];         // base cases        DP[0] = DP[1] = DP[2] = 1;        DP[3] = 2;         // iterate for all values from 4 to n        for (int i = 4; i <= n; i++)            DP[i] = DP[i - 1] + DP[i - 3]                    + DP[i - 4];         return DP[n];    }     // driver code    public static void main(String[] args)    {        int n = 10;        System.out.println(countWays(n));    }} // This code is contributed// by prerna saini.
# Python program to illustrate the number of# ways to represent N as sum of 1, 3 and 4. # Function to count the number of# ways to represent n as sum of 1, 3 and 4def countWays(n):     DP = [0 for i in range(0, n + 1)]         # base cases    DP[0] = DP[1] = DP[2] = 1    DP[3] = 2     # Iterate for all values from 4 to n    for i in range(4, n + 1):        DP[i] = DP[i - 1] + DP[i - 3] + DP[i - 4]         return DP[n]      # Driver coden = 10print (countWays(n)) # This code is contributed by Gitanjali.
// C# program to illustrate// the number of ways to represent// N as sum of 1, 3 and 4.using System; class GFG {     // Function to count the    // number of ways to represent    // n as sum of 1, 3 and 4    static int countWays(int n)    {        int []DP = new int[n + 1];         // base cases        DP[0] = DP[1] = DP[2] = 1;        DP[3] = 2;         // iterate for all values from 4 to n        for (int i = 4; i <= n; i++)            DP[i] = DP[i - 1] + DP[i - 3]                    + DP[i - 4];         return DP[n];    }     // Driver code    public static void Main()    {        int n = 10;        Console.WriteLine(countWays(n));    }} // This code is contributed// by vt_m.
<?php// PHP program to illustrate// the number of ways to// represent N as sum of// 1, 3 and 4. // function to count the// number of ways to// represent n as sum of// 1, 3 and 4function countWays($n){    $DP = array();     // base cases    $DP[0] = $DP[1] = $DP[2] = 1;    $DP[3] = 2;     // iterate for all    // values from 4 to n    for ($i = 4; $i <= $n; $i++)        $DP[$i] = $DP[$i - 1] +                  $DP[$i - 3] +                  $DP[$i - 4];         return $DP[$n];} // Driver Code$n = 10;echo countWays($n); // This code is contributed// by Sam007?>
<script> // Javascript program to illustrate// the number of ways to represent// N as sum of 1, 3 and 4. // Function to count the// number of ways to represent// n as sum of 1, 3 and 4function countWays(n){    var DP = [];    DP.length = 10;    DP.fill(0);     // Base cases    DP[0] = DP[1] = DP[2] = 1;    DP[3] = 2;     // Iterate for all values from 4 to n    for(var i = 4; i <= n; i++)        DP[i] = DP[i - 1] + DP[i - 3] +                DP[i - 4];     return DP[n];} // Driver codevar n = 10; document.write(countWays(n)); // This code is contributed by bunnyram19   </script>
Output: 
64
Time Complexity : O(n) Auxiliary Space : O(n) 
Sam007
bunnyram19
Combinatorial
Dynamic Programming
Dynamic Programming
Combinatorial
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