Given an integer N, the task is to find out the number of permutations in which N can be represented as a sum of 2s, 4s, and 6s only.
Note: The different permutations of the same combination will also be counted.
Examples:
Input: N = 8
Output: 7
Explanation: The possible combinations are:
2 2 2 2
4 4
4 2 2
2 2 4
2 4 2
2 6
6 2Input: N = 6
Output: 4
Approach: In order to solve this problem, we are using a Dynamic Programming approach:
- As the possible numbers that can be added to reach N are 2, 4, and 6, they can be considered as base cases. By making use of the base cases further cases can be computed.
- Let us consider a dp[] array of size N + 1 which computes and stores at every index i the possible ways to form a value i using only 2, 4, 6.
-
dp[2] = 1
dp[4] = 2 { 4 can be represented as 2+2, 4}
dp[6] = 4 { 6 can be represented as 2+2+2, 2+4, 4+2, 6}
dp[i] = dp[i-2]+dp[i-4] + dp[i – 6] for all even values of i in the range [8, N] - Hence, dp[N] contains the required answer.
Below is the implementation of the above approach:
C++
// C++ code for above implementation #include <bits/stdc++.h> using namespace std; // Returns number of ways // to reach score n int count(int n) { // table[i] will store count // of solutions for value i. if (n == 2) return 1; else if (n == 4) return 2; else if (n == 6) return 4; int table[n + 1], i; // Initialize all table // values as 0 for (i = 0; i < n + 1; i++) table[i] = 0; // Base case (If given value // is 0, 1, 2, or 4) table[0] = 0; table[2] = 1; table[4] = 2; table[6] = 4; for (i = 8; i <= n; i = i + 2) { table[i] = table[i - 2] + table[i - 4] + table[i - 6]; } return table[n]; } // Driver Code int main(void) { int n = 8; cout << count(n); return 0; } |
Java
// Java code for above implementation import java.util.*; class GFG{ // Returns number of ways // to reach score n static int count(int n) { // table[i] will store count // of solutions for value i. if (n == 2) return 1; else if (n == 4) return 2; else if (n == 6) return 4; int table[] = new int[n + 1]; int i; // Initialize all table // values as 0 for (i = 0; i < n + 1; i++) table[i] = 0; // Base case (If given value // is 0, 1, 2, or 4) table[0] = 0; table[2] = 1; table[4] = 2; table[6] = 4; for (i = 8; i <= n; i = i + 2) { table[i] = table[i - 2] + table[i - 4] + table[i - 6]; } return table[n]; } // Driver Code public static void main(String args[]) { int n = 8; System.out.print(count(n)); } } // This code is contributed by Akanksha_Rai |
Python3
# Python3 code for above implementation # Returns number of ways # to reach score n def count(n) : # table[i] will store count # of solutions for value i. if (n == 2) : return 1; elif (n == 4) : return 2; elif (n == 6) : return 4; table = [0] * (n + 1); # Initialize all table # values as 0 for i in range(n + 1) : table[i] = 0; # Base case (If given value # is 0, 1, 2, or 4) table[0] = 0; table[2] = 1; table[4] = 2; table[6] = 4; for i in range(8 , n + 1, 2) : table[i] = table[i - 2] + \ table[i - 4] + \ table[i - 6]; return table[n]; # Driver Code if __name__ == "__main__" : n = 8; print(count(n)); # This code is contributed by AnkitRai01 |
C#
// C# code for above implementation using System; class GFG{ // Returns number of ways // to reach score n static int count(int n) { // table[i] will store count // of solutions for value i. if (n == 2) return 1; else if (n == 4) return 2; else if (n == 6) return 4; int []table = new int[n + 1]; int i; // Initialize all table // values as 0 for (i = 0; i < n + 1; i++) table[i] = 0; // Base case (If given value // is 0, 1, 2, or 4) table[0] = 0; table[2] = 1; table[4] = 2; table[6] = 4; for (i = 8; i <= n; i = i + 2) { table[i] = table[i - 2] + table[i - 4] + table[i - 6]; } return table[n]; } // Driver Code public static void Main() { int n = 8; Console.Write(count(n)); } } // This code is contributed by Code_Mech |
7
Time Complexity: O(N)
Auxiliary Space Complexity: O(N)
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