Given an integer N, the task is to count the number of subsets formed from an array of integers from 1 to N which doesn’t contain adjacent elements. A subset can’t be chosen if it satisfies the non-adjacent element condition but it is possible to add more elements.
Examples:
Input: N = 4
Output: 3
Explanation:
Array is {1, 2, 3, 4}
So to satisfy the condition, the subsets formed are :
{1, 3}, {2, 4}, {1, 4}
Input: N = 5
Output: 4
Approach:
This problem can be solved by using Dynamic Programming. For the last element, we have two choices, either we include it, or we exclude it. Let DP[i] be the number of our desirable subsets ending at index i.
- If we select ith index, then we can’t select (i-1)th index, but we can select (i-2)th index. This makes DP[i-2] our new subproblem.
- If we don’t select ith index, then we have to select (i-1)th index, and further cannot select (i-2)th index, but can select (i-3)rd index.
So next Subproblem becomes DP[i-3]
So the DP relation becomes :
DP[i] = DP[i-2] + DP[i-3]
But, we need to observe the base cases:
- When N=0, we cannot form any subset with 0 numbers.
- When N=1, we can form 1 subset, {1}
- When N=2, we can form 2 subsets, {1}, {2}
- When N=3, we can form 2 subsets, {1, 3}, {2}
Below is the implementation of above approach:
C++
// C++ Code to count subsets not containing// adjacent elements from 1 to N #include <bits/stdc++.h>using namespace std; // Function to count subsetsint countSubsets(int N){ if(N <= 2) return N; if(N == 3) return 2; int DP[N + 1] = {0}; DP[0] = 0, DP[1] = 1, DP[2] = 2, DP[3] = 2; for (int i = 4; i <= N; i++) { DP[i] = DP[i - 2] + DP[i - 3]; } return DP[N];} // Driver Codeint main(){ int N = 20; cout << countSubsets(N); return 0;} |
Java
// Java code to count subsets not containing// adjacent elements from 1 to Nclass GFG{ // Function to count subsetsstatic int countSubsets(int N){ if(N <= 2) return N; if(N == 3) return 2; int []DP = new int[N + 1]; DP[0] = 0; DP[1] = 1; DP[2] = 2; DP[3] = 2; for(int i = 4; i <= N; i++) { DP[i] = DP[i - 2] + DP[i - 3]; } return DP[N];} // Driver codepublic static void main(String[] args){ int N = 20; System.out.print(countSubsets(N));}} // This code is contributed by sapnasingh4991 |
Python3
# Python3 Code to count subsets # not containing adjacent elements# from 1 to N # Function to count subsetsdef countSubsets(N): if(N <= 2): return N if(N == 3): return 2 DP = [0] * (N + 1) DP[0] = 0 DP[1] = 1 DP[2] = 2 DP[3] = 2 for i in range(4, N + 1): DP[i] = DP[i - 2] + DP[i - 3] return DP[N] # Driver Codeif __name__ == '__main__': N = 20 print(countSubsets(N)) # This code is contributed by Mohit Kumar |
C#
// C# code to count subsets not containing// adjacent elements from 1 to Nusing System;class GFG{ // Function to count subsetsstatic int countSubsets(int N){ if(N <= 2) return N; if(N == 3) return 2; int []DP = new int[N + 1]; DP[0] = 0; DP[1] = 1; DP[2] = 2; DP[3] = 2; for(int i = 4; i <= N; i++) { DP[i] = DP[i - 2] + DP[i - 3]; } return DP[N];} // Driver codepublic static void Main(String[] args){ int N = 20; Console.Write(countSubsets(N));}} // This code is contributed by sapnasingh4991 |
265
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.
