Given an integer N, the task is to find the number of permutations of integers from the range [1, N] that can form an acyclic graph according to the following conditions:
- For every 1 ? i ? N, find the largest j such that 1 ? j < i and A[j] > A[i], and add an undirected edge between node i and node j.
- For every 1 ? i ? N, find the smallest j such that i < j ? N and A[j] > A[i], and add an undirected edge between node i and node j.
Examples:
Input: 3
Output: 4
Explanation:
{1, 2, 3}: Possible (Graph Edges : { [1, 2], [2, 3] }. Therefore, no cycle exists)
{1, 3, 2}: Possible (Graph Edges : { [1, 3], [2, 3] }. Therefore, no cycle exists)
{2, 1, 3}: Not possible (Graph Edges : { [2, 3], [1, 3], [1, 2] }. Therefore, cycle exists)
{2, 3, 1}: Possible (Graph Edges : { [2, 3], [1, 3] }. Therefore, no cycle exists)
{3, 1, 2}: Not possible (Graph Edges : { [1, 2], [1, 3], [2, 3] }. Therefore, cycle exists)
{3, 2, 1}: Possible (Graph Edges : { [2, 3], [1, 2] }. Therefore, no cycle exists)Input : 4
Output : 8
Approach: Follow the steps below to solve the problem:
- There are a total of N! possible arrays that can be generated consisting of permutations of integers from the range [1, N].
- According to the given conditions, if i, j, k (i < j < k) are indices from the array, then if A[i] > A[j] and A[j] < A[k], then there will be a cycle consisting of edges {[A[j], A[i]], [A[j], A[k]], [A[i], A[k]]}.
- Removing these permutations, there are 2(N – 1) possible permutations remaining.
- Remaining permutations gives only two possible positions for integers from the range [1, N – 1] and 1 possible position for N.
Illustration:
For N = 3:
Permutation {2, 1, 3} contains a cycle as A[0] > A[1] and A[1] < A[2].
Permutation {3, 1, 2} contains a cycle as A[0] > A[1] and A[1] < A[2].
All remaining permutations can generate an acyclic graph.
Therefore, the count of valid permutations = 3! – 2 = 4 = 23 – 1
- Therefore, print 2N – 1 as the required answer.
Below is the implementation of the above approach :
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Find the count of possible graphsvoid possibleAcyclicGraph(int N){ cout << pow(2, N - 1); return;}// Driver Codeint main(){ int N = 4; possibleAcyclicGraph(N); return 0;} |
Java
// Java implementation of // the above approachimport java.util.*;class GFG{// Find the count of // possible graphsstatic void possibleAcyclicGraph(int N){ System.out.print((int)Math.pow(2, N - 1)); return;}// Driver Codepublic static void main(String[] args){ int N = 4; possibleAcyclicGraph(N);}}// This code is contributed by Princi Singh |
Python3
# Python3 implementation of above approach# Find the count of possible graphsdef possibleAcyclicGraph(N): print(pow(2, N - 1)) return# Driver codeif __name__ == '__main__': N = 4 possibleAcyclicGraph(N)# This code is contributed by mohit kumar 29 |
C#
// C# implementation of // the above approachusing System;class GFG{ // Find the count of // possible graphsstatic void possibleAcyclicGraph(int N){ Console.Write((int)Math.Pow(2, N - 1)); return;} // Driver Codepublic static void Main(){ int N = 4; possibleAcyclicGraph(N);}}// This code is contributed by sanjoy_62 |
Javascript
<script>// Javascript implementation of above approach// Find the count of// possible graphsfunction possibleAcyclicGraph(N){ document.write(Math.pow(2, N - 1)); return;} // Driver Codelet N = 4;possibleAcyclicGraph(N);// This code is contributed by target_2</script> |
8
Time Complexity: O(logN)
Space Complexity: O(1)
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