Given an array A of integers. Index i of A is said to be connected to index j if j = (i + A[i]) % n + 1 (Assume 1-based indexing). Start traversing array from index i and jump to its next connected index. If on traversing array in the described order, index i is again visited then index i is a magical index. Count the number of magical indexes in the array. Assume that array A consists of non-negative integers.
Examples :
Input : A = {1, 1, 1, 1}
Output : 4
Possible traversals:
1 -> 3 -> 1
2 -> 4 -> 2
3 -> 1 -> 3
4 -> 2 -> 4
Clearly all the indices are magical
Input : A = {0, 0, 0, 2}
Output : 2
Possible traversals:
1 -> 2 -> 3 -> 4 -> 3...
2 -> 3 -> 4 -> 3...
3 -> 4 -> 3
4 -> 3 ->4
Magical indices = 3, 4
Approach: The problem is of counting number of nodes in all the cycles present in the graph. Each index represents a single node of the graph. Each node has a single directed edge as described in the problem statement. This graph has a special property: On starting a traversal from any vertex, a cycle is always detected. This property will be helpful in reducing the time complexity of the solution.
Read this post on how to detect cycle in a directed graph: Detect Cycle in directed graph
Let the traversal begins from node i. Node i will be called parent node of this traversal and this parent node will be assigned to all the nodes visited during traversal. While traversing the graph if we discover a node that is already visited and parent node of that visited node is same as parent node of the traversal then a new cycle is detected. To count number of nodes in this cycle, start another dfs from this node until this same node is not visited again. This procedure is repeated for every node i of the graph. In worst case every node will be traversed at most 3 times. Hence solution has linear time complexity.
The stepwise algorithm is:
1. For each node in the graph:
if node i is not visited then:
for every adjacent node j:
if node j is not visited:
par[j] = i
else:
if par[j]==i
cycle detected
count nodes in cycle
2. return count
Implementation:
C++
// C++ program to find number of magical // indices in the given array. #include <bits/stdc++.h> using namespace std; #define mp make_pair #define pb push_back #define mod 1000000007 // Function to count number of magical indices. int solve(int A[], int n) { int i, cnt = 0, j; // Array to store parent node of traversal. int parent[n + 1]; // Array to determine whether current node // is already counted in the cycle. int vis[n + 1]; // Initialize the arrays. memset(parent, -1, sizeof(parent)); memset(vis, 0, sizeof(vis)); for (i = 0; i < n; i++) { j = i; // Check if current node is already // traversed or not. If node is not // traversed yet then parent value // will be -1. if (parent[j] == -1) { // Traverse the graph until an // already visited node is not // found. while (parent[j] == -1) { parent[j] = i; j = (j + A[j] + 1) % n; } // Check parent value to ensure // a cycle is present. if (parent[j] == i) { // Count number of nodes in // the cycle. while (!vis[j]) { vis[j] = 1; cnt++; j = (j + A[j] + 1) % n; } } } } return cnt; } int main() { int A[] = { 0, 0, 0, 2 }; int n = sizeof(A) / sizeof(A[0]); cout << solve(A, n); return 0; } |
Java
// Java program to find number of magical // indices in the given array. import java.io.*; import java.util.*; public class GFG { // Function to count number of magical // indices. static int solve(int []A, int n) { int i, cnt = 0, j; // Array to store parent node of // traversal. int []parent = new int[n + 1]; // Array to determine whether current node // is already counted in the cycle. int []vis = new int[n + 1]; // Initialize the arrays. for (i = 0; i < n+1; i++) { parent[i] = -1; vis[i] = 0; } for (i = 0; i < n; i++) { j = i; // Check if current node is already // traversed or not. If node is not // traversed yet then parent value // will be -1. if (parent[j] == -1) { // Traverse the graph until an // already visited node is not // found. while (parent[j] == -1) { parent[j] = i; j = (j + A[j] + 1) % n; } // Check parent value to ensure // a cycle is present. if (parent[j] == i) { // Count number of nodes in // the cycle. while (vis[j]==0) { vis[j] = 1; cnt++; j = (j + A[j] + 1) % n; } } } } return cnt; } // Driver code public static void main(String args[]) { int []A = { 0, 0, 0, 2 }; int n = A.length; System.out.print(solve(A, n)); } } // This code is contributed by Manish Shaw // (manishshaw1) |
Python3
# Python3 program to find number of magical # indices in the given array. # Function to count number of magical # indices. def solve(A, n) : cnt = 0 # Array to store parent node of # traversal. parent = [None] * (n + 1) # Array to determine whether current node # is already counted in the cycle. vis = [None] * (n + 1) # Initialize the arrays. for i in range(0, n+1): parent[i] = -1 vis[i] = 0 for i in range(0, n): j = i # Check if current node is already # traversed or not. If node is not # traversed yet then parent value # will be -1. if (parent[j] == -1) : # Traverse the graph until an # already visited node is not # found. while (parent[j] == -1) : parent[j] = i j = (j + A[j] + 1) % n # Check parent value to ensure # a cycle is present. if (parent[j] == i) : # Count number of nodes in # the cycle. while (vis[j]==0) : vis[j] = 1 cnt = cnt + 1 j = (j + A[j] + 1) % n return cnt # Driver code A = [ 0, 0, 0, 2 ] n = len(A) print (solve(A, n)) # This code is contributed by Manish Shaw # (manishshaw1) |
C#
// C# program to find number of magical // indices in the given array. using System; using System.Collections.Generic; using System.Linq; class GFG { // Function to count number of magical // indices. static int solve(int []A, int n) { int i, cnt = 0, j; // Array to store parent node of // traversal. int []parent = new int[n + 1]; // Array to determine whether current node // is already counted in the cycle. int []vis = new int[n + 1]; // Initialize the arrays. for (i = 0; i < n+1; i++) { parent[i] = -1; vis[i] = 0; } for (i = 0; i < n; i++) { j = i; // Check if current node is already // traversed or not. If node is not // traversed yet then parent value // will be -1. if (parent[j] == -1) { // Traverse the graph until an // already visited node is not // found. while (parent[j] == -1) { parent[j] = i; j = (j + A[j] + 1) % n; } // Check parent value to ensure // a cycle is present. if (parent[j] == i) { // Count number of nodes in // the cycle. while (vis[j]==0) { vis[j] = 1; cnt++; j = (j + A[j] + 1) % n; } } } } return cnt; } // Driver code public static void Main() { int []A = { 0, 0, 0, 2 }; int n = A.Length; Console.WriteLine(solve(A, n)); } } // This code is contributed by Manish Shaw // (manishshaw1) |
PHP
<?php // PHP program to find // number of magical // indices in the given // array. // Function to count number // of magical indices. function solve($A, $n) { $i = 0; $cnt = 0; $j = 0; // Array to store parent // node of traversal. $parent = array(); // Array to determine // whether current node // is already counted // in the cycle. $vis = array(); // Initialize the arrays. for ($i = 0; $i < $n + 1; $i++) { $parent[$i] = -1; $vis[$i] = 0; } for ($i = 0; $i < $n; $i++) { $j = $i; // Check if current node is // already traversed or not. // If node is not traversed // yet then parent value will // be -1. if ($parent[$j] == -1) { // Traverse the graph until // an already visited node // is not found. while ($parent[$j] == -1) { $parent[$j] = $i; $j = ($j + $A[$j] + 1) % $n; } // Check parent value to ensure // a cycle is present. if ($parent[$j] == $i) { // Count number of // nodes in the cycle. while ($vis[$j] == 0) { $vis[$j] = 1; $cnt++; $j = ($j + $A[$j] + 1) % $n; } } } } return $cnt; } // Driver code $A = array( 0, 0, 0, 2 ); $n = count($A); echo (solve($A, $n)); // This code is contributed by // Manish Shaw (manishshaw1) ?> |
2
Time Complexity: O(n)
Space Complexity: O(n)
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