# Total number of components in the index Array

Given an array arr[] of N integers of value from 0 to N, the task is to count the number of components in Index Array.

Index array means if we are at ith index then it leads to arr[i].
The component of an index array is counted when it forms a cycle. If no cycle persists or the array contains a single element then also we consider it as a component.
For Example:
Let array arr[] = {1, 2, 0, 3}
{1, 2, 0} will form one component as starting from index 0 we reach the index 0 again as:
1 -> 2(arr) -> 0(arr) -> 1(arr)

Examples:

Input: arr[] = {1, 2, 3, 5, 0, 4, 6}
Output: 2
Explanation:
Below is the traversal of the 2 components:
Component 1: Start traversal from 0, then the path of traversal is given by:
1 -> 2(arr) -> 3(arr) -> 5(arr) -> 4(arr) -> 0(arr) -> 1(arr).
Component 2: Only 6 is unvisited it creates one more component.
So, the total components = 2.

Input: arr[] = {1, 2, 0, 3}
Output: 2
Explanation:
Below is the traversal of the 2 components:
Component 1: Start traversal from 0, then the path of traversal is given by:
1 -> 2(arr) -> 0(arr) -> 1(arr)
Component 2: Only 3 is unvisited it creates one more component.
So, the total components = 2.

Approach: The idea is to use the concept of DFS traversal. Below are the steps:

1. Start from the first unvisited index which will be index with integer 0 in it.
2. During DFS Traversal mark the visited elements in the array until the elements form a cycle.
3. If a cycle is formed then it means that we have got one component and hence increase the component count.
4. Repeat all the above steps for all the unvisited index in the array and count the total components in the given index array.
5. If all the index of the array are visited, then print the total count of connected components.

Below is the implementation of the above approach:

 `// C++ program for the above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function for DFS traversal ` `void` `dfs(``int` `i, ``int` `a[], ` `         ``map<``int``, ``int``>& m) ` `{ ` `    ``// Check if not visited then ` `    ``// recurr for the next index ` `    ``if` `(m[i] == 0) { ` `        ``m[i] = 1; ` ` `  `        ``// DFS Traversal for next index ` `        ``dfs(a[i], a, m); ` `    ``} ` ` `  `    ``return``; ` `} ` ` `  `// Function for checking which ` `// indexes are remaining ` `int` `allvisited(``int` `a[], ``int` `n, ` `               ``map<``int``, ``int``>& m) ` `{ ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Marks that the ith ` `        ``// index is not visited ` `        ``if` `(m[i] == 0) ` `            ``return` `i; ` `    ``} ` `    ``return` `-1; ` `} ` ` `  `// Function for counting components ` `int` `count(``int` `a[], ``int` `n) ` `{ ` `    ``int` `c = 0; ` ` `  `    ``// To mark the visited index ` `    ``// during DFS Traversal ` `    ``map<``int``, ``int``> m; ` ` `  `    ``// Function call ` `    ``int` `x = allvisited(a, n, m); ` ` `  `    ``while` `(x != -1) { ` ` `  `        ``// Count number of components ` `        ``c++; ` ` `  `        ``// DFS call ` `        ``dfs(x, a, m); ` ` `  `        ``x = allvisited(a, n, m); ` `    ``} ` ` `  `    ``// Print the total count of components ` `    ``cout << c << endl; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Given array arr[] ` `    ``int` `arr[] = { 1, 4, 3, 5, 0, 2, 6 }; ` ` `  `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``// Function Call ` `    ``count(arr, N); ` ` `  `    ``return` `0; ` `} `

Output:

```3
```

Time Complexity: O(N)
Auxiliary Space: O(N) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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