Check if the end of the Array can be reached from a given position

/*
https://www.geeksforgeeks.org/check-if-the-end-of-the-array-can-be-reached-from-a-given-position/
This whole code is GFG article code so don't write this code anywhere,Modify the code.

Logic: BFS
We are starting from S and there is two situation only either go s+arr[i] or s-arr[i].so now,we are having two new starting points if any of the above two reaches to end,then answer will be yes.Suppose we again reach at s+arr[i] after some steps then the results will be same.So keep a visited array where we don't need to come to the same point because nothing is going to change,you will follow the same path again you went earlier.
T.C : O(N) as we move all i once.
S.C : O(N) for queue for BFS
*/

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if we can reach to
// the end of the arr[] with possible moves
void solve(int arr[], int n, int start)
{

    // Queue to perform BFS
    queue<int> q;

    // Initially all nodes(index)
    // are not visited.
    bool visited[n] = { false };

    // Initially the end of
    // the array is not reached
    bool reached = false;

    // Push start index in queue
    q.push(start);

    // Until queue becomes empty
    while (!q.empty()) {

        // Get the top element
        int temp = q.front();

        // Delete popped element
        q.pop();

        // If the index is already
        // visited. No need to
        // traverse it again.
        if (visited[temp] == true)
            continue;

        // Mark temp as visited
        // if not
        visited[temp] = true;
        if (temp == n - 1) {

            // If reached at the end
            // of the array then break
            reached = true;
            break;
        }

        // If temp + arr[temp] and
        // temp - arr[temp] are in
        // the index of array
        if (temp + arr[temp] < n) {
            q.push(temp + arr[temp]);
        }

        if (temp - arr[temp] >= 0) {
            q.push(temp - arr[temp]);
        }
    }

    // If reaches the end of the array,
    // Print "Yes"
    if (reached == true) {
        cout << "Yes";
    }

    // Else print "No"
    else {
        cout << "No";
    }
}

// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 4, 1, 3, 2, 5 };

    // Starting index
    int S = 1;
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    solve(arr, N, S);

    return 0;
}