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

  • Difficulty Level : Medium
  • Last Updated : 09 Nov, 2021

Given an array arr[] of N positive integers and a number S, the task is to reach the end of the array from index S. We can only move from current index i to index (i + arr[i]) or (i – arr[i]). If there is a way to reach the end of the array then print “Yes” else print “No”.

Examples: 

Input: arr[] = {4, 1, 3, 2, 5}, S = 1 
Output: Yes 
Explanation: 
initial position: arr[S] = arr[1] = 1. 
Jumps to reach the end: 1 -> 4 -> 5 
Hence end has been reached. 

Input: arr[] = {2, 1, 4, 5}, S = 2 
Output: No 
Explanation: 
initial position: arr[S] = arr[2] = 2. 
Possible Jumps to reach the end: 4 -> (index 7) or 4 -> (index -2) 
Since both are out of bounds, Hence end can’t be reached. 
 

Approach 1: This problem can be solved using Breadth First Search. Below are the steps:  

  • Consider start index S as the source node and insert it into the queue.
  • While the queue is not empty do the following: 
    1. Pop the element from the top of the queue say temp.
    2. If temp is already visited or it is array out of bound index then, go to step 1.
    3. Else mark it as visited.
    4. Now if temp is the last index of the array, then print “Yes”.
    5. Else take two possible destinations from temp to (temp + arr[temp]), (temp – arr[temp]) and push it into the queue.
  • If the end of the array is not reached after the above steps then print “No”.

Below is the implementation of the above approach: 

C++




// 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;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to check if we can reach to
// the end of the arr[] with possible moves
static void solve(int arr[], int n, int start)
{
 
    // Queue to perform BFS
    Queue<Integer> q = new LinkedList<>();
 
    // Initially all nodes(index)
    // are not visited.
    boolean []visited = new boolean[n];
 
    // Initially the end of
    // the array is not reached
    boolean reached = false;
 
    // Push start index in queue
    q.add(start);
 
    // Until queue becomes empty
    while (!q.isEmpty())
    {
 
        // Get the top element
        int temp = q.peek();
 
        // Delete popped element
        q.remove();
 
        // 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.add(temp + arr[temp]);
        }
 
        if (temp - arr[temp] >= 0)
        {
            q.add(temp - arr[temp]);
        }
    }
 
    // If reaches the end of the array,
    // Print "Yes"
    if (reached == true)
    {
        System.out.print("Yes");
    }
 
    // Else print "No"
    else
    {
        System.out.print("No");
    }
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array arr[]
    int arr[] = { 4, 1, 3, 2, 5 };
 
    // Starting index
    int S = 1;
    int N = arr.length;
 
    // Function call
    solve(arr, N, S);
}
}
 
// This code is contributed by gauravrajput1

Python3




# Python3 program for the above approach
from queue import Queue
  
# Function to check if we can reach to
# the end of the arr[] with possible moves
def solve(arr, n, start):
  
    # Queue to perform BFS
    q = Queue()
  
    # Initially all nodes(index)
    # are not visited.
    visited = [False] * n
    
    # Initially the end of
    # the array is not reached
    reached = False
    
    # Push start index in queue
    q.put(start);
  
    # Until queue becomes empty
    while (not q.empty()):
  
        # Get the top element
        temp = q.get()
  
        # 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.put(temp + arr[temp])
  
        if (temp - arr[temp] >= 0):
            q.put(temp - arr[temp])
  
    # If reaches the end of the array,
    # Print "Yes"
    if (reached == True):
        print("Yes")
  
    # Else print "No"
    else:
        print("No")
  
# Driver code
if __name__ == '__main__':
  
    # Given array arr[]
    arr = [ 4, 1, 3, 2, 5 ]
  
    # starting index
    S = 1
    N = len(arr)
  
    # Function call
    solve(arr, N, S)
  
# This code is contributed by himanshu77

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to check if we can reach to
// the end of the []arr with possible moves
static void solve(int []arr, int n,
                  int start)
{
 
    // Queue to perform BFS
    Queue<int> q = new Queue<int>();
 
    // Initially all nodes(index)
    // are not visited.
    bool []visited = new bool[n];
 
    // Initially the end of
    // the array is not reached
    bool reached = false;
 
    // Push start index in queue
    q.Enqueue(start);
 
    // Until queue becomes empty
    while (q.Count != 0)
    {
 
        // Get the top element
        int temp = q.Peek();
 
        // Delete popped element
        q.Dequeue();
 
        // 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.Enqueue(temp + arr[temp]);
        }
 
        if (temp - arr[temp] >= 0)
        {
            q.Enqueue(temp - arr[temp]);
        }
    }
 
    // If reaches the end of the array,
    // Print "Yes"
    if (reached == true)
    {
        Console.Write("Yes");
    }
 
    // Else print "No"
    else
    {
        Console.Write("No");
    }
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given array []arr
    int []arr = { 4, 1, 3, 2, 5 };
 
    // Starting index
    int S = 1;
    int N = arr.Length;
 
    // Function call
    solve(arr, N, S);
}
}
 
// This code is contributed by gauravrajput1

Javascript




<script>
 
// Javascript program for the above approach
 
// Function to check if we can reach to
// the end of the arr[] with possible moves
function solve(arr, n, start)
{
     
    // Queue to perform BFS
    let q = [];
 
    // Initially all nodes(index)
    // are not visited.
    let visited = new Array(n);
    visited.fill(false);
 
    // Initially the end of
    // the array is not reached
    let reached = false;
 
    // Push start index in queue
    q.push(start);
 
    // Until queue becomes empty
    while (q.length > 0)
    {
         
        // Get the top element
        let temp = q[0];
 
        // Delete popped element
        q.shift();
 
        // 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)
    {
        document.write("Yes");
    }
 
    // Else print "No"
    else
    {
        document.write("No");
    }
}
 
// Driver code
 
// Given array arr[]
let arr = [ 4, 1, 3, 2, 5 ];
 
// Starting index
let S = 1;
let N = arr.length;
 
// Function Call
solve(arr, N, S);
 
// This code is contributed by divyeshrabadiya07 
 
</script>
Output: 
Yes

 

Time Complexity: O(N)  
Auxiliary Space: O(N)

Approach 2: 

  1. Another approach to the same can be by using recursion.
  2. Use recursion to jump to the i + arr[i] and i – arr[i] position of the array and check if we have reached the end or not.
  3. The advantage of using the recursion approach is that it simplifies the code a lot. Below is the implementation.

C++




// C++ program for the above approach
#include <iostream>
using namespace std;
bool check_the_end(int arr[], int n, int i)
{
   
    // If we have reached out of bounds
    // of the array then return False
    if (i < 0 or i >= n)
        return false;
 
    // If we have reached the end then return True
    if (i == n - 1)
        return true;
 
    // Either of the condition return true solved the
    // problem
    return check_the_end(arr, n, i - arr[i])
           or check_the_end(arr, n, i + arr[i]);
}
 
// Driver Code
int main()
{
    int arr[] = { 4, 1, 3, 2, 5 };
    int S = 1;
    int n = sizeof(arr) / sizeof(arr[0]);
    bool result = check_the_end(arr, n, S);
    cout << result;
 
}
 
    // This Code is Contributed by Harshit Srivastava

Java




// Java program for the above approach
import java.io.*;
 
class GFG {
    static boolean check_the_end(int arr[], int n, int i)
    {
       
        // If we have reached out of bounds
        // of the array then return False
        if (i < 0 || i >= n)
            return false;
     
        // If we have reached the end then return True
        if (i == n - 1)
            return true;
     
        // Either of the condition return true solved the
        // problem
        return check_the_end(arr, n, i - arr[i])
               || check_the_end(arr, n, i + arr[i]);
    }
     
    // Driver Code
    public static void main (String[] args) {
        int arr[] = { 4, 1, 3, 2, 5 };
        int S = 1;
        int n = arr.length;
        boolean result = check_the_end(arr, n, S);
        System.out.println(result);
     
    }
}
 
// This Code is Contributed by Shubhamsingh10

Python3




# Python program for the above approach
def check_the_end(arr, i):
   
    # If we have reached out of bounds
    # of the array then return False
    if i < 0 or i >= len(arr):
        return False
       
    # If we have reached the end then return True
    if i == len(arr) - 1:
        return True
       
    # Either of the condition return true solved the problem
    return check_the_end(arr, i - arr[i]) or
            check_the_end(arr, i + arr[i])
 
 
# Driver Code
arr = [4, 1, 3, 2, 5]
S = 1
result = check_the_end(arr, S)
print(result)

C#




// C# program for the above approach
using System;
 
public class GFG{
    static bool check_the_end(int[] arr, int n, int i)
    {
       
        // If we have reached out of bounds
        // of the array then return False
        if (i < 0 || i >= n)
            return false;
     
        // If we have reached the end then return True
        if (i == n - 1)
            return true;
     
        // Either of the condition return true solved the
        // problem
        return check_the_end(arr, n, i - arr[i])
               || check_the_end(arr, n, i + arr[i]);
    }
     
    // Driver Code
    static public void Main (){
        int[] arr = { 4, 1, 3, 2, 5 };
        int S = 1;
        int n = arr.Length;
        bool result = check_the_end(arr, n, S);
        Console.WriteLine(result);
     
    }
}
 
// This Code is Contributed by Shubhamsingh10

Javascript




<script>
 
// JavaScript program for the above approach
function check_the_end(arr, n, i)
{
    // If we have reached out of bounds
    // of the array then return False
    if (i < 0 || i >= n)
        return false;
  
    // If we have reached the end then return True
    if (i == n - 1)
        return true;
  
    // Either of the condition return true solved the
    // problem
    return check_the_end(arr, n, i - arr[i])
           || check_the_end(arr, n, i + arr[i]);
}
 
// Driver Code
var arr = [4, 1, 3, 2, 5];
var S = 1
var n = arr.length;
var result = check_the_end(arr,n, S);
document.write(result)
 
// This code is contributed by Shivani
 
</script>

Output:

True

Time Complexity: O(N)  
Auxiliary Space: O(N)


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