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

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)

Recursive approach:

Approach:

In this approach, we recursively check if we can reach the end of the array by making valid jumps from the current position.

Define a function can_reach_end that takes an array arr and a starting position pos.
Check the base case: If the current position is the last index of the array, then we can reach the end, so return True.
Find the maximum number of steps we can take from the current position without going out of bounds, i.e., max_jump.
Loop through all possible next positions, starting from pos + 1 up to max_jump.
For each next position, recursively call can_reach_end with the next position as the new starting position.
If any of the recursive calls return True, then we can reach the end, so return True.
If we can’t reach the end from any of the possible jumps, return False.

C++

#include <iostream>
#include <vector>
using namespace std;

bool canReachEnd(vector<int>& arr, int pos) {
    // Base case: If the current position is the last index of the array, then we can reach the end.
    if (pos == arr.size() - 1) {
        return true;
    }

    // Recursive case: Try all possible jumps from the current position.
    int maxJump = min(pos + arr[pos], static_cast<int>(arr.size() - 1));
    for (int nextPos = pos + 1; nextPos <= maxJump; nextPos++) {
        if (canReachEnd(arr, nextPos)) {
            return true;
        }
    }

    // If we can't reach the end from any of the possible jumps, return false.
    return false;
}

int main() {
    vector<int> arr = {4, 1, 3, 2, 5};
    int startPos = 1;

    // Example usage
    if (canReachEnd(arr, startPos)) {
        cout << "Yes" << endl;
    } else {
        cout << "No" << endl;
    }

    return 0;
}

Java

import java.io.*;
public class GFG
{
    public static boolean canReachEnd(int[] arr, int pos) {
        // Base case: If the current position is the last index of the array then we can reach the end.
        if (pos == arr.length - 1) {
            return true;
        }
        // Recursive case: Try all possible jumps from the current position.
        int maxJump = Math.min(pos + arr[pos], arr.length - 1);
        for (int nextPos = pos + 1; nextPos <= maxJump; nextPos++) {
            if (canReachEnd(arr, nextPos)) {
                return true;
            }
        }
        // If we can't reach the end from any of the possible jumps return false.
        return false;
    }
    public static void main(String[] args) {
        int[] arr = {4, 1, 3, 2, 5};
        int startPos = 1;
        // Example usage
        if (canReachEnd(arr, startPos)) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }
}

Python3

def can_reach_end(arr, pos):
    # Base case: If the current position is the last index of the array, then we can reach the end.
    if pos == len(arr) - 1:
        return True

    # Recursive case: Try all possible jumps from the current position.
    max_jump = min(pos + arr[pos], len(arr) - 1)
    for next_pos in range(pos + 1, max_jump + 1):
        if can_reach_end(arr, next_pos):
            return True

    # If we can't reach the end from any of the possible jumps, return False.
    return False

# Example usage
arr = [4, 1, 3, 2, 5]
start_pos = 1
if can_reach_end(arr, start_pos):
    print("Yes")
else:
    print("No")

C#

using System;

public class GFG
{
    public static bool CanReachEnd(int[] arr, int pos)
    {
        // Base case: If the current position is the last
        // index of the array then we can reach the end.
        if (pos == arr.Length - 1)
        {
            return true;
        }

        // Recursive case: Try all possible jumps from the current position.
        int maxJump = Math.Min(pos + arr[pos], arr.Length - 1);
        for (int nextPos = pos + 1; nextPos <= maxJump; nextPos++)
        {
            if (CanReachEnd(arr, nextPos))
            {
                return true;
            }
        }

        // If we can't reach the end from any of the possible jumps, return false.
        return false;
    }

    public static void Main(string[] args)
    {
        int[] arr = { 4, 1, 3, 2, 5 };
        int startPos = 1;

        // Example usage
        if (CanReachEnd(arr, startPos))
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }
    }
}

Javascript

function canReachEnd(arr, pos) {
    // Base case: If the current position is the last index of the array then we can reach the end.
    if (pos == arr.length - 1) {
        return true;
    }
    // Recursive case: Try all possible jumps from the current position.
    let maxJump = Math.min(pos + arr[pos], arr.length - 1);
    for (let nextPos = pos + 1; nextPos <= maxJump; nextPos++) {
        if (canReachEnd(arr, nextPos)) {
            return true;
        }
    }
    // If we can't reach the end from any of the possible jumps return false.
    return false;
}

let arr = [4, 1, 3, 2, 5];
let startPos = 1;
// Example usage
if (canReachEnd(arr, startPos)) {
    console.log("Yes");
} else {
    console.log("No");
}

Output

Yes

Time Complexity: O(2^n) where n is the length of the array.
Space Complexity: O(n) where n is the length of the array.