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Check if the end of the Array can be reached from a given position
• Difficulty Level : Medium
• Last Updated : 19 Mar, 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.
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.
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 ``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);` `    ``// Untill 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);` `    ``// 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 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);` `    ``// Untill 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);`` ` `    ``# Untill 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);` `    ``// Untill 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`
Output:

`Yes`

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.

## 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)`

Output:

`True`

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