# Minimize the number of steps required to reach the end of the array | Set 2

• Difficulty Level : Hard
• Last Updated : 13 May, 2021

Given an integer array arr[] of length N consisting of positive integers, the task is to minimize the number of steps required to reach the arr[N – 1] starting from arr. At a given step if we are at index i we can go to index i – arr[i] or i + arr[i] given we have not visited those indexes before. Also, we cannot go outside the bounds of the array. Print -1 if there is no possible way.

Examples:

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Input: arr[] = {1, 1, 1}
Output:
The path will be 0 -> 1 -> 2.
Input: arr[] = {2, 1}
Output: -1

Approach: We have already discussed a dynamic programming based approach in this article which has a time complexity of O(n * 2n)
Here we’re going to discuss a BFS based solution:

1. This problem can be visualized as a directed graph where ith cell is connected with cells i + arr[i] and i – arr[i].
2. And the graph is un-weighted.

Due to the above, BFS can be used to find the shortest path between 0th and the (N – 1)th index. We will use the following algorithm:

1. Push index 0 in a queue.
2. Push all the adjacent cells to 0 in the queue.
3. Repeat the above steps i.e. traverse all the elements in the queue individually again if they have not been visited/traversed before.
4. Repeat till we don’t reach the index N – 1.
5. The depth of this traversal will give the minimum steps required to reach the end.

Remember to mark a cell visited after it has been traversed. For this, we will use a boolean array.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the minimum steps``// required to reach the end``// of the given array``int` `minSteps(``int` `arr[], ``int` `n)``{``    ``// Array to determine whether``    ``// a cell has been visited before``    ``bool` `v[n] = { 0 };` `    ``// Queue for bfs``    ``queue<``int``> q;` `    ``// Push the source i.e. index 0``    ``q.push(0);` `    ``// Variable to store``    ``// the depth of search``    ``int` `depth = 0;` `    ``// BFS algorithm``    ``while` `(q.size() != 0) {` `        ``// Current queue size``        ``int` `x = q.size();``        ``while` `(x--) {` `            ``// Top-most element of queue``            ``int` `i = q.front();``            ``q.pop();` `            ``// Base case``            ``if` `(v[i])``                ``continue``;` `            ``// If we reach the destination``            ``// i.e. index (n - 1)``            ``if` `(i == n - 1)``                ``return` `depth;` `            ``// Marking the cell visited``            ``v[i] = 1;` `            ``// Pushing the adjacent nodes``            ``// i.e. indices reachable``            ``// from the current index``            ``if` `(i + arr[i] < n)``                ``q.push(i + arr[i]);``            ``if` `(i - arr[i] >= 0)``                ``q.push(i - arr[i]);``        ``}``        ``depth++;``    ``}` `    ``return` `-1;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 1, 1, 1, 1, 1 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``);` `    ``cout << minSteps(arr, n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{` `// Function to return the minimum steps``// required to reach the end``// of the given array``static` `int` `minSteps(``int` `arr[], ``int` `n)``{``    ``// Array to determine whether``    ``// a cell has been visited before``    ``boolean``[] v = ``new` `boolean``[n];` `    ``// Queue for bfs``    ``Queue q = ``new` `LinkedList<>();` `    ``// Push the source i.e. index 0``    ``q.add(``0``);` `    ``// Variable to store``    ``// the depth of search``    ``int` `depth = ``0``;` `    ``// BFS algorithm``    ``while` `(q.size() > ``0``)``    ``{` `        ``// Current queue size``        ``int` `x = q.size();``        ``while` `(x-- > ``0``)``        ``{` `            ``// Top-most element of queue``            ``int` `i = q.peek();``            ``q.poll();` `            ``// Base case``            ``if` `(v[i])``                ``continue``;` `            ``// If we reach the destination``            ``// i.e. index (n - 1)``            ``if` `(i == n - ``1``)``                ``return` `depth;` `            ``// Marking the cell visited``            ``v[i] = ``true``;` `            ``// Pushing the adjacent nodes``            ``// i.e. indices reachable``            ``// from the current index``            ``if` `(i + arr[i] < n)``                ``q.add(i + arr[i]);``            ``if` `(i - arr[i] >= ``0``)``                ``q.add(i - arr[i]);``        ``}``        ``depth++;``    ``}` `    ``return` `-``1``;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``1``, ``1``, ``1``, ``1``, ``1``, ``1` `};``    ``int` `n = arr.length;``    ``System.out.println(minSteps(arr, n));``}``}` `/* This code contributed by PrinciRaj1992 */`

## Python3

 `# Python 3 implementation of the approach` `# Function to return the minimum steps``# required to reach the end``# of the given array``def` `minSteps(arr,n):``    ` `    ``# Array to determine whether``    ``# a cell has been visited before``    ``v ``=` `[``0` `for` `i ``in` `range``(n)]` `    ``# Queue for bfs``    ``q ``=` `[]` `    ``# Push the source i.e. index 0``    ``q.append(``0``)` `    ``# Variable to store``    ``# the depth of search``    ``depth ``=` `0` `    ``# BFS algorithm``    ``while` `(``len``(q) !``=` `0``):``        ``# Current queue size``        ``x ``=` `len``(q)``        ``while` `(x >``=` `1``):``            ``# Top-most element of queue``            ``i ``=` `q[``0``]``            ``q.remove(i)``            ` `            ``x ``-``=` `1` `            ``# Base case``            ``if` `(v[i]):``                ``continue``;` `            ``# If we reach the destination``            ``# i.e. index (n - 1)``            ``if` `(i ``=``=` `n ``-` `1``):``                ``return` `depth` `            ``# Marking the cell visited``            ``v[i] ``=` `1` `            ``# Pushing the adjacent nodes``            ``# i.e. indices reachable``            ``# from the current index``            ``if` `(i ``+` `arr[i] < n):``                ``q.append(i ``+` `arr[i])``            ``if` `(i ``-` `arr[i] >``=` `0``):``                ``q.append(i ``-` `arr[i])``                ` `        ` `        ``depth ``+``=` `1` `    ``return` `-``1` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``1``, ``1``, ``1``, ``1``, ``1``, ``1``]``    ``n ``=` `len``(arr)` `    ``print``(minSteps(arr, n))` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// A C# implementation of the approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `// Function to return the minimum steps``// required to reach the end``// of the given array``static` `int` `minSteps(``int` `[]arr, ``int` `n)``{``    ``// Array to determine whether``    ``// a cell has been visited before``    ``Boolean[] v = ``new` `Boolean[n];` `    ``// Queue for bfs``    ``Queue<``int``> q = ``new` `Queue<``int``>();` `    ``// Push the source i.e. index 0``    ``q.Enqueue(0);` `    ``// Variable to store``    ``// the depth of search``    ``int` `depth = 0;` `    ``// BFS algorithm``    ``while` `(q.Count > 0)``    ``{` `        ``// Current queue size``        ``int` `x = q.Count;``        ``while` `(x-- > 0)``        ``{` `            ``// Top-most element of queue``            ``int` `i = q.Peek();``            ``q.Dequeue();` `            ``// Base case``            ``if` `(v[i])``                ``continue``;` `            ``// If we reach the destination``            ``// i.e. index (n - 1)``            ``if` `(i == n - 1)``                ``return` `depth;` `            ``// Marking the cell visited``            ``v[i] = ``true``;` `            ``// Pushing the adjacent nodes``            ``// i.e. indices reachable``            ``// from the current index``            ``if` `(i + arr[i] < n)``                ``q.Enqueue(i + arr[i]);``            ``if` `(i - arr[i] >= 0)``                ``q.Enqueue(i - arr[i]);``        ``}``        ``depth++;``    ``}` `    ``return` `-1;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]arr = { 1, 1, 1, 1, 1, 1 };``    ``int` `n = arr.Length;``    ``Console.WriteLine(minSteps(arr, n));``}``}` `// This code contributed by Rajput-Ji`

## Javascript

 ``
Output:
`5`

Time complexity: O(N)

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