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Minimize the number of steps required to reach the end of the array
• Difficulty Level : Expert
• Last Updated : 10 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 index ‘N-1’. 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 not possible way.
Examples:

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

Input : {2, 1}
Output : -1```

This problem can be solved using dynamic programming approach.
Let’s discuss an approach that might seem correct at start. Let’s suppose we are at ith index. Can we directly say that dp[i] = 1 + min( dp[i-arr[i]], dp[i+arr[i]] ) ?
No, we cannot. The path we took to reach the index ‘i’ also matters as indexes we used before won’t be available to visit anymore.
Thus, the only approach we are left with is trying out all the possible combinations which can be really large. In this article, we will use bit-masking approach to reduce the complexity to exponential. Our mask will be an integer value with the following features.

```1) If, a index 'i' is visited, ith bit
will be set 1 in the mask.
2) Else that bit will be set 0.```

The required recurrence relation will be.

```dp[i][mask] = 1 + min(dp[i+arr[i]][mask|(1<<i)],
dp[i-arr[i]][mask|(1<<i)])```

Below is the implementation of the above approach:

## CPP

 `// C++ implementation of the above approach` `#include ``#define maxLen 10``#define maskLen 130``using` `namespace` `std;` `// variable to store states of dp``int` `dp[maxLen][maskLen];` `// variable to check if a given state``// has been solved``bool` `v[maxLen][maskLen];` `// Function to find the minimum number of steps``// required to reach the end of the array``int` `minSteps(``int` `arr[], ``int` `i, ``int` `mask, ``int` `n)``{``    ``// base case``    ``if` `(i == n - 1)``        ``return` `0;` `    ``if` `(i > n - 1 || i < 0)``        ``return` `9999999;``    ``if` `((mask >> i) & 1)``        ``return` `9999999;` `    ``// to check if a state has``    ``// been solved``    ``if` `(v[i][mask])``        ``return` `dp[i][mask];``    ``v[i][mask] = 1;` `    ``// required recurrence relation``    ``dp[i][mask] = 1 + min(minSteps(arr, i - arr[i], (mask | (1 << i)), n),``                          ``minSteps(arr, i + arr[i], (mask | (1 << i)), n));` `    ``// returning the value``    ``return` `dp[i][mask];``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 2, 2, 2, 1, 1 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``);` `    ``int` `ans = minSteps(arr, 0, 0, n);``    ``if` `(ans >= 9999999)``        ``cout << -1;``    ``else``        ``cout << ans;``}`

## Java

 `// Java implementation of the above approach` `class` `GFG``{` `    ``static` `int` `maxLen = ``10``;``    ``static` `int` `maskLen = ``130``;` `    ``// variable to store states of dp``    ``static` `int``[][] dp = ``new` `int``[maxLen][maskLen];` `    ``// variable to check if a given state``    ``// has been solved``    ``static` `boolean``[][] v = ``new` `boolean``[maxLen][maskLen];` `    ``// Function to find the minimum number of steps``    ``// required to reach the end of the array``    ``static` `int` `minSteps(``int` `arr[], ``int` `i, ``int` `mask, ``int` `n)``    ``{``        ``// base case``        ``if` `(i == n - ``1``)``        ``{``            ``return` `0``;``        ``}` `        ``if` `(i > n - ``1` `|| i < ``0``)``        ``{``            ``return` `9999999``;``        ``}``        ``if` `((mask >> i) % ``2` `== ``1``)``        ``{``            ``return` `9999999``;``        ``}` `        ``// to check if a state has``        ``// been solved``        ``if` `(v[i][mask])``        ``{``            ``return` `dp[i][mask];``        ``}``        ``v[i][mask] = ``true``;` `        ``// required recurrence relation``        ``dp[i][mask] = ``1` `+ Math.min(minSteps(arr, i - arr[i], (mask | (``1` `<< i)), n),``                                ``minSteps(arr, i + arr[i], (mask | (``1` `<< i)), n));` `        ``// returning the value``        ``return` `dp[i][mask];``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = {``1``, ``2``, ``2``, ``2``, ``1``, ``1``};` `        ``int` `n = arr.length;` `        ``int` `ans = minSteps(arr, ``0``, ``0``, n);``        ``if` `(ans >= ``9999999``)``        ``{``            ``System.out.println(-``1``);``        ``}``        ``else``        ``{``            ``System.out.println(ans);``        ``}``    ``}``}` `/* This code contributed by PrinciRaj1992 */`

## Python

 `# Python3 implementation of the above approach``maxLen ``=` `10``maskLen ``=` `130`  `# variable to store states of dp``dp ``=` `[[ ``0` `for` `i ``in` `range``(maskLen)] ``for` `i ``in` `range``(maxLen)]` `# variable to check if a given state``# has been solved``v ``=` `[[``False` `for` `i ``in` `range``(maskLen)] ``for` `i ``in` `range``(maxLen)]` `# Function to find the minimum number of steps``# required to reach the end of the array``def` `minSteps(arr, i, mask, n):` `    ``# base case``    ``if` `(i ``=``=` `n ``-` `1``):``        ``return` `0` `    ``if` `(i > n ``-` `1` `or` `i < ``0``):``        ``return` `9999999` `    ``if` `((mask >> i) & ``1``):``        ``return` `9999999` `    ``# to check if a state has``    ``# been solved``    ``if` `(v[i][mask] ``=``=` `True``):``        ``return` `dp[i][mask]``    ``v[i][mask] ``=` `True` `    ``# required recurrence relation``    ``dp[i][mask] ``=` `1` `+` `min``(minSteps(arr, i ``-` `arr[i], (mask | (``1` `<< i)), n),``                        ``minSteps(arr, i ``+` `arr[i], (mask | (``1` `<< i)), n))` `    ``# returning the value``    ``return` `dp[i][mask]`  `# Driver code` `arr``=``[``1``, ``2``, ``2``, ``2``, ``1``, ``1``]` `n ``=` `len``(arr)` `ans ``=` `minSteps(arr, ``0``, ``0``, n)` `if` `(ans >``=` `9999999``):``    ``print``(``-``1``)``else``:``    ``print``(ans)``    ` `# This code is contributed by mohit kumar 29   `

## C#

 `// C# implementation of the above approach``using` `System;` `class` `GFG``{``    ` `    ``static` `int` `maxLen = 10;``    ``static` `int` `maskLen = 130;` `    ``// variable to store states of dp``    ``static` `int``[,] dp = ``new` `int``[maxLen, maskLen];` `    ``// variable to check if a given state``    ``// has been solved``    ``static` `bool``[,] v = ``new` `bool``[maxLen, maskLen];` `    ``// Function to find the minimum number of steps``    ``// required to reach the end of the array``    ``static` `int` `minSteps(``int` `[]arr, ``int` `i, ``int` `mask, ``int` `n)``    ``{``        ``// base case``        ``if` `(i == n - 1)``        ``{``            ``return` `0;``        ``}` `        ``if` `(i > n - 1 || i < 0)``        ``{``            ``return` `9999999;``        ``}``        ``if` `((mask >> i) % 2 == 1)``        ``{``            ``return` `9999999;``        ``}` `        ``// to check if a state has``        ``// been solved``        ``if` `(v[i, mask])``        ``{``            ``return` `dp[i, mask];``        ``}``        ``v[i, mask] = ``true``;` `        ``// required recurrence relation``        ``dp[i,mask] = 1 + Math.Min(minSteps(arr, i - arr[i], (mask | (1 << i)), n),``                                ``minSteps(arr, i + arr[i], (mask | (1 << i)), n));` `        ``// returning the value``        ``return` `dp[i,mask];``    ``}` `    ``// Driver code``    ``static` `public` `void` `Main ()``    ``{``        ``int` `[]arr = {1, 2, 2, 2, 1, 1};` `        ``int` `n = arr.Length;` `        ``int` `ans = minSteps(arr, 0, 0, n);``        ``if` `(ans >= 9999999)``        ``{``            ``Console.WriteLine(-1);``        ``}``        ``else``        ``{``            ``Console.WriteLine(ans);``        ``}``    ``}``}` `/* This code contributed by ajit. */`

## Javascript

 ``
Output:
`3`

Time Complexity: O(N*(2N))

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