# Minimum steps to reach end of array under constraints

Given an array containing one digit numbers only, assuming we are standing at first index, we need to reach to end of array using minimum number of steps where in one step, we can jump to neighbor indices or can jump to a position with same value.

In other words, if we are at index i, then in one step you can reach to, arr[i-1] or arr[i+1] or arr[K] such that arr[K] = arr[i] (value of arr[K] is same as arr[i])

Examples:

```Input : arr[] = {5, 4, 2, 5, 0}
Output : 2
Explanation : Total 2 step required.
and in second step we move to value 0 (End of arr[]).

Input  : arr[] = [0, 1, 2, 3, 4, 5, 6, 7, 5, 4,
3, 6, 0, 1, 2, 3, 4, 5, 7]
Output : 5
Explanation : Total 5 step required.
0(0) -> 0(12) -> 6(11) -> 6(6) -> 7(7) ->
(18)
(inside parenthesis indices are shown)
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

This problem can be solved using BFS. We can consider the given array as unweighted graph where every vertex has two edges to next and previous array elements and more edges to array elements with same values. Now for fast processing of third type of edges, we keep 10 vectors which store all indices where digits 0 to 9 are present. In above example, vector corresponding to 0 will store [0, 12], 2 indices where 0 has occurred in given array.
Another Boolean array is used, so that we don’t visit same index more than once. As we are using BFS and BFS proceeds level by level, optimal minimum steps are guaranteed.

 `// C++ program to find minimum jumps to reach end ` `// of array ` `#include ` `using` `namespace` `std; ` ` `  `//  Method returns minimum step to reach end of array ` `int` `getMinStepToReachEnd(``int` `arr[], ``int` `N) ` `{ ` `    ``// visit boolean array checks whether current index ` `    ``// is previously visited ` `    ``bool` `visit[N]; ` ` `  `    ``// distance array stores distance of current ` `    ``// index from starting index ` `    ``int` `distance[N]; ` ` `  `    ``// digit vector stores indicies where a ` `    ``// particular number resides ` `    ``vector<``int``> digit; ` ` `  `    ``//  In starting all index are unvisited ` `    ``memset``(visit, ``false``, ``sizeof``(visit)); ` ` `  `    ``//  storing indices of each number in digit vector ` `    ``for` `(``int` `i = 1; i < N; i++) ` `        ``digit[arr[i]].push_back(i); ` ` `  `    ``//  for starting index distance will be zero ` `    ``distance = 0; ` `    ``visit = ``true``; ` ` `  `    ``// Creating a queue and inserting index 0. ` `    ``queue<``int``> q; ` `    ``q.push(0); ` ` `  `    ``//  loop untill queue in not empty ` `    ``while``(!q.empty()) ` `    ``{ ` `        ``// Get an item from queue, q. ` `        ``int` `idx = q.front();        q.pop(); ` ` `  `        ``//  If we reached to last index break from loop ` `        ``if` `(idx == N-1) ` `            ``break``; ` ` `  `        ``// Find value of dequeued index ` `        ``int` `d = arr[idx]; ` ` `  `        ``// looping for all indices with value as d. ` `        ``for` `(``int` `i = 0; i= 0 && !visit[idx - 1]) ` `        ``{ ` `            ``visit[idx - 1] = ``true``; ` `            ``q.push(idx - 1); ` `            ``distance[idx - 1] = distance[idx] + 1; ` `        ``} ` ` `  `        ``//  checking condition for next index ` `        ``if` `(idx + 1 < N && !visit[idx + 1]) ` `        ``{ ` `            ``visit[idx + 1] = ``true``; ` `            ``q.push(idx + 1); ` `            ``distance[idx + 1] = distance[idx] + 1; ` `        ``} ` `    ``} ` ` `  `    ``//  N-1th position has the final result ` `    ``return` `distance[N - 1]; ` `} ` ` `  `//  driver code to test above methods ` `int` `main() ` `{ ` `    ``int` `arr[] = {0, 1, 2, 3, 4, 5, 6, 7, 5, ` `                 ``4, 3, 6, 0, 1, 2, 3, 4, 5, 7}; ` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(``int``); ` `    ``cout << getMinStepToReachEnd(arr, N); ` `    ``return` `0; ` `} `

Output:

```5
```

## tags:

Arrays Graph BFS Arrays Graph BFS