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# Program to find number of minimum steps to reach last index in Python

Suppose we have a list of numbers called nums and we are placed currently at nums[0]. On each step, we can either jump from the current index i to i + 1 or i - 1 or j where nums[i] == nums[j]. We have to find the minimum number of steps required to reach the final index.

So, if the input is like nums = [4, 8, 8, 5, 4, 6, 5], then the output will be 3, as we can jump from index 0 to index 4 as their values are both 4. And then we jump back to index 3. Finally, we can jump from index 3 to 6 since both of their values are 5.

To solve this, we will follow these steps −

- pos := an empty map
- for each index i, and value n in nums, do
- insert i at the end of pos[n]

- n := size of nums
- visited := make a list of size n, and fill this with False
- visited[0] := True
- while q is not empty, do
- (u, d) := left element of q, and delete left element
- if u is same as n - 1, then
- return d

- for each v in the lists pos[nums[u]] and [u - 1, u + 1], do
- if 0 <= v < n and visited[v] is false, then
- visited[v] := True
- insert pair (v, d + 1) at the end of q

- if 0 <= v < n and visited[v] is false, then
- remove pos[nums[u]]

Let us see the following implementation to get better understanding −

## Example

class Solution: def solve(self, nums): from collections import defaultdict, deque pos = defaultdict(list) for i, n in enumerate(nums): pos[n].append(i) q = deque([(0, 0)]) n = len(nums) visited = [False] * n visited[0] = True while q: u, d = q.popleft() if u == n - 1: return d for v in pos[nums[u]] + [u - 1, u + 1]: if 0 <= v < n and not visited[v]: visited[v] = True q.append((v, d + 1)) del pos[nums[u]] ob = Solution() nums = [4, 8, 8, 5, 4, 6, 5] print(ob.solve(nums))

## Input

[4, 8, 8, 5, 4, 6, 5]

## Output

3

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