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

PythonServer Side ProgrammingProgramming

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
    • remove pos[nums[u]]

Let us see the following implementation to get better understanding −

Example

 Live Demo

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
raja
Published on 19-Nov-2020 06:18:54
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