Syntax error Program to Find Out the Number of Moves to Reach the Finish Line in Python

Program to Find Out the Number of Moves to Reach the Finish Line in Python

Suppose, we have a car and are driving it on a one−dimensional road. Currently we are at position = 0 and with speed = 1. We can perform any of these two operations.

  • Acceleration: position := position + speed and speed := speed * 2 Reverse Gear: speed := −1 when speed > 0 otherwise speed := 1.

We have to find the number of moves needed at least to reach the target.

So, if the input is like target = 10, then the output will be 7.

To solve this, we will follow these steps −

  • Define a function dfs() . This will take digit, cost, pos, neg, target

    • tot := cost + maximum of 2 *(pos − 1) and 2 * (neg − 1)

    • if tot >= ans, then

      • return

    • if target is same as 0, then

      • ans := minimum of ans and tot

      • return

    • step := (2^digit) − 1

    • if step * 2 < |target|, then

      • return

    • dfs(digit − 1, cost, pos, neg, target)

    • dfs(digit − 1, cost + digit, pos + 1, neg, target − step)

    • dfs(digit − 1, cost + digit * 2, pos + 2, neg, target − step * 2)

    • dfs(digit − 1, cost + digit, pos, neg + 1, target + step)

    • dfs(digit − 1, cost + digit * 2, pos, neg + 2, target + step * 2)

  • From the main function, do the following −

  • ans := infinity

  • hi := 1

  • while 2^hi < target, do

    • hi := hi + 1

  • dfs(hi, 0, 0, 0, target)

  • return ans

Let us see the following implementation to get better understanding −

Example

 Live Demo

class Solution:
   def solve(self, target):
      self.ans = int(1e9)
      hi = 1
      while (1 << hi) < target:
         hi += 1
      self.dfs(hi, 0, 0, 0, target)
      return self.ans
   def dfs(self, digit, cost, pos, neg, target):
      tot = cost + max(2 * (pos − 1), 2 * neg − 1)
      if tot >= self.ans:
         return
      if target == 0:
         self.ans = min(self.ans, tot)
         return
      step = (1 << digit) − 1
      if step * 2 < abs(target):
         return
      self.dfs(digit − 1, cost, pos, neg, target)
      self.dfs(digit − 1, cost + digit, pos + 1, neg, target − step)
      self.dfs(digit − 1, cost + digit * 2, pos + 2, neg, target − step * 2)
      self.dfs(digit − 1, cost + digit, pos, neg + 1, target + step)
      self.dfs(digit − 1, cost + digit * 2, pos, neg + 2, target + step * 2)
ob = Solution()
print(ob.solve(10))

Input

10

Output

7
Updated on: 2020-12-26T11:26:06+05:30

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