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

PythonServer Side ProgrammingProgramming

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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