# Race to Finish Line - Google Top Interview Questions

### Problem Statement :

```You are driving a car in a one-dimensional line and are currently at position = 0 with speed = 1. You can make one of two moves:

Accelerate: position += speed and speed *= 2

Reverse: speed = -1 if speed > 0 otherwise speed = 1.

Return the minimum number of moves it would take to reach target.

Constraints

1 ≤ target ≤ 100,000

Example 1

Input

target = 7

Output

3

Explanation

We can accelerate 3 times to reach 7. 0 -> 1 -> 3 -> 7

Example 2

Input

target = 6

Output

5

Explanation

We can accelerate 3 times to reach 7. 0 -> 1 -> 3 -> 7. Then we reverse to change our speed to -1. Then we accelerate to reach 6.```

### Solution :

```                        ```Solution in C++ :

int dp;
// dp[i] is the minimum time needed to travel exactly a distance of i

int solve(int target) {
if (dp[target]) {
return dp[target];
}
int time = 0;
int speed = 1;
int position = 0;
// this is achievable by accelerating, then reversing twice, repeating "target" times until we
// get to the target
int besttime = 3 * target;
while (true) {
position += speed;
time++;
if (position == target) {
besttime = min(besttime, time);
break;
}
// reverse once, but since we're after the target we don't continue traveling further
// forward
if (position > target) {
besttime = min(besttime, time + 1 + solve(position - target));
break;
}
// reverse twice to reset speed to 1
besttime = min(besttime, time + 2 + solve(target - position));
// reverse in the middle, but don't reverse all the way
if (speed > 1) {
int candtime = time + 2;
int candposition = target - position;
for (int revspeed = 1; candposition + revspeed < target; revspeed *= 2) {
candposition += revspeed;
candtime++;
besttime = min(besttime, candtime + solve(candposition));
}
}
speed *= 2;
}
dp[target] = besttime;
return besttime;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(int target) {
if (target == 0)
return 0;
boolean[][] ff = new boolean[target * 2 + 1];
ff = true;
int lay = 0, count = 1;
while (!ll1.isEmpty()) {
int p = ll1.poll(), s = ll2.poll();
int speed = s < 19 ? 1 << s : -(1 << (s - 19));
if (p + speed == target)
return lay + 1;
if (p + speed >= 0 && p + speed < ff.length && !ff[s + 1][p + speed]) {
ff[s + 1][p + speed] = true;
}
s = s < 19 ? 19 : 0;
if (!ff[s][p]) {
ff[s][p] = true;
}
--count;
if (count == 0) {
++lay;
count = ll1.size();
}
}
return -1;
}
}```
```

```                        ```Solution in Python :

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)```
```

## Super Maximum Cost Queries

Victoria has a tree, T , consisting of N nodes numbered from 1 to N. Each edge from node Ui to Vi in tree T has an integer weight, Wi. Let's define the cost, C, of a path from some node X to some other node Y as the maximum weight ( W ) for any edge in the unique path from node X to Y node . Victoria wants your help processing Q queries on tree T, where each query contains 2 integers, L and

## Contacts

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## No Prefix Set

There is a given list of strings where each string contains only lowercase letters from a - j, inclusive. The set of strings is said to be a GOOD SET if no string is a prefix of another string. In this case, print GOOD SET. Otherwise, print BAD SET on the first line followed by the string being checked. Note If two strings are identical, they are prefixes of each other. Function Descriptio

## Cube Summation

You are given a 3-D Matrix in which each block contains 0 initially. The first block is defined by the coordinate (1,1,1) and the last block is defined by the coordinate (N,N,N). There are two types of queries. UPDATE x y z W updates the value of block (x,y,z) to W. QUERY x1 y1 z1 x2 y2 z2 calculates the sum of the value of blocks whose x coordinate is between x1 and x2 (inclusive), y coor

## Direct Connections

Enter-View ( EV ) is a linear, street-like country. By linear, we mean all the cities of the country are placed on a single straight line - the x -axis. Thus every city's position can be defined by a single coordinate, xi, the distance from the left borderline of the country. You can treat all cities as single points. Unfortunately, the dictator of telecommunication of EV (Mr. S. Treat Jr.) do

## Subsequence Weighting

A subsequence of a sequence is a sequence which is obtained by deleting zero or more elements from the sequence. You are given a sequence A in which every element is a pair of integers i.e A = [(a1, w1), (a2, w2),..., (aN, wN)]. For a subseqence B = [(b1, v1), (b2, v2), ...., (bM, vM)] of the given sequence : We call it increasing if for every i (1 <= i < M ) , bi < bi+1. Weight(B) =