Problem Description

In this problem, we are given a "broken" calculator that can perform only two operations:

1. Multiply the current number displayed by 2.
2. Subtract 1 from the current number displayed.

The calculator initially shows a number called startValue. The task is to transform this startValue into another given number called target using the fewest number of operations. We need to find out the minimum operations required to achieve this.

Intuition

The intuitive approach to solve this problem might be to start from startValue and try to reach target using a series of multiplications and subtractions. However, this can be very inefficient because the number of possibilities can explode, leading to a high time complexity solution.

Instead, we reverse our thinking process and start from target and try to reach startValue. We take advantage of the fact that the reverse operations (dividing by 2 and adding 1) are more restricted since we can divide by 2 only when the current number is even. This gives us a direction in our decision-making process and reduces the number of choices at each step, making the problem much simpler.

Here's the step-by-step intuition:

1. If target is greater than startValue, we can only reach it by performing the reverse operations because multiplying startValue may overshoot target.
2. If target is odd, the last operation performed must have been subtracting 1 (since we cannot divide an odd number by 2). So, we add 1 to target.
3. If target is even, the last operation could have been a division by 2, so we divide the target by 2.
4. We count each operation performed, and once target is less than or equal to startValue, we stop.
5. The remaining difference between startValue and target represents the number of times we'd need to subtract 1 from startValue to reach target.

By following these steps, we can ensure that we use the minimum operations to transform startValue into target on the broken calculator.

Learn more about Greedy and Math patterns.

Solution Approach

The Solution provided is a direct implementation of the thought process described in the Intuition section. It's a linear approach, where the algorithm goes through a series of steps to transform the target back to the startValue. The crucial insight is that working backward from the target value is more efficient than trying to approach the target starting from the startValue. This is because multiplying can lead to rapidly overshooting the goal, but working backward constrains the choices.

Here's a step-by-step walkthrough of the implementation:

• A variable ans is initialized to 0 to count the number of operations needed.
• While startValue is less than target, a loop continues to perform the reverse operations to bring target closer to startValue.
  • Inside the loop, first, there is a check to see if target is odd using target & 1. This is a bitwise AND operation, which is equivalent to checking if the last bit of target is 1. If it is odd (true), 1 is added to target to simulate the reverse of a subtract operation.
  • If target is even (false), the target is right-shifted by 1 bit using target >>= 1, which is equivalent to dividing the target by 2.
  • The ans counter is then incremented for each operation performed, whether it's an addition or a division.
• When target is less than or equal to startValue, the loop ends. The final difference between startValue and target indicates how many subtractions would be necessary if we were working forwards from startValue. Thus, startValue - target is added to the ans counter to reflect these operations.
• Finally, ans is returned as the minimum number of operations needed to display target on the calculator from startValue.

This approach is efficient because each iteration reduces target significantly (either by a factor of 2 or approaching the even number when odd), and there are no unneeded complexities in performing the operations, leading to a time complexity that is linear in terms of the number of operations needed.

By continuously halving the target, we ensure that we're using the most significant reduction at each step when possible, and only when target becomes odd, we perform an addition. This way, despite simulating an inefficient calculator, the algorithm efficiently reaches the optimal solution.

Example Walkthrough

Let's walk through an example to illustrate the solution approach using the broken calculator problem.

Example

• startValue = 5
• target = 8

Now we need to find the minimum operations to transform startValue into target.

Solution Steps

1. Since target > startValue, we know we have to work backwards from target to startValue.
2. Initial ans is 0 because no operations have been performed yet.

Now we enter the loop:

• First iteration: target is 8, which is even.
  • We perform the reverse of multiplication by dividing by 2. So, target = target / 2 = 8 / 2 = 4.
  • Increment ans to 1.
• Second iteration: target is 4, which is even.
  • Again, we divide by 2. So, target = target / 2 = 4 / 2 = 2.
  • Increment ans to 2.
• Third iteration: target is 2, which is even.
  • Continue with division. So, target = target / 2 = 2 / 2 = 1.
  • Increment ans to 3.

The loop ends here because target = 1 is less than startValue = 5. We now add startValue - target to ans.

• startValue - target = 5 - 1 = 4. This means we need to perform 4 subtractions if we were moving forwards from startValue.
• Add 4 to ans, which is 3 from earlier. Now ans = 3 + 4 = 7.

Conclusion

The minimum number of operations required for the broken calculator starting from 5 to reach 8 is 7. These operations are three divisions (halving the number three times) and four subtractions.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given code is O(log(target)). The while loop runs until startValue is greater than or equal to target. At each iteration of the loop, if target is even, it is halved (which significantly decreases the target in logarithmic steps), or if it's odd, it is incremented by 1, which eventually makes it even for the next step. Since target is divided by 2 in potentially every other iteration, the loop runs in O(log(target)) time with respect to the target value.

Space Complexity

The space complexity of the code is O(1). The solution does not use any additional storage that grows with the size of the input. It uses a fixed amount of space for the variables ans, startValue, and target irrespective of the input size.

Learn more about how to find time and space complexity quickly using problem constraints.