What is the time complexity of Broken Calculator?

The optimal solution has a time complexity of O(log(target)).

What data structures are used to solve Broken Calculator?

This problem uses Math, Greedy. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Broken Calculator?

Insight 1: Working backward from the target is more efficient. Instead of trying to reach the target from the start, we can reverse the operations: divide the target by 2 (if even) or add 1 to the target. Insight 2: Since startValue and target are positive, we can deduce that when target is greater than startValue, we should try to reach to target from the startValue. Doubling results in a larger number and subtracting reduces to a smaller number. If startValue is greater than target then the only way to reach target is by subtracting 1.

Which companies ask LeetCode 991. Broken Calculator in interviews?

This problem is frequently asked at Millennium, Nutanix.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Broken Calculator - LeetCode 991 Solution

Broken Calculator - Complete Solution Guide

Broken Calculator is LeetCode problem 991, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There is a broken calculator that has the integer startValue on its display initially. In one operation, you can: multiply the number on display by 2 , or subtract 1 from the number on display. Given two integers startValue and target , return the minimum number of operations needed to display target on the calculator . Example 1: Input: startValue = 2, target = 3 Output: 2 Explanation: Use double operation and then decrement operation {2 -> 4 -> 3}. Example 2: Input: startValue = 5, target = 8

Detailed Explanation

The problem presents a broken calculator with only two operations: doubling the displayed number or subtracting one. Given an initial value `startValue` and a target value `target`, the goal is to determine the minimum number of operations required to transform `startValue` into `target` using only these two operations. The constraints specify that both `startValue` and `target` are positive integers between 1 and 10^9.

Solution Approach

The provided solution employs a greedy approach by working backward from the `target` to the `startValue`. It iteratively applies reverse operations: if the `target` is even, it divides it by 2 (undoing the doubling); otherwise, it adds 1 to the `target` (undoing the subtraction). This process continues until the `target` becomes less than or equal to the `startValue`. Finally, it adds the difference between `startValue` and `target` to the operation count to account for the remaining subtractions needed to reach the `startValue` from `target` (if `target` is less than `startValue` at the end).

Step-by-Step Algorithm

Step 1: Initialize a variable `operations` to 0. This variable will store the count of operations.
Step 2: Enter a `while` loop that continues as long as `target` is greater than `startValue`.
Step 3: Inside the loop, increment `operations` by 1 for each operation performed.
Step 4: Check if `target` is even (i.e., `target % 2 == 0`).
Step 5: If `target` is even, divide it by 2 (`target //= 2`). This simulates reversing the doubling operation.
Step 6: If `target` is odd, add 1 to it (`target += 1`). This simulates reversing the subtraction operation.
Step 7: After the loop finishes (when `target` is no longer greater than `startValue`), add the difference `startValue - target` to `operations`. This accounts for the number of subtraction operations needed to reach `startValue` from the remaining `target` if target is less than startValue. Note this will be zero if target is startValue after the loop.
Step 8: Return the final value of `operations`.

Key Insights

Insight 1: Working backward from the target is more efficient. Instead of trying to reach the target from the start, we can reverse the operations: divide the target by 2 (if even) or add 1 to the target.
Insight 2: Since startValue and target are positive, we can deduce that when target is greater than startValue, we should try to reach to target from the startValue. Doubling results in a larger number and subtracting reduces to a smaller number. If startValue is greater than target then the only way to reach target is by subtracting 1.
Insight 3: The optimal strategy involves prioritizing doubling operations when possible, which translates to halving when working backward. We only add 1 when the target is odd, as halving can only be applied to even numbers.

Complexity Analysis

Time Complexity: O(log(target))

Space Complexity: O(1)

Topics

This problem involves: Math, Greedy.

Companies

Asked at: Millennium, Nutanix.

Broken Calculator | LeetCode 991

Medium

Math Greedy

Time: O(log(target))Space: O(1)

Solution Code

Available in:

class Solution {
    public int brokenCalc(int startValue, int target) {
        int operations = 0;
        while (target > startValue) {
            operations += 1;
            if (target % 2 == 1) {
                target += 1;
            } else {
                target /= 2;
            }
        }
        
        return operations + startValue - target;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 991: Broken Calculator - Detailed explanation and solution approaches.

Topics

Math Greedy

Companies

MillenniumNutanix

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:991

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize a variable `operations` to 0. This variable will store the count of operations.
2
Step 2: Enter a `while` loop that continues as long as `target` is greater than `startValue`.
3
Step 3: Inside the loop, increment `operations` by 1 for each operation performed.
4
Step 4: Check if `target` is even (i.e., `target % 2 == 0`).
5
Step 5: If `target` is even, divide it by 2 (`target //= 2`). This simulates reversing the doubling operation.
6
Step 6: If `target` is odd, add 1 to it (`target += 1`). This simulates reversing the subtraction operation.
7
Step 7: After the loop finishes (when `target` is no longer greater than `startValue`), add the difference `startValue - target` to `operations`. This accounts for the number of subtraction operations needed to reach `startValue` from the remaining `target` if target is less than startValue. Note this will be zero if target is startValue after the loop.
8
Step 8: Return the final value of `operations`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(log(target))

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is logarithmic because in the best case target keeps getting halved until it is less than startValue. The number of divisions is logarithmic in nature. Space complexity is constant because the algorithm requires only a few variables to store integers, and their size is independent of the input values.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Breadth-First Search (BFS)

One could consider BFS to explore the possible operations from `startValue` towards `target`, but this would lead to Time Limit Exceeded (TLE) due to the exponential growth of the search space. Also, the range of possible values is very large (up to 10^9), requiring a huge queue.

Time: O(2^n) in the worst case where n is the number of operations.Space: O(2^n) in the worst case.

Trade-offs: Not efficient due to exponential nature. Useful for much smaller input ranges, but not for the given constraints.

✨

Key Insights

1Insight 1: Working backward from the target is more efficient. Instead of trying to reach the target from the start, we can reverse the operations: divide the target by 2 (if even) or add 1 to the target.
2Insight 2: Since startValue and target are positive, we can deduce that when target is greater than startValue, we should try to reach to target from the startValue. Doubling results in a larger number and subtracting reduces to a smaller number. If startValue is greater than target then the only way to reach target is by subtracting 1.
3Insight 3: The optimal strategy involves prioritizing doubling operations when possible, which translates to halving when working backward. We only add 1 when the target is odd, as halving can only be applied to even numbers.

📚

Key Terms Explained

For Students

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: `startValue == target`. The while loop doesn't execute and the function returns 0, which is the correct answer.
Edge case 2: `startValue > target`. The while loop doesn't execute, and the result will be `startValue - target`, representing only subtractions, which is the correct answer. For example, `startValue=5, target=1`. The answer is `5-1=4`
Edge case 3: target = 1. The while loop operates until target is less than startValue then `startValue - target` is returned.
Edge case 4: startValue = 1. Consider the case when startValue = 1 and target is a large number. The iterations continue to halve the target if its even or add one if its odd. The number of operations are properly tracked.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Trying to work forward from `startValue` to `target` is generally much less efficient. It is much harder to decide whether to multiply or subtract at each step, making it difficult to design a greedy algorithm. Reversing the process simplifies the decision making.

✗Common mistake 2: Incorrectly handling the edge case when `startValue` is greater than `target`. The correct approach is to return `startValue - target` in this case.

✗Common mistake 3: Not understanding that we must deal with the possibility of target being less than the startValue after the loop ends, by adding the `startValue-target` as the final operations.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimization problems where reversing the objective function simplifies the solution strategy.

🏦Real-world scenario 2: Compiler optimizations. Some compiler optimizations, like common subexpression elimination, are simpler when approached from the back end of the compiler's pipeline.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 31, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 29

❓

Frequently Asked Questions

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Broken Calculator"?

For this medium-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?