What is the time complexity of Minimum Moves to Reach Target Score?

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

What data structures are used to solve Minimum Moves to Reach Target Score?

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

What are the key insights for solving Minimum Moves to Reach Target Score?

Insight 1: Working backwards from the target is more efficient. Instead of trying to reach the target by adding or doubling, we can try to reach 1 from the target by subtracting or halving. Insight 2: Prioritize doubling when possible (and allowed). If the target is even and we have remaining doubles, divide it by 2. Otherwise, subtract 1.

Which companies ask LeetCode 2139. Minimum Moves to Reach Target Score in interviews?

This problem is frequently asked at Wayfair.

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Minimum Moves to Reach Target Score - LeetCode 2139 Solution

Minimum Moves to Reach Target Score - Complete Solution Guide

Minimum Moves to Reach Target Score is LeetCode problem 2139, 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

You are playing a game with integers. You start with the integer 1 and you want to reach the integer target . In one move, you can either: Increment the current integer by one (i.e., x = x + 1 ). Double the current integer (i.e., x = 2 * x ). You can use the increment operation any number of times, however, you can only use the double operation at most maxDoubles times. Given the two integers target and maxDoubles , return the minimum number of moves needed to reach target starting with 1 . Exam

Detailed Explanation

The problem asks us to find the minimum number of moves to transform the integer '1' to a given 'target' integer. We have two possible moves: increment the current number by 1 or double the current number. The doubling operation can only be performed a maximum of 'maxDoubles' times. The objective is to find the fewest moves required, considering both types of operations and the constraint on the number of doubling operations.

Solution Approach

The provided code implements a greedy approach by working backwards from the 'target' towards 1. It prioritizes halving the 'target' when possible (if it's even and the 'maxDoubles' count is greater than 0). Otherwise, it decrements the 'target' by 1. This process continues until either the 'target' reaches 1 or 'maxDoubles' reaches 0. If 'target' is still greater than 1 after using all allowed doubles, the remaining difference is decremented until target becomes 1.

Step-by-Step Algorithm

Step 1: Initialize a variable 'moves' to 0 to store the total number of moves.
Step 2: Enter a 'while' loop that continues as long as 'target' is greater than 1 and 'maxDoubles' is greater than 0.
Step 3: Inside the loop, check if 'target' is even. If it is, divide 'target' by 2 and decrement 'maxDoubles' by 1. Increment 'moves' by 1.
Step 4: If 'target' is odd, decrement 'target' by 1. Increment 'moves' by 1.
Step 5: After the loop, if 'target' is still greater than 1, increment 'moves' by the difference between 'target' and 1 (representing the remaining decrements needed).
Step 6: Return the final value of 'moves'.

Key Insights

Insight 1: Working backwards from the target is more efficient. Instead of trying to reach the target by adding or doubling, we can try to reach 1 from the target by subtracting or halving.
Insight 2: Prioritize doubling when possible (and allowed). If the target is even and we have remaining doubles, divide it by 2. Otherwise, subtract 1.
Insight 3: Once maxDoubles is 0, we can only increment (or decrement when working backwards). Therefore, we simply need to decrement the remaining value down to 1.

Complexity Analysis

Time Complexity: O(log(target))

Space Complexity: O(1)

Topics

This problem involves: Math, Greedy.

Companies

Asked at: Wayfair.

Minimum Moves to Reach Target Score | LeetCode 2139

Medium

Math Greedy

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

Solution Code

Available in:

class Solution {
    public int minMoves(int target, int maxDoubles) {
        int moves = 0;
        while (target > 1 && maxDoubles > 0) {
            if (target % 2 == 0) {
                target /= 2;
                maxDoubles--;
            } else {
                target--;
            }
            moves++;
        }
        
        if (target > 1) {
            moves += (target - 1);
        }
        
        return (int)moves;
    }
}

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Problem Info

Description

Problem 2139: Minimum Moves to Reach Target Score - Detailed explanation and solution approaches.

Topics

Math Greedy

Companies

Wayfair

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2139

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

6 steps

1
Step 1: Initialize a variable 'moves' to 0 to store the total number of moves.
2
Step 2: Enter a 'while' loop that continues as long as 'target' is greater than 1 and 'maxDoubles' is greater than 0.
3
Step 3: Inside the loop, check if 'target' is even. If it is, divide 'target' by 2 and decrement 'maxDoubles' by 1. Increment 'moves' by 1.
4
Step 4: If 'target' is odd, decrement 'target' by 1. Increment 'moves' by 1.
5
Step 5: After the loop, if 'target' is still greater than 1, increment 'moves' by the difference between 'target' and 1 (representing the remaining decrements needed).
6
Step 6: Return the final value of 'moves'.

💡

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 (when the target is a power of 2 and maxDoubles is sufficient), the target is halved in each iteration of the while loop. This continues until the target reaches 1. The space complexity is constant because the algorithm only uses a few integer variables, regardless of the size of the input.

📖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

Dynamic Programming

A dynamic programming approach could be used, but it is not as efficient. Create a DP table where dp[i] stores the minimum moves to reach 'i' from 1. Fill the table from 1 to target. The state transition would be dp[i] = min(dp[i-1] + 1, dp[i/2] + 1) if i is even and doubling is possible.

Time: O(target)Space: O(target)

Trade-offs: The DP approach uses significantly more space than the greedy approach. It's also slower, especially for large target values. The greedy approach is preferred for its simplicity and efficiency.

✨

Key Insights

1Insight 1: Working backwards from the target is more efficient. Instead of trying to reach the target by adding or doubling, we can try to reach 1 from the target by subtracting or halving.
2Insight 2: Prioritize doubling when possible (and allowed). If the target is even and we have remaining doubles, divide it by 2. Otherwise, subtract 1.
3Insight 3: Once maxDoubles is 0, we can only increment (or decrement when working backwards). Therefore, we simply need to decrement the remaining value down to 1.

📚

Key Terms Explained

For Students

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: target = 1, maxDoubles = 0. The while loop will not execute, and 'moves' remains 0, which is the correct answer.
Edge case 2: target = a large number, maxDoubles = 0. The while loop won't execute because maxDoubles is 0. The if statement after the loop will add (target - 1) to moves, correctly calculating the moves.
Edge case 3: target = a large number, maxDoubles = a large number. The while loop will execute halving operations until target becomes odd or maxDoubles becomes zero, at which point decrements are applied. The logic correctly handles both conditions.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Trying to solve the problem by starting at 1 and incrementing and doubling to reach the target. This would require exploring many possibilities and is significantly harder to optimize.

✗Common mistake 2: Not handling the case where 'maxDoubles' becomes 0 before 'target' reaches 1. In this case, only increment operations are allowed for the rest of the process.

✗Common mistake 3: Assuming 'maxDoubles' will always be sufficient to reduce the target to a small value. We need to handle the case where the target is significantly larger than 1 even after using up all doubling 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: Resource optimization. Imagine minimizing the number of steps to manufacture a product where you can either add one unit of raw material or double the existing quantity. 'maxDoubles' would represent constraints on using the doubling process (e.g., limited capacity of the doubling machine).

🏦Real-world scenario 2: Optimizing network bandwidth usage. If you need to send data to a recipient, you can either increase the size of each packet by one unit or double the packet size. There might be limits on how many times you can double the packet size before it becomes inefficient due to packet loss or other factors.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: November 7, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

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

@AlgoMasterJun 30

❓

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 "Minimum Moves to Reach Target Score"?

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?