What is the time complexity of Minimum Operations to Reduce an Integer to 0?

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

What data structures are used to solve Minimum Operations to Reduce an Integer to 0?

This problem uses Dynamic Programming, Greedy, Bit Manipulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Operations to Reduce an Integer to 0?

Insight 1: Each operation either adds or subtracts a power of 2. The goal is to efficiently represent the number as a sum or difference of powers of 2. Insight 2: When the last two bits of 'n' are '11', adding 1 reduces the number of operations compared to subtracting 1. This is because adding 1 turns '11' into '00' and carries over, potentially eliminating several 1s at once, reducing subsequent operations. Similarly, if last two bits are '01' subtracting 1 reduces the number of operations.

Which companies ask LeetCode 2571. Minimum Operations to Reduce an Integer to 0 in interviews?

This problem is frequently asked at IBM, Nvidia, Two Sigma.

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Minimum Operations to Reduce an Integer to 0 - LeetCode 2571 Solution

Q: How to solve Minimum Operations to Reduce an Integer to 0?

The solution uses a greedy approach based on bit manipulation. It iteratively reduces 'n' to 0. In each iteration, it checks the least significant bit (LSB). If the LSB is 1, it increments the operation count and decides whether to add or subtract based on the next bit. Specifically, if the two least significant bits are '11' (n & 3 == 3), it's better to add 1 to 'n' because it sets these bits to '00', potentially leading to faster convergence. Otherwise, if the last two bits are '01', it subtracts 1 from 'n'. If the LSB is 0, it right-shifts 'n' by 1 (equivalent to dividing by 2). This process continues until 'n' becomes 0.

Minimum Operations to Reduce an Integer to 0 - Complete Solution Guide

Minimum Operations to Reduce an Integer to 0 is LeetCode problem 2571, 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 given a positive integer n , you can do the following operation any number of times: Add or subtract a power of 2 from n . Return the minimum number of operations to make n equal to 0 . A number x is power of 2 if x == 2 i where i >= 0 . Example 1: Input: n = 39 Output: 3 Explanation: We can do the following operations: - Add 2 0 = 1 to n, so now n = 40. - Subtract 2 3 = 8 from n, so now n = 32. - Subtract 2 5 = 32 from n, so now n = 0. It can be shown that 3 is the minimum number of ope

Detailed Explanation

The problem asks to find the minimum number of operations needed to reduce a given positive integer 'n' to 0. The only allowed operation is to add or subtract a power of 2 from 'n'. We want to find the fewest such additions and subtractions to achieve n = 0. For example, if n = 39, we can add 1 to get 40, subtract 8 to get 32, and subtract 32 to get 0, requiring 3 operations. The constraint is 1 <= n <= 10^5.

Solution Approach

The solution uses a greedy approach based on bit manipulation. It iteratively reduces 'n' to 0. In each iteration, it checks the least significant bit (LSB). If the LSB is 1, it increments the operation count and decides whether to add or subtract based on the next bit. Specifically, if the two least significant bits are '11' (n & 3 == 3), it's better to add 1 to 'n' because it sets these bits to '00', potentially leading to faster convergence. Otherwise, if the last two bits are '01', it subtracts 1 from 'n'. If the LSB is 0, it right-shifts 'n' by 1 (equivalent to dividing by 2). This process continues until 'n' becomes 0.

Step-by-Step Algorithm

Step 1: Initialize a variable 'ops' to 0. This variable will store the number of operations.
Step 2: While 'n' is greater than 0, repeat the following steps:
Step 3: Check if the least significant bit of 'n' is 1 (n & 1 == 1).
Step 4: If the LSB is 1, increment 'ops' by 1.
Step 5: If the two least significant bits of 'n' are '11' (n & 3 == 3), add 1 to 'n'. This prepares for the next right shift by setting the two LSBs to 00.
Step 6: Otherwise (if the two least significant bits are '01'), subtract 1 from 'n'.
Step 7: If the LSB is 0, right-shift 'n' by 1 (n >>= 1). This is equivalent to dividing 'n' by 2 and moving to the next bit.
Step 8: Return the value of 'ops'.

Key Insights

Insight 1: Each operation either adds or subtracts a power of 2. The goal is to efficiently represent the number as a sum or difference of powers of 2.
Insight 2: When the last two bits of 'n' are '11', adding 1 reduces the number of operations compared to subtracting 1. This is because adding 1 turns '11' into '00' and carries over, potentially eliminating several 1s at once, reducing subsequent operations. Similarly, if last two bits are '01' subtracting 1 reduces the number of operations.
Insight 3: We can iteratively process the number from the least significant bit, deciding whether to add or subtract a power of 2 based on the last two bits.

Complexity Analysis

Time Complexity: O(log n)

Space Complexity: O(1)

Topics

This problem involves: Dynamic Programming, Greedy, Bit Manipulation.

Companies

Asked at: IBM, Nvidia, Two Sigma.

Minimum Operations to Reduce an Integer to 0 | LeetCode 2571

Medium

Dynamic Programming Greedy Bit Manipulation

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

Solution Code

Available in:

class Solution {
    public int minOperations(int n) {
        int ops = 0;
        while (n > 0) {
            if ((n & 1) == 1) {
                ops += 1;
                if ((n & 3) == 3) {
                    n += 1;
                } else {
                    n -= 1;
                }
            } else {
                n >>= 1;
            }
        }
        return ops;
    }
}

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Problem Info

Description

Problem 2571: Minimum Operations to Reduce an Integer to 0 - Detailed explanation and solution approaches.

Topics

Dynamic Programming Greedy Bit Manipulation

Companies

IBMNvidiaTwo Sigma

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:2571

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 'ops' to 0. This variable will store the number of operations.
2
Step 2: While 'n' is greater than 0, repeat the following steps:
3
Step 3: Check if the least significant bit of 'n' is 1 (n & 1 == 1).
4
Step 4: If the LSB is 1, increment 'ops' by 1.
5
Step 5: If the two least significant bits of 'n' are '11' (n & 3 == 3), add 1 to 'n'. This prepares for the next right shift by setting the two LSBs to 00.
6
Step 6: Otherwise (if the two least significant bits are '01'), subtract 1 from 'n'.
7
Step 7: If the LSB is 0, right-shift 'n' by 1 (n >>= 1). This is equivalent to dividing 'n' by 2 and moving to the next bit.
8
Step 8: Return the value of 'ops'.

💡

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 n)

🚀 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 O(log n) because in each iteration, either 'n' is divided by 2 (right shift) or 'n' is incremented or decremented. Since 'n' is at most 10^5, the number of iterations is proportional to the number of bits in 'n', which is log base 2 of 'n'. The space complexity is O(1) because the algorithm uses only a constant amount of extra space, regardless of the input size.

📖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 where dp[i] stores the minimum number of operations to reduce 'i' to 0. We can calculate dp[i] by iterating through all powers of 2 less than or equal to 'i' and taking the minimum of dp[i - power_of_2] + 1 and dp[i + power_of_2] + 1.

Time: O(n log n)Space: O(n)

Trade-offs: The dynamic programming approach has a higher time and space complexity than the greedy approach. It is less efficient but can be helpful for gaining intuition about the problem.

Recursion with Memoization

We can recursively define the minimum operations to reduce `n` to 0 as `minOperations(n) = 1 + min(minOperations(n - 2^k), minOperations(n + 2^k))` for all 2^k <= n. Using memoization will prevent recalculations of the same values

Time: O(n log n)Space: O(n)

Trade-offs: Recursion with memoization is similar to dynamic programming in terms of complexity. Can be more concise in code but with function call overhead.

✨

Key Insights

1Insight 1: Each operation either adds or subtracts a power of 2. The goal is to efficiently represent the number as a sum or difference of powers of 2.
2Insight 2: When the last two bits of 'n' are '11', adding 1 reduces the number of operations compared to subtracting 1. This is because adding 1 turns '11' into '00' and carries over, potentially eliminating several 1s at once, reducing subsequent operations. Similarly, if last two bits are '01' subtracting 1 reduces the number of operations.
3Insight 3: We can iteratively process the number from the least significant bit, deciding whether to add or subtract a power of 2 based on the last two bits.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: n = 1. The algorithm correctly identifies that one operation (subtract 1) is required.
Edge case 2: n is a power of 2 (e.g., n = 32). The algorithm will directly right-shift until n becomes 0, requiring only one operation (subtract n from itself).
Edge case 3: n = 3. The algorithm will increment the number leading to n=4 and then right shift twice (increment, shift, shift) leading to two operations (add 1 and subtract 4).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the '11' case in the least significant bits and always subtracting 1. This leads to a suboptimal number of operations.

✗Common mistake 2: Using a loop to find the nearest power of 2 in each iteration. This results in a higher time complexity (O(n log n)) than necessary. The bitwise operation is much more efficient.

✗Common mistake 3: Integer overflow when adding to n. While constraints don't specify large input, always be cautious when incrementing variables inside loops.

✅

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: This type of problem can be seen as a simplified model for resource allocation, where resources come in powers of 2 and the goal is to fulfill a certain requirement using the fewest possible resources.

🏦Real-world scenario 2: Minimizing the number of operations can be analogous to minimizing the number of gates used in digital circuit design where addition and subtraction of powers of two are fundamental operations.

💡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

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

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

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

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: July 22, 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 14

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

@InterviewProJul 7

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

@CodeNinjaJun 30

❓

Frequently Asked Questions

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

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 Operations to Reduce an Integer to 0"?

For this medium-level Dynamic Programming 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.

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