What is the time complexity of Integer Replacement?

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

What data structures are used to solve Integer Replacement?

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

What are the key insights for solving Integer Replacement?

Insight 1: The core idea is to repeatedly apply the given operations until 'n' becomes 1. Insight 2: When 'n' is odd, the choice between adding 1 and subtracting 1 is crucial. It's generally better to choose the operation that leads to more factors of 2 in the subsequent steps, allowing more efficient divisions.

Which companies ask LeetCode 397. Integer Replacement in interviews?

This problem is frequently asked at Baidu.

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Integer Replacement - LeetCode 397 Solution

Q: How to solve Integer Replacement?

The provided code uses a `while` loop to repeatedly apply the operations until 'n' becomes 1. Inside the loop, it first checks if 'n' is even. If it is, it divides 'n' by 2. If 'n' is odd, it checks if 'n' is equal to 3. If it is, it adds 2 to the count and breaks the loop. Otherwise, if the last two bits of 'n' are equal to 01 (n&3 == 1), then it decrements 'n'. If the last two bits of 'n' are not equal to 01 (n&3 != 1), it increments 'n'. The operations are guided by the idea that we want to produce even number after the addition or subtraction. The count of operations is incremented in each iteration.

Integer Replacement - Complete Solution Guide

Integer Replacement is LeetCode problem 397, 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

Given a positive integer n , you can apply one of the following operations: If n is even, replace n with n / 2 . If n is odd, replace n with either n + 1 or n - 1 . Return the minimum number of operations needed for n to become 1 . Example 1: Input: n = 8 Output: 3 Explanation: 8 -> 4 -> 2 -> 1 Example 2: Input: n = 7 Output: 4 Explanation: 7 -> 8 -> 4 -> 2 -> 1 or 7 -> 6 -> 3 -> 2 -> 1 Example 3: Input: n = 4 Output: 2 Constraints: 1 <= n <= 2 31 - 1

Detailed Explanation

The problem requires you to find the minimum number of operations to transform a given positive integer 'n' to 1. The allowed operations are: if 'n' is even, divide it by 2; if 'n' is odd, either add 1 or subtract 1 from it. The input is a positive integer 'n' and the output is the minimum number of operations required.

Solution Approach

The provided code uses a `while` loop to repeatedly apply the operations until 'n' becomes 1. Inside the loop, it first checks if 'n' is even. If it is, it divides 'n' by 2. If 'n' is odd, it checks if 'n' is equal to 3. If it is, it adds 2 to the count and breaks the loop. Otherwise, if the last two bits of 'n' are equal to 01 (n&3 == 1), then it decrements 'n'. If the last two bits of 'n' are not equal to 01 (n&3 != 1), it increments 'n'. The operations are guided by the idea that we want to produce even number after the addition or subtraction. The count of operations is incremented in each iteration.

Step-by-Step Algorithm

Step 1: Initialize a counter 'count' to 0. This will store the number of operations.
Step 2: Start a `while` loop that continues as long as 'n' is greater than 1.
Step 3: Inside the loop, check if 'n' is even using the bitwise AND operator (`n & 1 == 0`).
Step 4: If 'n' is even, right-shift 'n' by 1 (equivalent to dividing by 2) using `n >>= 1`.
Step 5: If 'n' is odd, check if 'n' is equal to 3. If so, increment count by 2 and break from the loop. Note: this must come before the next conditional.
Step 6: If 'n' is odd and not equal to 3, use bitwise AND to check the last two bits of 'n' (`n & 3`).
Step 7: If the last two bits are 01 (n & 3 == 1), decrement 'n' by 1.
Step 8: Otherwise (if the last two bits are 11), increment 'n' by 1.
Step 9: Increment 'count' by 1 to reflect the operation performed.
Step 10: After the loop finishes, return the 'count'.

Key Insights

Insight 1: The core idea is to repeatedly apply the given operations until 'n' becomes 1.
Insight 2: When 'n' is odd, the choice between adding 1 and subtracting 1 is crucial. It's generally better to choose the operation that leads to more factors of 2 in the subsequent steps, allowing more efficient divisions.
Insight 3: The special case when n=3 needs explicit handling. It's better to transform 3 -> 2 -> 1 in 2 steps, instead of 3 -> 4 -> 2 -> 1 which requires 3 steps.

Complexity Analysis

Time Complexity: O(log n)

Space Complexity: O(1)

Topics

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

Companies

Asked at: Baidu.

Integer Replacement | LeetCode 397

Medium

Dynamic Programming Greedy Bit Manipulation Memoization

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

Solution Code

Available in:

class Solution {
    public int integerReplacement(int n) {
        int count = 0;
        while (n > 1) {
            if ((n & 1) == 0) {
                n >>>= 1;
            } else if (n == 3) {
                count += 2;
                break;
            } else if ((n & 3) == 1) {
                n -= 1;
            } else {
                n += 1;
            }
            count += 1;
        }
        return count;
    }
}

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Problem Info

Description

Problem 397: Integer Replacement - Detailed explanation and solution approaches.

Topics

Dynamic Programming Greedy Bit Manipulation Memoization

Companies

Baidu

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:397

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

10 steps

1
Step 1: Initialize a counter 'count' to 0. This will store the number of operations.
2
Step 2: Start a `while` loop that continues as long as 'n' is greater than 1.
3
Step 3: Inside the loop, check if 'n' is even using the bitwise AND operator (`n & 1 == 0`).
4
Step 4: If 'n' is even, right-shift 'n' by 1 (equivalent to dividing by 2) using `n >>= 1`.
5
Step 5: If 'n' is odd, check if 'n' is equal to 3. If so, increment count by 2 and break from the loop. Note: this must come before the next conditional.
6
Step 6: If 'n' is odd and not equal to 3, use bitwise AND to check the last two bits of 'n' (`n & 3`).
7
Step 7: If the last two bits are 01 (n & 3 == 1), decrement 'n' by 1.
8
Step 8: Otherwise (if the last two bits are 11), increment 'n' by 1.
9
Step 9: Increment 'count' by 1 to reflect the operation performed.
10
Step 10: After the loop finishes, return the 'count'.

💡

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 value of 'n' is halved in each iteration of the `while` loop when 'n' is even. When 'n' is odd, we add or subtract 1, which eventually makes it even. Therefore, the number of iterations is proportional to the number of times 'n' can be divided by 2 until it reaches 1, which is log₂(n). The space complexity is O(1) because we are only using a constant amount of extra space for the 'count' variable, regardless of the input 'n'.

📖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

Recursive Memoization

A recursive approach with memoization (dynamic programming) can also solve this problem. It involves creating a memo table to store the results of subproblems, which avoids redundant computations.

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

Trade-offs: The recursive approach with memoization has a similar time complexity, but it uses O(log n) space due to the memoization table and recursion stack. The iterative approach is generally preferred due to its lower space complexity.

✨

Key Insights

1Insight 1: The core idea is to repeatedly apply the given operations until 'n' becomes 1.
2Insight 2: When 'n' is odd, the choice between adding 1 and subtracting 1 is crucial. It's generally better to choose the operation that leads to more factors of 2 in the subsequent steps, allowing more efficient divisions.
3Insight 3: The special case when n=3 needs explicit handling. It's better to transform 3 -> 2 -> 1 in 2 steps, instead of 3 -> 4 -> 2 -> 1 which requires 3 steps.

📚

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 while loop condition (n > 1) prevents the code from entering the loop and executes nothing, so it immediately returns 0, which is the correct answer.
Edge case 2: n = 3. The algorithm correctly handles this edge case by recognizing and completing the operation with 2 additional operations and breaking the while loop ( 3 -> 2 -> 1).
Edge case 3: n = Integer.MAX_VALUE. This is handled implicitly by the bitwise operations. Since bitwise operations in Java and C++ use modulo 2^32 for ints, adding one to Integer.MAX_VALUE results in a negative number, but the algorithm continues to function correctly to find a path from the resulting value to 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to consider the special case of 'n' = 3. Directly applying the odd number rules may lead to a longer path than necessary.

✗Common mistake 2: Integer overflow. When adding 1 to a large odd number, you might cause integer overflow, leading to incorrect results if not handled carefully. The bitwise operations implicitly handle this.

✗Common mistake 3: Incorrectly prioritizing adding or subtracting 1 without considering the two least significant bits.

✅

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 problem's concept of minimizing operations can be applied to optimization problems where resources are limited and efficient transitions are needed.

🏦Real-world scenario 2: In compiler design, instruction scheduling aims to minimize the number of clock cycles required to execute a program. The integer replacement problem is a simplified version of such optimization problems.

🎮Real-world scenario 3: In game development, finding the optimal path between two states, given a set of possible actions, relates to the core concept of the integer replacement problem.

💡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

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

⚠️ 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 8, 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

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

@InterviewProJul 14

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

@CodeNinjaJul 7

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

@TechLeadJun 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 "Integer Replacement"?

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.

🤖 Need help with this problem?