What is the time complexity of Non-negative Integers without Consecutive Ones?

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

What data structures are used to solve Non-negative Integers without Consecutive Ones?

This problem uses Dynamic Programming. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Non-negative Integers without Consecutive Ones?

Insight 1: The problem can be efficiently solved using dynamic programming combined with the Fibonacci sequence. The number of binary strings of length 'k' without consecutive 1s can be derived from Fibonacci numbers. Insight 2: We can build the solution by iterating through the bits of the binary representation of the given number 'n'. If we encounter a '1', we can add the number of valid strings of shorter length to our count. We must also handle the case of consecutive 1's.

Which companies ask LeetCode 600. Non-negative Integers without Consecutive Ones in interviews?

This problem is frequently asked at Pocket Gems.

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Non-negative Integers without Consecutive Ones - LeetCode 600 Solution

Q: How to solve Non-negative Integers without Consecutive Ones?

The solution leverages the Fibonacci sequence to count the number of valid binary strings of different lengths that are less than or equal to the binary representation of 'n'. The algorithm iterates through the bits of 'n' from left to right (most significant bit to least significant bit). If the current bit is '1', we add fib[k] to the answer, where k is the number of remaining bits. A crucial condition is checked if the previous bit was also '1'. If so, it means any further numbers generated would be greater than 'n' (because of consecutive ones where they shouldn't be), so we return the answer immediately. Otherwise, if the current bit is '0', we simply set the 'prev_bit_is_one' flag to false.

Non-negative Integers without Consecutive Ones - Complete Solution Guide

Non-negative Integers without Consecutive Ones is LeetCode problem 600, a Hard 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 , return the number of the integers in the range [0, n] whose binary representations do not contain consecutive ones. Example 1: Input: n = 5 Output: 5 Explanation: Here are the non-negative integers <= 5 with their corresponding binary representations: 0 : 0 1 : 1 2 : 10 3 : 11 4 : 100 5 : 101 Among them, only integer 3 disobeys the rule (two consecutive ones) and the other 5 satisfy the rule. Example 2: Input: n = 1 Output: 2 Example 3: Input: n = 2 Output: 3 Constra

Detailed Explanation

The problem asks us to count the number of non-negative integers within the range [0, n] (inclusive) that do not have consecutive 1s in their binary representation. For instance, if n = 5, the binary representations are 0 (0), 1 (1), 2 (10), 3 (11), 4 (100), and 5 (101). Only 3 (11) has consecutive 1s, so the answer is 5 (the count of the others). The input n is a positive integer between 1 and 10^9.

Solution Approach

The solution leverages the Fibonacci sequence to count the number of valid binary strings of different lengths that are less than or equal to the binary representation of 'n'. The algorithm iterates through the bits of 'n' from left to right (most significant bit to least significant bit). If the current bit is '1', we add fib[k] to the answer, where k is the number of remaining bits. A crucial condition is checked if the previous bit was also '1'. If so, it means any further numbers generated would be greater than 'n' (because of consecutive ones where they shouldn't be), so we return the answer immediately. Otherwise, if the current bit is '0', we simply set the 'prev_bit_is_one' flag to false.

Step-by-Step Algorithm

Step 1: Convert the integer 'n' to its binary string representation.
Step 2: Create an array 'fib' to store the Fibonacci sequence. Initialize fib[0] = 1 and fib[1] = 2. Then, calculate fib[i] = fib[i-1] + fib[i-2] for i from 2 to 30.
Step 3: Initialize a variable 'ans' to 0, which will store the count of valid integers.
Step 4: Initialize a boolean variable 'prev_bit_is_one' to false, to track if the previous bit was a '1'.
Step 5: Iterate through the binary string from left to right (most significant bit to least significant bit).
Step 6: For each bit at index 'i', calculate 'k' as length - 1 - i (number of remaining bits).
Step 7: If the bit at index 'i' is '1':
a. Add fib[k] to 'ans'.
b. If 'prev_bit_is_one' is true, return 'ans' (because having consecutive 1's means we've exceeded n and all remaining numbers will exceed n as well)
c. Set 'prev_bit_is_one' to true.
Step 8: If the bit at index 'i' is '0', set 'prev_bit_is_one' to false.
Step 9: After the loop finishes, return 'ans + 1'. Adding 1 accounts for the integer 0, which always satisfies the condition.

Key Insights

Insight 1: The problem can be efficiently solved using dynamic programming combined with the Fibonacci sequence. The number of binary strings of length 'k' without consecutive 1s can be derived from Fibonacci numbers.
Insight 2: We can build the solution by iterating through the bits of the binary representation of the given number 'n'. If we encounter a '1', we can add the number of valid strings of shorter length to our count. We must also handle the case of consecutive 1's.
Insight 3: Pre-computing the Fibonacci sequence up to a certain length (31, since n <= 10^9 fits within 31 bits) helps to optimize the calculation during the iteration.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Dynamic Programming.

Companies

Asked at: Pocket Gems.

Non-negative Integers without Consecutive Ones | LeetCode 600

Hard

Dynamic Programming

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

Solution Code

Available in:

class Solution {
    public int findIntegers(int n) {
        String s = Integer.toBinaryString(n);
        int length = s.length();
        
        int[] fib = new int[31];
        fib[0] = 1;
        fib[1] = 2;
        for (int i = 2; i < 31; i++) {
            fib[i] = fib[i - 1] + fib[i - 2];
        }
        
        int ans = 0;
        boolean prev_bit_is_one = false;
        
        for (int i = 0; i < length; i++) {
            int k = length - 1 - i;
            
            if (s.charAt(i) == '1') {
                ans += fib[k];
                if (prev_bit_is_one) {
                    return ans;
                }
                prev_bit_is_one = true;
            } else {
                prev_bit_is_one = false;
            }
        }
        
        return ans + 1;
    }
}

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Problem Info

Description

Problem 600: Non-negative Integers without Consecutive Ones - Detailed explanation and solution approaches.

Topics

Dynamic Programming

Companies

Pocket Gems

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:600

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

12 steps

1
Step 1: Convert the integer 'n' to its binary string representation.
2
Step 2: Create an array 'fib' to store the Fibonacci sequence. Initialize fib[0] = 1 and fib[1] = 2. Then, calculate fib[i] = fib[i-1] + fib[i-2] for i from 2 to 30.
3
Step 3: Initialize a variable 'ans' to 0, which will store the count of valid integers.
4
Step 4: Initialize a boolean variable 'prev_bit_is_one' to false, to track if the previous bit was a '1'.
5
Step 5: Iterate through the binary string from left to right (most significant bit to least significant bit).
6
Step 6: For each bit at index 'i', calculate 'k' as length - 1 - i (number of remaining bits).
7
Step 7: If the bit at index 'i' is '1':
8
a. Add fib[k] to 'ans'.
9
b. If 'prev_bit_is_one' is true, return 'ans' (because having consecutive 1's means we've exceeded n and all remaining numbers will exceed n as well)
10
c. Set 'prev_bit_is_one' to true.
11
Step 8: If the bit at index 'i' is '0', set 'prev_bit_is_one' to false.
12
Step 9: After the loop finishes, return 'ans + 1'. Adding 1 accounts for the integer 0, which always satisfies the condition.

💡

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(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the loop that iterates through the bits of the binary representation of n. The space complexity is constant because the 'fib' array has a fixed size, independent of the input value. No other dynamic memory allocation occurs (other than the fixed-size array and the string in C, which are both O(1)).

📖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

Brute Force

Iterate through all numbers from 0 to n. For each number, convert it to its binary representation and check if it contains consecutive 1s. If it doesn't, increment a counter.

Time: O(n * log(n)), where n is the input number. Converting each number to binary takes O(log(n)) time, and this operation happens n times.Space: O(log(n)) for the binary conversion string of each number.

Trade-offs: The brute force approach is simple to understand but is significantly less efficient than the dynamic programming approach for larger values of n. It's only suitable for very small input values.

✨

Key Insights

1Insight 1: The problem can be efficiently solved using dynamic programming combined with the Fibonacci sequence. The number of binary strings of length 'k' without consecutive 1s can be derived from Fibonacci numbers.
2Insight 2: We can build the solution by iterating through the bits of the binary representation of the given number 'n'. If we encounter a '1', we can add the number of valid strings of shorter length to our count. We must also handle the case of consecutive 1's.
3Insight 3: Pre-computing the Fibonacci sequence up to a certain length (31, since n <= 10^9 fits within 31 bits) helps to optimize the calculation during the iteration.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: n = 0. The solution correctly handles this because the 'ans' is initialized to 0 and then incremented by 1 at the end, yielding 1. The empty string has no consecutive ones.
Edge case 2: n = 1. The binary representation is "1". The loop will run once, add fib[0] = 1 to ans, and set prev_bit_is_one to true. It returns ans + 1 = 2, which is correct (0 and 1 are valid).
Edge case 3: n = 2. The binary representation is "10". The loop runs twice. The first bit is 1, so it adds fib[0] = 1 to ans and sets prev_bit_is_one to true. The second bit is 0, so it sets prev_bit_is_one to false. It returns ans + 1 = 2, however, the correct answer is 3 (0, 1, and 2 (10) are valid). This off-by-one error occurs in many solutions if 'ans + 1' is not added to account for the integer 0 if all other conditions pass. This edge case tests the addition of '1' at the end.
Edge case 4: n = a number with consecutive 1s, such as 3 or 7. The algorithm is designed to terminate the loop early if consecutive ones are found, returning the count of valid numbers up to that point.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to precompute the Fibonacci sequence. Calculating the Fibonacci numbers within the loop would lead to redundant calculations and a higher time complexity.

✗Common mistake 2: Not handling the consecutive ones case correctly. If consecutive ones are encountered during iteration, the algorithm needs to terminate early to avoid counting invalid numbers.

✗Common mistake 3: Not adding 1 to the final answer to account for the number 0. 0 always satisfies the condition and needs to be included in the count.

✗Common mistake 4: Incorrectly converting the integer to binary. Proper handling of bitwise operations or string manipulation is required.

✅

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: Constraints in communication protocols. In some network protocols, certain bit patterns are reserved or have special meanings. This problem could be adapted to count the number of valid data patterns that avoid certain forbidden sequences of bits.

🏦Real-world scenario 2: Error detection and correction codes. Encoding schemes might be designed to avoid consecutive identical bits to improve robustness against certain types of transmission errors.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

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

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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 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

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

@DevExpert92Jul 14

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

@AlgoMasterJul 7

❓

Frequently Asked Questions

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Non-negative Integers without Consecutive Ones"?

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

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