How do you solve Non-negative Integers without Consecutive Ones on LeetCode?

Non-negative Integers without Consecutive Ones (LeetCode #600) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(1) space complexity. Key insight: The Fibonacci sequence helps in counting valid configurations without consecutive ones.

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

The optimal solution for Non-negative Integers without Consecutive Ones runs in O(log n) time and uses O(1) space. The time complexity is O(log n) because we only iterate through the bits of n, which is at most 32 for the given constraints.

What is the brute force solution for Non-negative Integers without Consecutive Ones?

The brute force approach for Non-negative Integers without Consecutive Ones is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each number from 0 to n and determining if its binary representation contains consecutive ones. This is straightforward but inefficient for large n. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve Non-negative Integers without Consecutive Ones?

Non-negative Integers without Consecutive Ones uses the following concepts: Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Bit Manipulation.

What are the interview tips for Non-negative Integers without Consecutive Ones?

Explain your thought process clearly while coding. Discuss the trade-offs of different approaches before jumping into coding. Practice converting numbers to binary and understanding bit manipulation.

What are common mistakes when solving Non-negative Integers without Consecutive Ones?

Not considering edge cases like n being a power of 2. Overlooking the importance of bit manipulation in the optimal solution.

#600

Non-negative Integers without Consecutive Ones

Hard

Dynamic ProgrammingDynamic ProgrammingBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log n)
Space	O(1)	O(1)

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Intuition

Time O(log n)Space O(1)

The optimal solution uses dynamic programming to count valid numbers based on their binary representation. We can leverage the Fibonacci sequence to determine how many valid numbers exist up to a certain bit length without consecutive ones.

⚙️

Algorithm

5 steps

1Step 1: Create an array dp where dp[i] represents the count of valid numbers with i bits.
2Step 2: Initialize dp[0] = 1 (the empty number) and dp[1] = 2 (0 and 1).
3Step 3: Fill dp[i] using the relation dp[i] = dp[i-1] + dp[i-2] for i from 2 to the number of bits in n.
4Step 4: Count valid numbers less than or equal to n by checking each bit of n from the most significant to the least significant, using dp to count valid configurations.
5Step 5: Return the total count.

solution.py16 lines

1# Full working Python code
2def findIntegers(n):
3    dp = [0] * 32
4    dp[0], dp[1] = 1, 2
5    for i in range(2, 32):
6        dp[i] = dp[i - 1] + dp[i - 2]
7    count, prev_bit = 0, 0
8    for i in range(31, -1, -1):
9        if (n & (1 << i)):
10            count += dp[i]
11            if prev_bit:
12                return count
13            prev_bit = 1
14        else:
15            prev_bit = 0
16    return count + 1

ℹ

Complexity note: The time complexity is O(log n) because we only iterate through the bits of n, which is at most 32 for the given constraints.

1The Fibonacci sequence helps in counting valid configurations without consecutive ones.
2Understanding binary representation is crucial for efficiently solving this problem.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.