Leetcode #600: Non-negative Integers without Consecutive Ones
In this guide, we solve Leetcode #600 Non-negative Integers without Consecutive Ones in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
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.
Python Solution
class Solution:
def findIntegers(self, n: int) -> int:
def dfs(i: int, pre: int, limit: bool) -> int:
if i < 0:
return 1
up = (n >> i & 1) if limit else 1
ans = 0
for j in range(up + 1):
if pre and j:
continue
ans += dfs(i - 1, j, limit and j == up)
return ans
return dfs(n.bit_length() - 1, 0, True)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.