Leetcode #306: Additive Number
In this guide, we solve Leetcode #306 Additive Number 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
An additive number is a string whose digits can form an additive sequence. A valid additive sequence should contain at least three numbers.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: String, Backtracking
Intuition
We must explore combinations of choices, but many branches can be pruned early.
Backtracking enumerates valid candidates while keeping the search space under control.
Approach
Use DFS to build candidates step by step, and backtrack when constraints are violated.
Pruning keeps the exploration practical for typical constraints.
Steps:
- Define the decision tree.
- DFS through choices and backtrack.
- Prune invalid paths early.
Example
Input: "112358"
Output: true
Explanation:
The digits can form an additive sequence: 1, 1, 2, 3, 5, 8.
1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8
Python Solution
class Solution:
def isAdditiveNumber(self, num: str) -> bool:
def dfs(a, b, num):
if not num:
return True
if a + b > 0 and num[0] == '0':
return False
for i in range(1, len(num) + 1):
if a + b == int(num[:i]):
if dfs(b, a + b, num[i:]):
return True
return False
n = len(num)
for i in range(1, n - 1):
for j in range(i + 1, n):
if i > 1 and num[0] == '0':
break
if j - i > 1 and num[i] == '0':
continue
if dfs(int(num[:i]), int(num[i:j]), num[j:]):
return True
return False
Complexity
The time complexity is Exponential (worst case). The space complexity is O(depth).
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.