Algorithm
Problem Name: 306. Additive Number
An additive number is a string whose digits can form an additive sequence.
A valid additive sequence should contain at least three numbers. Except for the first two numbers, each subsequent number in the sequence must be the sum of the preceding two.
Given a string containing only digits, return true if it is an additive number or false otherwise.
Note: Numbers in the additive sequence cannot have leading zeros, so sequence 1, 2, 03 or 1, 02, 3 is invalid.
Example 1:
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
Example 2:
Input: "199100199" Output: true Explanation: The additive sequence is: 1, 99, 100, 199. 1 + 99 = 100, 99 + 100 = 199
Constraints:
1 <= num.length <= 35numconsists only of digits.
Code Examples
#1 Code Example with Javascript Programming
Code -
Javascript Programming
const isAdditiveNumber = function(num) {
const n = num.length
for (let i = 1; i < = (n / 2) >> 0; ++i) {
if (num.charAt(0) === '0' && i > 1) return false
const x1 = +num.slice(0, i)
for (let j = 1; Math.max(j, i) < = n - i - j; ++j) {
if (num.charAt(i) == '0' && j > 1) break
const x2 = +num.slice(i, i + j)
if (isValid(x1, x2, j + i, num)) return true
}
}
return false
}
function isValid(x1, x2, start, num) {
if (start === num.length) return true
x2 = x2 + x1
x1 = x2 - x1
const sum = x2 + ''
return num.startsWith(sum, start) && isValid(x1, x2, start + sum.length, num)
}
Input
"112358"
Output
true
#2 Code Example with Python Programming
Code -
Python Programming
class Solution:
def isAdditiveNumber(self, num):
def getStarter():
arr = []
for i in range(1, len(num) - 1):
for j in range(i + 1, len(num)):
s1, s2 = num[:i], num[i:j]
if (s1[0] == "0" and len(s1) > 1) or (s2[0] == "0" and len(s2) > 1):
continue
arr.append((int(s1), int(s2), j))
return arr
def dfs(pre1, pre2, i):
if i == len(num):
return True
sm = pre1 + pre2
l = len(str(sm))
new = int(num[i:i + l])
return new == sm and dfs(pre2, new, i + l)
q = getStarter()
return any(dfs(p1, p2, i) for p1, p2, i in q)
Input
"112358"
Output
true