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Convert Integer to the Sum of Two No-Zero Integers

No-Zero integer is a positive integer that does not contain any 0 in its decimal representation.

Given an integer n, return a list of two integers [a, b] where:

  • a and b are No-Zero integers.
  • a + b = n

The test cases are generated so that there is at least one valid solution. If there are many valid solutions, you can return any of them.

 

Example 1:

Input: n = 2
Output: [1,1]
Explanation: Let a = 1 and b = 1.
Both a and b are no-zero integers, and a + b = 2 = n.

Example 2:

Input: n = 11
Output: [2,9]
Explanation: Let a = 2 and b = 9.
Both a and b are no-zero integers, and a + b = 11 = n.
Note that there are other valid answers as [8, 3] that can be accepted.

 

Constraints:

  • 2 <= n <= 104

Solution Explanation for Converting Integer to Sum of Two No-Zero Integers

This problem asks to find two positive integers, a and b, such that their sum is equal to a given integer n, and neither a nor b contains the digit '0' in their decimal representation. We'll explore two approaches: direct enumeration with string conversion and direct enumeration with a helper function.

Approach 1: Direct Enumeration with String Conversion

This approach iterates through possible values of a from 1 to n-1. For each a, it calculates b = n - a. Then, it converts both a and b to strings and checks if either string contains '0'. If neither contains '0', the pair [a, b] is a valid solution, and it's returned.

Code (Python):

class Solution:
    def getNoZeroIntegers(self, n: int) -> List[int]:
        for a in range(1, n):
            b = n - a
            if "0" not in str(a) + str(b):
                return [a, b]

Time Complexity: O(n log n) - The outer loop iterates up to n times. The string conversion and checking within the loop take O(log n) time in the worst case (where n is large, requiring more digits to represent).

Space Complexity: O(log n) - The space used is mainly for string conversion, which is proportional to the number of digits in n.

Approach 2: Direct Enumeration with Helper Function

This approach is similar to the first, but it uses a helper function, f(x), to efficiently check if an integer x contains the digit '0'. The helper function iterates through the digits of x using modular arithmetic. If a '0' is found, it returns False; otherwise, it returns True.

Code (Python):

class Solution:
    def getNoZeroIntegers(self, n: int) -> List[int]:
        def f(x):
            while x:
                if x % 10 == 0:
                    return False
                x //= 10
            return True
 
        for a in range(1, n):
            b = n - a
            if f(a) and f(b):
                return [a, b]

Time Complexity: O(n log n) - Similar to Approach 1, the outer loop iterates up to n times. The helper function f(x) takes O(log n) time in the worst case (proportional to the number of digits).

Space Complexity: O(1) - This approach uses constant extra space, regardless of the input size. The helper function's space is not dependent on the input n.

Comparison:

Both approaches have the same time complexity, but Approach 2 is slightly more efficient in terms of space complexity because it avoids string conversions. The choice depends on preference and coding style; Approach 2 might be considered slightly more elegant due to its modularity. For smaller values of n, the difference will be negligible. For extremely large values of n, the space efficiency of Approach 2 might become more noticeable.