LeetCode 1317 - Convert Integer to the Sum of Two No-Zero Integers

Difficulty: Easy
Topic: Array, Math
Daily Question: 8th September 2025

Video Explanation

Watch the complete solution walkthrough on my YouTube channel!

Intuition

We need two positive integers a and b such that a + b = n and neither a nor b contains the digit 0 anywhere in its decimal representation.

Important clarification: “No-Zero integer” does not mean “the number is not equal to 0”; it means no digit '0' is allowed in the number.

Examples: 7, 19, 345 ✅; 10, 101, 203 ❌ (because they contain the digit 0).

Given 2 ≤ n ≤ 10^4, a simple linear scan over a (and thus b = n - a) is sufficient. As soon as we find a pair where both contain no '0', we can return it (the problem guarantees at least one valid answer).

Approach

  • Loop a from 1 to n - 1 (ensures both parts are positive).
  • Let b = n - a.
  • Check the “No-Zero” property by verifying '0' not in str(a) and '0' not in str(b).
  • Return the first valid pair [a, b] immediately (early exit).

Why this works:

  • The constraints are small; the worst-case scan is at most n-1 iterations.
  • String containment check is straightforward and unambiguous for the “no digit 0“ requirement.
  • A valid pair is guaranteed to exist, so the loop will terminate with a result.

Complexity Analysis

  • Time complexity: O(n) — we may try each a once until we find a valid pair.
  • Space complexity: O(1) — constant extra space.

Solution

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from typing import List

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

Example Walkthrough

Example 1: n = 11

  • Try a = 1, b = 10 → '0' in str(10) ❌
  • Try a = 2, b = 9 → Both contain no '0' ✅
  • Return [2, 9]

Example 2: n = 10000

  • Try a = 1, b = 9999 → Both contain no '0' ✅
  • Return [1, 9999]

Key Takeaways

  1. String-based digit checking is the most straightforward approach for “no digit 0” requirement
  2. Linear search is sufficient given the small constraint bounds
  3. Early termination optimizes performance when valid pair is found
  4. Problem guarantees ensure a valid solution always exists

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