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Leetcode 1317. Convert Integer to the Sum of Two No

Problem Statement

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

Given an integer n. You need to find two integers a and b such that:

  1. a + b == n
  2. a * b > 0 (both a and b must be positive)
  3. Neither a nor b contains any zero digits.

Return the two integers in the form of an array [a, b]. There may be multiple valid solutions. You can return any of them.

Clarifying Questions

  1. Can n be negative?
    • The problem statement implies n is a positive integer.
  2. Are there constraints on n?
    • Constraints are typically present in the full problem statement on LeetCode but are usually positive integers up to a given limit.
  3. What should we do if no valid pair exists?
    • In this problem, there will always be at least one valid pair for positive integers.

Strategy

  1. Incremental Search:
    • Start with the smallest possible pairs.
    • Increment one of the numbers while decrementing the other to keep the sum constant.
  2. No-Zero Check:
    • A utility function to check if a number contains any zero digits.

Time Complexity

The algorithm will primarily involve iterating through the numbers up to n, making the time complexity approximately O(n), but since we find a solution quickly, it is often much faster.

Code

#include <vector>
#include <string>

// Function to check if a number contains any zero digits
bool hasZero(int num) {
    while (num > 0) {
        if (num % 10 == 0) {
            return true;
        }
        num /= 10;
    }
    return false;
}

std::vector<int> getNoZeroIntegers(int n) {
    for (int a = 1; a < n; ++a) {
        int b = n - a;
        if (!hasZero(a) && !hasZero(b)) {
            return {a, b};
        }
    }
    return {}; // Should never reach here for valid positive n.
}

Explanation

  1. hasZero Function:
    • We define a utility function, hasZero, to check if a given integer contains the digit ‘0’.
  2. getNoZeroIntegers Function:
    • We loop through all values of a starting from 1 to n-1.
    • For each value of a, compute b as n - a.
    • We then check if both a and b do not contain the digit ‘0’ using the hasZero function.
    • If both numbers are valid, we return them as a vector.

This approach ensures that we find a valid pair {a, b} that meet the criteria, with reasonable efficiency given the constraints of the problem.

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