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

table of contents

First, I defined a function that checked whether an integer contains any 0 in its decimal representation:

bool checkZeroIntegers(int n) {
    for (; n > 0; n /= 10) {
        if (n % 10 == 0) {
            return false;
        }
    }
    return true;
}

Then, I simply iterated through all pairs of integers that sum to n, running checkZeroIntegers on both integers present in the pair. If both integers do not contain any zero integers, then we can simply return the pair.

for (int i = 1; i < n; ++i) {
    if (checkZeroIntegers(i) && checkZeroIntegers(n-i)) {
        return {i, n-i};
    }
}

code

class Solution {
public:
    vector<int> getNoZeroIntegers(int n) {
        for (int i = 1; i < n; ++i) {
            if (checkZeroIntegers(i) && checkZeroIntegers(n-i)) {
                return {i, n-i};
            }
        }
        return {};
    }
private:
    bool checkZeroIntegers(int n) {
        for (; n > 0; n /= 10) {
            if (n % 10 == 0) {
                return false;
            }
        }
        return true;
    }
};

complexity

learnings

time taken