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

Description

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

Solutions

Solution 1: Direct Enumeration

Starting from $1$, we enumerate $a$, then $b = n - a$. For each $a$ and $b$, we convert them to strings and concatenate them, then check if they contain the character '0'. If they do not contain '0', we have found the answer and return $[a, b]$.

The time complexity is $O(n \times \log n)$, where $n$ is the integer given in the problem. The space complexity is $O(\log n)$.

Python3

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]

Java

class Solution {
    public int[] getNoZeroIntegers(int n) {
        for (int a = 1;; ++a) {
            int b = n - a;
            if (!(a + "" + b).contains("0")) {
                return new int[] {a, b};
            }
        }
    }
}

C++

class Solution {
public:
    vector<int> getNoZeroIntegers(int n) {
        for (int a = 1;; ++a) {
            int b = n - a;
            if ((to_string(a) + to_string(b)).find('0') == -1) {
                return {a, b};
            }
        }
    }
};

Go

func getNoZeroIntegers(n int) []int {
	for a := 1; ; a++ {
		b := n - a
		if !strings.Contains(strconv.Itoa(a)+strconv.Itoa(b), "0") {
			return []int{a, b}
		}
	}
}

TypeScript

function getNoZeroIntegers(n: number): number[] {
    for (let a = 1; ; ++a) {
        const b = n - a;
        if (!`${a}${b}`.includes('0')) {
            return [a, b];
        }
    }
}

Solution 2: Direct Enumeration (Alternative Approach)

In Solution 1, we converted $a$ and $b$ into strings and concatenated them, then checked if they contained the character '0'. Here, we can use a function $f(x)$ to check whether $x$ contains the character '0', and then directly enumerate $a$, checking whether both $a$ and $b = n - a$ do not contain the character '0'. If they do not, we have found the answer and return $[a, b]$.

The time complexity is $O(n \times \log n)$, where $n$ is the integer given in the problem. The space complexity is $O(1)$.

Python3

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]

Java

class Solution {
    public int[] getNoZeroIntegers(int n) {
        for (int a = 1;; ++a) {
            int b = n - a;
            if (f(a) && f(b)) {
                return new int[] {a, b};
            }
        }
    }

    private boolean f(int x) {
        for (; x > 0; x /= 10) {
            if (x % 10 == 0) {
                return false;
            }
        }
        return true;
    }
}

C++

class Solution {
public:
    vector<int> getNoZeroIntegers(int n) {
        auto f = [](int x) {
            for (; x; x /= 10) {
                if (x % 10 == 0) {
                    return false;
                }
            }
            return true;
        };
        for (int a = 1;; ++a) {
            int b = n - a;
            if (f(a) && f(b)) {
                return {a, b};
            }
        }
    }
};

Go

func getNoZeroIntegers(n int) []int {
	f := func(x int) bool {
		for ; x > 0; x /= 10 {
			if x%10 == 0 {
				return false
			}
		}
		return true
	}
	for a := 1; ; a++ {
		b := n - a
		if f(a) && f(b) {
			return []int{a, b}
		}
	}
}

TypeScript

function getNoZeroIntegers(n: number): number[] {
    const f = (x: number): boolean => {
        for (; x; x = (x / 10) | 0) {
            if (x % 10 === 0) {
                return false;
            }
        }
        return true;
    };
    for (let a = 1; ; ++a) {
        const b = n - a;
        if (f(a) && f(b)) {
            return [a, b];
        }
    }
}