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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 = 9 = 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 $O(\log n)$.

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 $O(1)$.

• Java
• C++
• Python
• Go
• TypeScript
• Rust

• 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};
              }
          }
      }
  }
  
  
  // Solution 2
  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;
      }
  }
  
  

• 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};
              }
          }
      }
  };
  
  
  // Solution 2
  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};
              }
          }
      }
  };
  
  

• 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]
  
  
  # Solution 2
  class Solution:
      def getNoZeroIntegers(self, n: int) -> List[int]:
          def f(x: int) -> bool:
              while x:
                  if x % 10 == 0:
                      return False
                  x //= 10
              return True
  
          for a in count(1):
              b = n - a
              if f(a) and f(b):
                  return [a, b]
  
  

• 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}
  		}
  	}
  }
  
  
  // Solution 2
  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}
  		}
  	}
  }
  
  

• function getNoZeroIntegers(n: number): number[] {
      for (let a = 1; ; ++a) {
          const b = n - a;
          if (!`${a}${b}`.includes('0')) {
              return [a, b];
          }
      }
  }
  
  
  // Solution 2
  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];
          }
      }
  }
  
  

• impl Solution {
      pub fn get_no_zero_integers(n: i32) -> Vec<i32> {
          for a in 1..n {
              let b = n - a;
              if !a.to_string().contains('0') && !b.to_string().contains('0') {
                  return vec![a, b];
              }
          }
          vec![]
      }
  }
  
  
  // Solution 2
  impl Solution {
      pub fn get_no_zero_integers(n: i32) -> Vec<i32> {
          fn f(mut x: i32) -> bool {
              while x > 0 {
                  if x % 10 == 0 {
                      return false;
                  }
                  x /= 10;
              }
              true
          }
  
          for a in 1..n {
              let b = n - a;
              if f(a) && f(b) {
                  return vec![a, b];
              }
          }
          vec![]
      }
  }
  
  

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