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3483. Unique 3-Digit Even Numbers

Description

You are given an array of digits called digits. Your task is to determine the number of distinct three-digit even numbers that can be formed using these digits.

Note: Each copy of a digit can only be used once per number, and there may not be leading zeros.

 

Example 1:

Input: digits = [1,2,3,4]

Output: 12

Explanation: The 12 distinct 3-digit even numbers that can be formed are 124, 132, 134, 142, 214, 234, 312, 314, 324, 342, 412, and 432. Note that 222 cannot be formed because there is only 1 copy of the digit 2.

Example 2:

Input: digits = [0,2,2]

Output: 2

Explanation: The only 3-digit even numbers that can be formed are 202 and 220. Note that the digit 2 can be used twice because it appears twice in the array.

Example 3:

Input: digits = [6,6,6]

Output: 1

Explanation: Only 666 can be formed.

Example 4:

Input: digits = [1,3,5]

Output: 0

Explanation: No even 3-digit numbers can be formed.

 

Constraints:

• 3 <= digits.length <= 10
• 0 <= digits[i] <= 9

Solutions

Solution 1: Hash Set + Enumeration

We use a hash set $\mathit{\text{s}}$ to record all distinct three-digit even numbers, and then enumerate all possible three-digit even numbers to add them to the hash set.

Finally, we return the size of the hash set.

The time complexity is $O(n^3)$, and the space complexity is $O(n^3)$. Where $n$ is the length of the array $\mathit{\text{digits}}$.

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

• class Solution {
      public int totalNumbers(int[] digits) {
          Set<Integer> s = new HashSet<>();
          int n = digits.length;
          for (int i = 0; i < n; ++i) {
              if (digits[i] % 2 == 1) {
                  continue;
              }
              for (int j = 0; j < n; ++j) {
                  if (i == j) {
                      continue;
                  }
                  for (int k = 0; k < n; ++k) {
                      if (digits[k] == 0 || k == i || k == j) {
                          continue;
                      }
                      s.add(digits[k] * 100 + digits[j] * 10 + digits[i]);
                  }
              }
          }
          return s.size();
      }
  }
  
  

• class Solution {
  public:
      int totalNumbers(vector<int>& digits) {
          unordered_set<int> s;
          int n = digits.size();
          for (int i = 0; i < n; ++i) {
              if (digits[i] % 2 == 1) {
                  continue;
              }
              for (int j = 0; j < n; ++j) {
                  if (i == j) {
                      continue;
                  }
                  for (int k = 0; k < n; ++k) {
                      if (digits[k] == 0 || k == i || k == j) {
                          continue;
                      }
                      s.insert(digits[k] * 100 + digits[j] * 10 + digits[i]);
                  }
              }
          }
          return s.size();
      }
  };
  
  

• class Solution:
      def totalNumbers(self, digits: List[int]) -> int:
          s = set()
          for i, a in enumerate(digits):
              if a & 1:
                  continue
              for j, b in enumerate(digits):
                  if i == j:
                      continue
                  for k, c in enumerate(digits):
                      if c == 0 or k in (i, j):
                          continue
                      s.add(c * 100 + b * 10 + a)
          return len(s)
  
  

• func totalNumbers(digits []int) int {
  	s := make(map[int]struct{})
  	for i, a := range digits {
  		if a%2 == 1 {
  			continue
  		}
  		for j, b := range digits {
  			if i == j {
  				continue
  			}
  			for k, c := range digits {
  				if c == 0 || k == i || k == j {
  					continue
  				}
  				s[c*100+b*10+a] = struct{}{}
  			}
  		}
  	}
  	return len(s)
  }
  
  

• function totalNumbers(digits: number[]): number {
      const s = new Set<number>();
      const n = digits.length;
      for (let i = 0; i < n; ++i) {
          if (digits[i] % 2 === 1) {
              continue;
          }
          for (let j = 0; j < n; ++j) {
              if (i === j) {
                  continue;
              }
              for (let k = 0; k < n; ++k) {
                  if (digits[k] === 0 || k === i || k === j) {
                      continue;
                  }
                  s.add(digits[k] * 100 + digits[j] * 10 + digits[i]);
              }
          }
      }
      return s.size;
  }
  
  

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