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Posts tagged as “enumerate”

花花酱 LeetCode 2094. Finding 3-Digit Even Numbers

You are given an integer array digits, where each element is a digit. The array may contain duplicates.

You need to find all the unique integers that follow the given requirements:

  • The integer consists of the concatenation of three elements from digits in any arbitrary order.
  • The integer does not have leading zeros.
  • The integer is even.

For example, if the given digits were [1, 2, 3], integers 132 and 312 follow the requirements.

Return sorted array of the unique integers.

Example 1:

Input: digits = [2,1,3,0]
Output: [102,120,130,132,210,230,302,310,312,320]
Explanation: 
All the possible integers that follow the requirements are in the output array. 
Notice that there are no odd integers or integers with leading zeros.

Example 2:

Input: digits = [2,2,8,8,2]
Output: [222,228,282,288,822,828,882]
Explanation: 
The same digit can be used as many times as it appears in digits. 
In this example, the digit 8 is used twice each time in 288, 828, and 882. 

Example 3:

Input: digits = [3,7,5]
Output: []
Explanation: 
No even integers can be formed using the given digits.

Example 4:

Input: digits = [0,2,0,0]
Output: [200]
Explanation: 
The only valid integer that can be formed with three digits and no leading zeros is 200.

Example 5:

Input: digits = [0,0,0]
Output: []
Explanation: 
All the integers that can be formed have leading zeros. Thus, there are no valid integers.

Constraints:

  • 3 <= digits.length <= 100
  • 0 <= digits[i] <= 9

Solution: Enumerate all three digits even numbers

Check 100, 102, … 998. Use a hashtable to check whether all digits are covered by the given digits.

Time complexity: O(1000*lg(1000))
Space complexity: O(10)

C++

Python-Like enumerate() In C++17 – C++ Weekly EP1

C++