Finding 3-Digit Even Numbers LeetCode solution in Java

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/*
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As we know that we want unique numbers of 3 digits only that too only even.  so first we
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gather the frequency of all the digits we have, then we iterate from 100 to 999 ( all possible 3 digits numbers, 100,102,104...
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all possible even 3 digit numbers).  for ex  we are iterating and we are
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at 104  so we will see that if we have  digits
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1,0,4 in our database if yes then we can make this number from our
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available digits given to us.
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Time complexity : O(digits.length)  // due to making of frequency map
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Space Complexity : O(1) //fixed map array space for digits 0 to 9
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*/
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class Solution {
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  public int[] findEvenNumbers(int[] digits) {
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    int[] map = new int[10]; // for freq of 0 to 9 (digits are fixed)
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    for (int i = 0; i < digits.length; i++) { // make a frequency map of digits
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      map[digits[i]]++;
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    }
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    List<Integer> arr = new ArrayList<>();
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    for (int i = 100; i <= 999; i = i + 2) { // will always runs from 100 to 999
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      int num = i;
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      int[] freq = new int[10];
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      while (num > 0) { // will always run 3 times
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        int rem = num % 10;
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        freq[rem]++;
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        num = num / 10;
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      }
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      boolean res = findans(freq, map);
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      if (res) arr.add(i);
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    }
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    int[] ans = new int[arr.size()]; // logic for arraylist to array conversion
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    for (int i = 0; i < arr.size(); i++) { // at max we can have all num from 100 to 998 only
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      ans[i] = arr.get(i);
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    }
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    return ans;
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  }
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  private boolean findans(int[] currentNum, int[] database) {
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    for (int i = 0; i < 10; i++) { // it will always run for at max 10 times
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      if (currentNum[i] > database[i]) return false;
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    }
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    return true;
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  }
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}