2094: Finding-3-Digit-Even-Numbers
Easy
Since the solution space is quite small (only a maximum of
449 possible solutions), you can just
brute-force this with a recursive function.
I basically just iterated through all the possible 3 digit
numbers (using a hash-map to keep track of
numbers). Then, if it is even, I just add it to my
ans vector.
Finally, since my recursive function always
iterates from 0 to 9 at each stage, when
we end up pushing digits to the ans vector, they will
always be sorted.
Code:
class Solution {
public:
void recursiveDigits(vector<int>& digits, vector<int>& mp, vector<int>& ans, int curr) {
if (curr >= 100) {
if (curr % 2 == 0) ans.push_back(curr);
return;
}
for (int i = 0; i < 10; ++i) {
if (i == 0 && curr == 0) continue;
if (mp[i] > 0) {
--mp[i];
curr = curr * 10 + i;
recursiveDigits(digits, mp, ans, curr);
curr /= 10;
++mp[i];
}
}
}
vector<int> findEvenNumbers(vector<int>& digits) {
vector<int> ans;
vector<int> mp(10, 0);
for (int& i : digits) {
++mp[i];
}
recursiveDigits(digits, mp, ans, 0);
return ans;
}
};