Hard
Concatenated Divisibility — C++
Full explanation · Time O(nlogr + k * n * 2^n) · Space O(logr + k * 2^n)
// Time: O(nlogr + k * n * 2^n)
// Space: O(logr + k * 2^n)
// dp, backtracing
class Solution {
public:
vector<int> concatenatedDivisibility(vector<int>& nums, int k) {
const auto& length = [](int x) {
int l = 0;
for (; x > 0; x /= 10) {
++l;
}
return max(l, 1);
};
vector<int> lookup(size(nums));
for (int i = 0; i < size(nums); ++i) {
lookup[i] = length(nums[i]);
}
const int mx = ranges::max(lookup);
vector<int> pow10(mx + 1);
pow10[0] = 1 % k;
for (int i = 0; i + 1 < size(pow10); ++i) {
pow10[i + 1] = (pow10[i] * 10) % k;
}
vector<vector<bool>> dp(1 << size(nums), vector<bool>(k));
dp.back()[0] = true;
for (int mask = (size(dp) - 1) - 1; mask >= 0; --mask) {
for (int r = 0; r < k; ++r) {
for (int i = 0; i < size(nums); ++i) {
if (mask & (1 << i)) {
continue;
}
if (dp[mask | (1 << i)][(r * pow10[lookup[i]] + nums[i]) % k]) {
dp[mask][r] = true;
break;
}
}
}
}
vector<int> result;
if (!dp[0][0]) {
return result;
}
vector<pair<int,int>> order;
for (int i = 0; i < size(nums); ++i) {
order.emplace_back(nums[i], i);
}
sort(begin(order), end(order));
for (int _ = 0, mask = 0, r = 0; _ < size(nums); ++_) {
for (const auto& [_, i] : order) {
if (mask & (1 << i)) {
continue;
}
if (dp[mask | (1 << i)][(r * pow10[lookup[i]] + nums[i]) % k]) {
result.emplace_back(nums[i]);
mask |= (1 << i);
r = (r * pow10[lookup[i]] + nums[i]) % k;
break;
}
}
}
return result;
}
};