Medium
Balanced K-Factor Decomposition — C++
Full explanation · Time precompute: O(rlogr) runtime: O(k * (logn)^(k - 1)) · Space O(rlogr)
// Time: precompute: O(rlogr)
// runtime: O(k * (logn)^(k - 1))
// Space: O(rlogr)
// backtracking, number theory
const auto& factors = [](int n) {
vector<vector<int>> result(n + 1);
for (int i = 1; i <= n; ++i) {
for (int j = i; j <= n; j += i) {
result[j].emplace_back(i);
}
}
return result;
};
const int MAX_N = 1e5;
const auto& FACTORS = factors(MAX_N);
class Solution {
public:
vector<int> minDifference(int n, int k) {
vector<int> result, curr;
function<void (int)> backtracking = [&](int remain) {
const int start = !empty(curr) ? curr.back() : 1;
if (size(curr) == k - 1) {
if (remain >= start) {
curr.emplace_back(remain);
if (empty(result) || result.back() - result[0] > curr.back() - curr[0]) {
result = curr;
}
curr.pop_back();
}
return;
}
const auto& factors = FACTORS[remain];
for (auto it = lower_bound(cbegin(factors), cend(factors), start); it != cend(factors); ++it) {
curr.emplace_back(*it);
backtracking(remain / *it);
curr.pop_back();
}
};
backtracking(n);
return result;
}
};
// Time: O(k * (n^(1/2) * n^(1/4) * n^(1/8) * n^(1/6) + n^(1/2) * n^(1/4) * n^(1/8) + n^(1/2) * n^(1/4) + n^(1/2))) <= O(k^2 * n)
// Space: O(k)
// backtracking, number theory
class Solution2 {
public:
vector<int> minDifference(int n, int k) {
vector<int> result, curr;
function<void (int)> backtracking = [&](int remain) {
const int start = !empty(curr) ? curr.back() : 1;
if (size(curr) == k - 1) {
if (remain >= start) {
curr.emplace_back(remain);
if (empty(result) || result.back() - result[0] > curr.back() - curr[0]) {
result = curr;
}
curr.pop_back();
}
return;
}
for (int i = 1; i * i <= remain; ++i) {
if (remain % i) {
continue;
}
const int j = remain / i;
if (i >= start) {
curr.emplace_back(i);
backtracking(j);
curr.pop_back();
}
if (j == i) {
continue;
}
if (j >= start) {
curr.emplace_back(j);
backtracking(i);
curr.pop_back();
}
}
};
backtracking(n);
return result;
}
};
// Time: O(2^(k-1) * k * n)
// Space: O(k)
// backtracking, number theory
class Solution3 {
public:
vector<int> minDifference(int n, int k) {
vector<int> result, curr;
function<void (int)> backtracking = [&](int remain) {
if (size(curr) == k - 1) {
curr.emplace_back(remain);
if (empty(result) || ranges::max(result) - ranges::min(result) > ranges::max(curr) - ranges::min(curr)) {
result = curr;
}
curr.pop_back();
return;
}
for (int i = 1; i * i <= remain; ++i) {
if (remain % i) {
continue;
}
const int j = remain / i;
curr.emplace_back(i);
backtracking(j);
curr.pop_back();
if (j == i) {
continue;
}
curr.emplace_back(j);
backtracking(i);
curr.pop_back();
}
};
backtracking(n);
return result;
}
};