Easy
Minimum Sum of Four Digit Number After Splitting Digits — C++
Full explanation · Time O(1) · Space O(1)
// Time: O(d) = O(1), d is the number of digits
// Space: O(d) = O(1)
// greedy
class Solution {
public:
int minimumSum(int num) {
vector<int> nums;
for (; num; num /= 10) {
nums.emplace_back(num % 10);
}
inplace_counting_sort(&nums, false);
int a = 0, b = 0;
for (const auto& x: nums) {
a = a * 10 + x;
swap(a, b);
}
return a + b;
}
private:
void inplace_counting_sort(vector<int> *nums, bool is_reverse) {
const int max_num = *max_element(cbegin(*nums), cend(*nums));
vector<int> count(max_num + 1);
for (const auto& num : *nums) {
++count[num];
}
for (int i = 1; i < size(count); ++i) {
count[i] += count[i - 1];
}
for (int i = size(*nums) - 1; i >= 0; --i) { // inplace but unstable sort
while ((*nums)[i] >= 0) {
--count[(*nums)[i]];
const int j = count[(*nums)[i]];
tie((*nums)[i], (*nums)[j]) = pair((*nums)[j], ~(*nums)[i]);
}
}
for (auto& num : *nums) {
num = ~num; // restore values
}
if (is_reverse) { // unstable sort
reverse(begin(*nums), end(*nums));
}
}
};
// Time: O(dlogd) = O(1), d is the number of digits
// Space: O(d) = O(1)
// greedy
class Solution2 {
public:
int minimumSum(int num) {
vector<int> nums;
for (; num; num /= 10) {
nums.emplace_back(num % 10);
}
sort(begin(nums), end(nums));
int a = 0, b = 0;
for (const auto& x: nums) {
a = a * 10 + x;
swap(a, b);
}
return a + b;
}
};