#sort

Solutions The brute force approach is easy: just run two nested loops and for each value \( x \) find if it exists a value equals to \( target - x \). This solution is quadratic. Efficient solutions are based either on sorting or additional storage (e.g. frequency table, hash maps). Brute force solution 1 2 3 4 5 6 7 8 9 10 fn solve(m: usize, flavors: &[usize]) -> (usize, usize) { for (i, x) in flavors. [Read More]

Solutions This is a simple greedy algorithm. Naming \(t_i\) the time of order \(i\) and \(d_i\) the time needed to complete order \(i\), we sort all the orders by \(t_i + d_i\) and by \(i\). The idea implemented in C++: 1 2 3 4 5 6 7 8 9 10 int N, ti, di; cin >> N; vector<pair<int, int>> orders(N); for (auto i=1; i<=N; ++i) { cin >> ti >> di; orders[i-1] = {ti+di, i}; } sort(begin(orders), end(orders)); for (const auto& o : orders) cout << o. [Read More]

Solutions The sum is minimal when the array is sorted. Both descending and ascending order will yield the minimum sum of the absolute difference of adjacent elements. We need to count the number of swaps in both cases. To find the number of swaps we have to find the number of cycles in the array. For instance, given the following array: 1 1 2 4 3 We can construct a directed graph by making each value of the array point to its sorted position: [Read More]

Solutions Simple observation: there is no point in winning unimportant contests so we can lose them all to get their luck. If \(K\) is greater than the number of important contests, we can lose them all. In this case the luck balance is just the sum of all the contest values. Otherwise, we have to win some important contests but we don’t have to lose more than \(K\). So the number of important contests to win is: [Read More]

Solutions This easy problem can be solved by first sorting the calories in decreasing order: 1 2 3 4 5 6 7 8 9 10 long long n; cin >> n; vector<int> v(n); copy_n(istream_iterator<int>(cin), n, begin(v)); sort(begin(v), end(v), greater<>{}); auto miles = 0ll; for (auto i=0; i<n; ++i) { miles += v[i] * (1ll<<i); } cout << miles; The zip | map | reduce pattern emerges from the code: conceptually, it’s like zipping each calory with the ith power of 2, mapping the pair with product and summing along the way: [Read More]

Solutions Since we need to maximize the number of toys, we can sort the array in incresing order of price and pick them as long as the budget is available. A solution in C++: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 #include <iterator> #include <vector> #include <iostream> #include <algorithm> using namespace std; int main() { int n, k; cin >> n >> k; vector<int> prices(n); copy_n(istream_iterator<int>(cin), n, begin(prices)); sort(begin(prices), end(prices)); auto toys = 0, cost=0; for (auto i=0u; i<prices. [Read More]

Solutions The problem reduces to choosing K consecutive numbers in a sorted list for which the value of D is minimum. Which can be easily done by sorting the given number and calculating D for each group of K consecutive number. Here is a possible C++ Solution: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 #include <iterator> #include <cstdio> #include <vector> #include <iostream> #include <algorithm> #include <climits> using namespace std; int main() { int N, K; cin >> N >> K; vector<int> v(N); copy_n(istream_iterator<int>(cin), N, begin(v)); sort(begin(v), end(v)); auto tail = 0; auto head = K-1; auto unfairness = INT_MAX; while (head<N) { unfairness = min(unfairness, v[head]-v[tail]); tail++; head++; } cout << unfairness; } This results in a combination of zip | map | reduce patterns that in C++ can be written in terms of inner_product: [Read More]

Solutions This easy challenge has, actually, an interesting generalization. The problem can be quickly solved by noting that the number of elements to select is one less (4) than the length of the input array (5). So we can just sum all the numbers and subtract the result to the max and to the min of the array. The domain is subject to overflow so we have to use a 64-bit integer: [Read More]

Solutions The brute force solution consists in calculating the absolute difference of each pair and reducing to the minimum value. This takes \(O(N^2)\). We can find better solution, instead, by making a simple observation: the closest two numbers, the lowest their difference. It’s just like calculating the distance between points on the same axis. Indeed, the formula to calculate the distance between two points \(x1, x2\) on the same axis is simply \(|x1-x2|\). [Read More]

Solutions The naive solution is quadratic and it’s too slow for a few test cases. We show it here, for completeness: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 #include <iterator> #include <climits> #include <vector> #include <iostream> #include <algorithm> using namespace std; int main() { int n; cin >> n; vector<int> v(n); copy_n(istream_iterator<int>(cin), n, begin(v)); auto minLoss = INT_MAX; for (auto i=0; i<n; ++i) { for (auto j=i+1; j<n; ++j) { if (v[i] > v[j]) { minLoss = min(minLoss, v[i] - v[j]); } } } cout << minLoss; } Rust naive solution: [Read More]