#patterns

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 A tradeoff of this problem is about using a temporary array or not. In the latter case, the steps to do are simpler because we can just copy the two parts the array: all the elements before the rotation point and then all the element after. Here is a Java implementation: 1 2 3 4 5 6 7 8 9 10 11 public static int[] rotateArray(int[] arr, int d) { int n = arr. [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 This is an easy problem we can experiment with. The simplest solution consists in counting how many characters mismatch: 1 2 3 4 5 6 7 char c1, c2, c3; auto cnt = 0; while (cin >> c1 >> c2 >> c3) { cnt += (c1 != 'S') + (c2 != 'O') + (c3 != 'S'); } cout << cnt; We can use the reduce pattern: 1 2 3 4 static const char match[] = {'S', 'O', 'S'}; cout << accumulate(istream_iterator<char>(cin), istream_iterator<char>(), make_pair(0, 0), [](pair<int, int> s, char c) { return make_pair(s. [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 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]

Bob has an array \( a \) with \( n \) positive integer elements. Bob likes order, so he wants his array to be sorted in non-decreasing order. He decides to swap two elements in the array to make his array sorted. (A swap is defined as switching two elements at distinct locations in the array.) Your task is to determine if this can be done. If he can’t sort the array with one swap, print \( -1 \). [Read More]

Solutions If the string is already a palindrome, -1 is the answer since no character need be removed. If the given string is not a palindrome, we must find one character that, once removed, will make it a palindrome. We can do this by checking if str[i] == str[N - 1 - i] where N is the length of the string for all i starting from i=0. Once this condition fails, all we have to do is to check if str[0:i-1] + str[i+1:N-1] is a palindrome. [Read More]

Solutions We can find the answer by sorting one of the arrays in ascending order and the other in descending order. Then we check if the sum of all elements in pairs is greater than or equal to \(K\). It’s a greedy algorithm and its proof is left to the reader (see here). In C++: 1 2 3 4 5 6 7 8 sort(begin(v1), end(v1)); sort(begin(v2), end(v2), greater<int>()); bool ok = true; for (auto i=0; i<N && ok; ++i) { ok = (v1[i]+v2[i])>=K; } cout << (ok ? [Read More]