Exercise 10

Added Exercise 10 to Repo

Informatik_I/Exercise_10/Task_4/README.org

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+#+title: Task 4: Decomposing a Set into Intervals
+
+#+author: JirR02
+* Task
+:PROPERTIES:
+:CUSTOM_ID: task
+:END:
+Any finite set of integers can be uniquely decomposed as a union of
+disjoint maximal integer intervals, e.g as the union of non-overlapping
+intervals which are as large as possible. For example, the set
+\(X = \{ 1,2,3,5,6,8 \}\) decomposes as
+\(X = [1,3] cup [5,6] cup [8,8]\). Note that
+\(X = [1,2] cup [3,3] cup [5,6] cup [8,8]\) or
+\(X = [1,2] cup [5,6] cup [5,6] cup [8,9]\) are not valid
+decompositions, as \([1,2]\) and \([3,3]\) are not maximal interval
+subsets of , and intervals may not repeat.
+
+Write a program that inputs a set of integers, as the (possibly
+repeating) sequence of its members, and outputs the interval
+decomposition of the set in ascending order of boundaries.
+
+/Reminder/: ordered sets can be represented in C++ using the
+=std::set<...>= container, in which elements are added using the
+=insert= method. Iteration (using iterators) in ordered sets takes place
+in increasing order.
+
+=Hint=: You may first define a function that, from two =std::set=
+iterators =first= and =last=, finds the largest integer interval
+starting from =first= and ending before =last=, and returns an iterator
+just after this interval.
+
+* Input
+:PROPERTIES:
+:CUSTOM_ID: input
+:END:
+A set of non-negative integer, given as the sequence of its members
+followed by a negative integer. The sequence can be in any order and may
+feature repetitions.
+
+Example:
+
+#+begin_src shell
+3 5 8 6 5 3 2 1 -1
+#+end_src
+
+* Output
+:PROPERTIES:
+:CUSTOM_ID: output
+:END:
+The interval decomposition of the input set, with interval given in
+increasing order of boundaries. The interval decomposition must be given
+as a parenthesized sequence of intervals, with intervals separated by
+the character U. Intervals themselves must be given by their lower and
+upper bound, separated by a comma and parenthesized by brackets, with
+the lower bound coming first.
+
+Example:
+
+#+begin_src shell
+([1,3]U[5,6]U[8,8])
+#+end_src
+
+* Solution
+:PROPERTIES:
+:CUSTOM_ID: solution
+:END:
+#+begin_src cpp
+#include <iostream>
+#include <set>
+
+using namespace std;
+
+void find_interval(set<int>::iterator &first, set<int>::iterator &last,
+                   set<pair<int, int>> &Interval) {
+  if (first == last)
+    return;
+
+  int firstElement = *first;
+  int curElement = *first;
+
+  while ((++first != last) && (*first - curElement <= 1)) {
+    curElement = *first;
+  }
+
+  pair<int, int> curInterval = {firstElement, curElement};
+  Interval.insert(curInterval);
+
+  find_interval(first, last, Interval);
+}
+
+int main() {
+  int in;
+  set<int> Set;
+  set<pair<int, int>> Interval;
+
+  cin >> in;
+
+  while (in >= 0) {
+    Set.insert(in);
+    cin >> in;
+  }
+
+  set<int>::iterator first = Set.begin();
+  set<int>::iterator last = Set.end();
+
+  find_interval(first, last, Interval);
+
+  int size = Interval.size();
+  int count = 1;
+
+  cout << "(";
+  for (pair<int, int> h : Interval) {
+    cout << "[" << h.first << "," << h.second;
+    if (count < size)
+      cout << "]" << "U";
+    else
+      cout << "]";
+    ++count;
+  }
+  cout << ")";
+
+  return 0;
+}
+#+end_src