Problem Description
Given a 1-indexed array of distinct integers, distribute the elements into two arrays, arr1 and arr2, through n operations. Initially, add the first element to arr1 and the second element to arr2. For each subsequent element, compare the last elements of arr1 and arr2; if the last element of arr1 is greater than that of arr2, append the element to arr1, otherwise append it to arr2. Finally, return the result obtained by concatenating arr1 and arr2.
Key Insights
- The problem requires simulating the operations as described.
- Each operation depends solely on the current state (last elements) of arr1 and arr2.
- The constraints are small (n up to 50), allowing a straightforward solution with linear iteration.
- Use simple dynamic arrays (lists) for storing elements.
Space and Time Complexity
Time Complexity: O(n) - We perform a single pass through the input array. Space Complexity: O(n) - We use additional arrays to store the distributed elements.
Solution
We solve the problem by simulating the operations according to the rules provided. Start by initializing arr1 with the first element and arr2 with the second element. Then for each subsequent number, check the last element in arr1 and arr2. Append the number to arr1 if its last element is greater than that of arr2; otherwise, append it to arr2. Finally, merge arr1 and arr2 to form the resulting array. The approach is intuitive and leverages conditional checks along with dynamic array updates.