# Solutions in Rust for some competitive programming problems
# Solutions in Rust for some competitive programming problems
Created for the "Competitive Programming" course (academic year 23/24) at the Department of Computer Science of the University of Pisa.
Created for the "Competitive Programming" course (academic year 23/24) at the Department of Computer Science of the University of Pisa. It contains horrible code
* **Idea:** We want to find the maximum subarray sum in an array of integers. It works by _iterating through the array and keeping track of the maximum sum seen so far and the maximum sum ending at the current index_. At each index, the algorithm compares the current element with the maximum sum ending at the previous index plus the current element. If the current element is greater, then the maximum sum ending at the current index is just the current element. Otherwise, the maximum sum ending at the current index is the sum of the maximum sum ending at the previous index and the current element. The algorithm also keeps track of the maximum sum seen so far, which is the maximum of all the maximum sums ending at each index. The final result is the maximum sum seen so far.
* The strategy used to solve this problem is the two-pointer approach. We use _two pointers_, one at the beginning and one at the end of the array. We also keep track of _the maximum height_ of the left and right sides. We then move the pointers towards each other, updating the maximum height as we go. If the maximum height of the left side is less than the maximum height of the right side, we know that we can trap water on the left side. Similarly, if the maximum height of the right side is less than the maximum height of the left side, we know that we can trap water on the right side. We keep adding the trapped water to the result until the pointers meet.
* [x] [Search for a peak in an (unsorted) array](https://leetcode.com/problems/find-peak-element/)
* The idea is to use a modified version of the binary search and observe that in there always will be a peak element (the worst case is a monotonically increasing/decreasing sequence, and in that case the peak will be the last/first element). So we just check if `nums[mid]` is greater than `nums[mid+1]` or not. If it is, we can discard the right part of the array, otherwise we can discard the left part of the array. We will always move towards the direction of the peak. It's important to note that `nums[-1] = nums[n] = -∞` and that `nums[i] != nums[i+1]` for all `i`.
Luca Lombardo
1 year ago
committed by
GitHub