Freaks City

Problem T1: Geometry and Distances The first problem involves calculating two distances. The first is the distance from a given point to a line segment, and the second is the distance from the same point to a circle. The distance to the circle is straightforward: it's the distance from the point to the circle's center minus the radius. For the ...

Problems Data Partitioning Given an integer array nums of even length, split it into two arrays nums1 and nums2, such that: nums1.length == nums2.length == nums.length / 2 Both nums1 and nums2 contain unique elements. Return true if the array can be partitioned in this way, otherwise return false. Approach Use a hash map to count occurrences ...

This problem centers on dynamically maintaining and querying the minimum number of positive rating increments needed to offset a cumulative negative rating sum—effectively determining how many smallest available positive values must be selected (in sorted order) so their total is at least the absolute value of the negative sum. The core insight ...

Problem A Given three integers A, B, and C, determine whether C falls within the range [A, B], i.e., check if A ≤ C ≤ B. Solution #include <bits/stdc++.h> using namespace std; int main() { int first, last, target; cin >> first >> last >> target; if (first <= target && target <= last) ...

This problem requires counting the number of zeros between the first and last occurrence of 1 in an array. #include <bits/stdc++.h> using namespace std; typedef long long ll; void solve() { int n; cin >> n; vector<int> arr(n); int firstOne = -1, lastOne = -1; for (int i = 0; i < n; i++) { ...

Chino’s Take-Away Game The game ends when only one stone remains. If the initial pile contains an odd count, the first player can always remove exactly one stone, leaving an even number. Any subsequent move by the second player must remove an odd amount (co-prime with the even remainder), again leaving an odd count. By induction the first playe ...

A. Be Positive Count how many elements are negative. If the amount is odd, flip one more element so the product becomes positive. void solve() { int n; cin >> n; int neg = 0; for (int i = 0; i < n; ++i) { int v; cin >> v; if (v < 0) ++neg; } cout << (neg & 1 ? neg + 1 : neg) <&l ...

Understanding Arrays Array Definition An array represents a collection of elements with identical data types, occupying contiguous memory locations. All stored elements maintain consistent type Memory allocation occurs in sequential blocks Each location has an assigned identifier starting from zero (index) Array Creation and Initialization Cr ...

Binary search is a fundamental technique for efficiently locating an element in a sorted array or finding an optimal value in a monotonic space. The core idea is to maintain a search interval that shrinks until the desired position is identified. Rather than memorizing multiple templates, a single open-interval approach works for both ordinary ...

Merging Multiple Sorted Linked Lists Efficiently combining several pre-sorted linked structures requires a mechanism to consistently extract the minimum available element across all sources. A min-heap provides an optimal approach for this task, maintaining a pool of candidate nodes and guaranteeing logarithmic insertion and extraction times. B ...