LeetCode 3666 | Minimum Operations to Equalize Binary String | BFS + Math | Java | Hard скачать в хорошем качестве

LeetCode 3666 | Minimum Operations to Equalize Binary String | BFS + Math | Java | Hard

LeetCode 3666 – Minimum Operations to Equalize Binary String (Java) Today's LeetCode Daily You are given a binary string s and an integer k. In one operation: Choose exactly k different indices and flip each bit. Goal: Make all characters equal to 1 using minimum operations. Core Insight: Instead of tracking the full string, we only track the number of zeros. Why? Because each operation changes the zero count deterministically. If current zero count is curr, after flipping k bits, new zero count depends on: how many zeros we flipped how many ones we flipped This creates a range of reachable zero counts. Approach Used: ✔ Count initial zeros ✔ If zero count is 0 → return 0 ✔ Use BFS on zero count states ✔ steps array tracks minimum operations ✔ Maintain two TreeSets: one for even counts one for odd counts Parity matters because zero count parity changes based on k. For each state: Compute minReach and maxReach which defines reachable zero counts. Use TreeSet ceiling to efficiently visit valid next states. Stop when zero count becomes 0. Key Idea: Convert problem into graph traversal over zero count states. Use BFS to find shortest path. Time Complexity: Approximately O(n log n) Space Complexity: O(n) Problem Link: https://leetcode.com/problems/minimum... Java Solution (GitHub): https://github.com/Amandf/LeetCode-So... More daily Java solutions. Clean code. No mic. Pure logic. #leetcode #leetcodedailychallenge #java #bfs #graphtraversal #hardproblem #dsa #codinginterview #programming