cycle-detection

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LeetCode 957: Prison Cells After N Days

Problem Restatement We are given an array cells of length 8 . Each cell is either: 0 # vacant 1 # occupied Every day, the cells change using this rule: If a cell has two adjacent neighbors that are both occupied or both vacant, the cell becomes occupied. Otherwise, it becomes vacant. The first and last cells do not have two neighbors, so they always become vacant after one day....

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LeetCode 565: Array Nesting

Problem Restatement We are given an integer array nums of length n . The array is a permutation of the numbers from 0 to n - 1 . That means every value appears exactly once. For each starting index k , we build a set: s[k] = { nums[k], nums[nums[k]], nums[nums[nums[k]]], ... } We stop right before a duplicate value would be added. Return the largest possible size of s[k]...

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LeetCode 802: Find Eventual Safe States

Problem Restatement We are given a directed graph with n nodes labeled from 0 to n - 1 . The graph is represented as an adjacency list: graph[i] contains every node that node i points to. A terminal node has no outgoing edges. A safe node is a node where every possible path starting from that node eventually reaches a terminal node. If a node can reach a cycle, then...

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LeetCode 287: Find the Duplicate Number

Problem Restatement We are given an array nums containing n + 1 integers. Each integer is in the range: 1 <= nums[i] <= n Since there are n + 1 numbers but only n possible values, at least one number must repeat. The problem says there is exactly one repeated number, but it may appear more than twice. We need to return that repeated number. We must solve it without...

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LeetCode 457: Circular Array Loop

Problem Restatement We are given a circular array nums of non-zero integers. Each value tells us how far to move from the current index: nums[i] > 0 # move forward nums[i] < 0 # move backward Because the array is circular, moving past the last index wraps around to the first index, and moving before the first index wraps around to the last index. We need to determine whether there...