There is a directed graph consisting of n nodes numbered from 0 to n - 1 and n directed edges.
You are given a 0-indexed array edges where edges[i] indicates that there is an edge from node i to node edges[i].
Consider the following process on the graph:
\n\nx and keep visiting other nodes through edges until you reach a node that you have already visited before on this same process.Return an array answer where answer[i] is the number of different nodes that you will visit if you perform the process starting from node i.
\n
Example 1:
\n![\"\"](\"https://assets.leetcode.com/uploads/2023/08/31/graaphdrawio-1.png\")
\n\nInput: edges = [1,2,0,0]\nOutput: [3,3,3,4]\nExplanation: We perform the process starting from each node in the following way:\n- Starting from node 0, we visit the nodes 0 -> 1 -> 2 -> 0. The number of different nodes we visit is 3.\n- Starting from node 1, we visit the nodes 1 -> 2 -> 0 -> 1. The number of different nodes we visit is 3.\n- Starting from node 2, we visit the nodes 2 -> 0 -> 1 -> 2. The number of different nodes we visit is 3.\n- Starting from node 3, we visit the nodes 3 -> 0 -> 1 -> 2 -> 0. The number of different nodes we visit is 4.\n\n\n
Example 2:
\n![\"\"](\"https://assets.leetcode.com/uploads/2023/08/31/graaph2drawio.png\")
\n\nInput: edges = [1,2,3,4,0]\nOutput: [5,5,5,5,5]\nExplanation: Starting from any node we can visit every node in the graph in the process.\n\n\n
\n
Constraints:
\n\nn == edges.length2 <= n <= 1050 <= edges[i] <= n - 1edges[i] != iCompiled with clang 11 using the latest C++ 20 standard.
Your code is compiled with level two optimization (-O2). AddressSanitizer is also enabled to help detect out-of-bounds and use-after-free bugs.
Most standard library headers are already included automatically for your convenience.
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