You are given an undirected graph as an adjacency list `adj`, where `adj[i]` lists the indices of guests who know guest `i`. You must seat every guest into at most two groups so that no two guests in the same group know each other. Return `true` if such a two-group seating is possible, otherwise `false`. This is equivalent to deciding whether the friendship graph is **bipartite**. Your solution must handle **disconnected components** (run the check from every unvisited node). **Assumptions** (state these before coding): - Edges are bidirectional: if `j` appears in `adj[i]`, the pair `(i, j)` is an undirected acquaintance edge. (The reference treats edges that appear from either endpoint; for a well-formed input each edge is listed on both sides.) - A **self-loop** (`i` in `adj[i]`) means a guest "knows" themselves and they cannot be placed in any group consistently, so the answer is `false`. - **Duplicate entries** in an adjacency list are harmless and treated the same as a single edge. - Guests are indexed `0 .. n-1` where `n = len(adj)`. **Example 1:** `adj = [[1], [0]]` -> `true` (two guests who know each other go in different groups). **Example 2:** `adj = [[1,2],[0,2],[0,1]]` -> `false` (a triangle / odd cycle cannot be 2-colored). **Example 3:** `adj = []` -> `true` (no guests, vacuously seatable).