2097. Valid Arrangement of Pairs

Author: sergeyleschev

2001-2500/2097. Valid Arrangement of Pairs.ts

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+// Solution by Sergey Leschev
+// 2097. Valid Arrangement of Pairs
+
+// Time Complexity: O(m+n), where m == pairs.size(), and n is the number of nodes in the graph (the number of distinct numbers in pairs)
+// Space Complexity: O(m+n)
+
+function validArrangement(pairs: number[][]): number[][] {
+    const adj: Map<number, number[]> = new Map()
+    const inDegree: Map<number, number> = new Map()
+    const outDegree: Map<number, number> = new Map()
+
+    // Build adjacency list and in/out-degree maps
+    for (const [u, v] of pairs) {
+        if (!adj.has(u)) adj.set(u, [])
+        adj.get(u)!.push(v)
+
+        outDegree.set(u, (outDegree.get(u) || 0) + 1)
+        inDegree.set(v, (inDegree.get(v) || 0) + 1)
+    }
+
+    // Find starting node
+    let start = -1
+    for (const [node] of adj) {
+        const out = outDegree.get(node) || 0
+        const inD = inDegree.get(node) || 0
+        if (out - inD === 1) {
+            start = node
+            break
+        }
+    }
+
+    if (start === -1) {
+        // Eulerian circuit: start from any node
+        start = pairs[0][0]
+    }
+
+    const result: number[][] = []
+    const stack: number[] = [start]
+
+    // Perform Eulerian path
+    while (stack.length > 0) {
+        const curr = stack[stack.length - 1]
+
+        if (adj.has(curr) && adj.get(curr)!.length > 0) {
+            const next = adj.get(curr)!.pop()!
+            stack.push(next)
+        } else {
+            stack.pop()
+            if (stack.length > 0) {
+                result.push([stack[stack.length - 1], curr])
+            }
+        }
+    }
+
+    return result.reverse()
+}