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| 1 | +/** |
| 2 | + * Title: Largest Color Value in a Directed Graph |
| 3 | + * Description: There is a directed graph of n colored nodes and m edges. The nodes are numbered from 0 to n - 1. |
| 4 | + * Author: Hasibul Islam |
| 5 | + * Date: 10/04/2023 |
| 6 | + */ |
| 7 | + |
| 8 | +/** |
| 9 | + * @param {string} colors |
| 10 | + * @param {number[][]} edges |
| 11 | + * @return {number} |
| 12 | + */ |
| 13 | +const initializeGraph = (n) => { |
| 14 | + let G = []; |
| 15 | + for (let i = 0; i < n; i++) { |
| 16 | + G.push([]); |
| 17 | + } |
| 18 | + return G; |
| 19 | +}; |
| 20 | +const packDGDegree = (G, Edges, Deg) => { |
| 21 | + for (const [u, v] of Edges) { |
| 22 | + G[u].push(v); |
| 23 | + Deg[v]++; |
| 24 | + } |
| 25 | +}; // generate direct graph with indegree |
| 26 | + |
| 27 | +const largestPathValue = (colors, edges) => { |
| 28 | + let n = colors.length, |
| 29 | + m = edges.length, |
| 30 | + indegree = Array(n).fill(0); |
| 31 | + let g = initializeGraph(n); |
| 32 | + packDGDegree(g, edges, indegree); |
| 33 | + let order = topologicalSort(g, indegree); |
| 34 | + if (order.length == 0) return -1; |
| 35 | + let res = 0; |
| 36 | + for (let x = 97; x <= 122; x++) { |
| 37 | + let c = String.fromCharCode(x), |
| 38 | + dp = Array(n).fill(0); |
| 39 | + for (let i = 0; i < n; i++) { |
| 40 | + let cur = order[i]; |
| 41 | + if (colors[cur] == c) dp[cur]++; |
| 42 | + res = Math.max(res, dp[cur]); |
| 43 | + for (const child of g[cur]) dp[child] = Math.max(dp[child], dp[cur]); |
| 44 | + } |
| 45 | + } |
| 46 | + return res; |
| 47 | +}; |
| 48 | + |
| 49 | +const topologicalSort = (g, indegree) => { |
| 50 | + let res = [], |
| 51 | + q = [], |
| 52 | + n = g.length; |
| 53 | + for (let i = 0; i < n; i++) { |
| 54 | + // all nodes with no incoming edges |
| 55 | + if (indegree[i] == 0) q.push(i); |
| 56 | + } |
| 57 | + while (q.length) { |
| 58 | + let cur = q.shift(); |
| 59 | + res.push(cur); |
| 60 | + for (const child of g[cur]) { |
| 61 | + indegree[child]--; // remove an edge from cur to child |
| 62 | + if (indegree[child] == 0) q.push(child); // child has no other incoming edges, add to q for next bfs |
| 63 | + } |
| 64 | + } |
| 65 | + for (let i = 0; i < n; i++) { |
| 66 | + if (indegree[i] > 0) return []; |
| 67 | + } |
| 68 | + return res; |
| 69 | +}; |
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