1857. Largest Color Value in a Directed Graph

Description

There is a directed graph of n colored nodes and m edges. The nodes are numbered from 0 to n - 1.

You are given a string colors where colors[i] is a lowercase English letter representing the color of the ith node in this graph (0-indexed). You are also given a 2D array edges where edges[j] = [aj, bj] indicates that there is a directed edge from node aj to node bj.

A valid path in the graph is a sequence of nodes x1 -> x2 -> x3 -> ... -> xk such that there is a directed edge from xi to xi+1 for every 1 <= i < k. The color value of the path is the number of nodes that are colored the most frequently occurring color along that path.

Return the largest color value of any valid path in the given graph, or -1 if the graph contains a cycle.

 

Example 1:

Input: colors = "abaca", edges = [[0,1],[0,2],[2,3],[3,4]]
Output: 3
Explanation: The path 0 -> 2 -> 3 -> 4 contains 3 nodes that are colored "a" (red in the above image).

Example 2:

Input: colors = "a", edges = [[0,0]]
Output: -1
Explanation: There is a cycle from 0 to 0.

 

Constraints:

• n == colors.length
• m == edges.length
• 1 <= n <= 105
• 0 <= m <= 105
• colors consists of lowercase English letters.
• 0 <= aj, bj < n

Solutions

Solution 1: Topological Sort + Dynamic Programming

Calculate the in-degree of each node and perform a topological sort.

Define a 2D array \(dp\), where \(dp[i][j]\) represents the number of nodes with color \(j\) on the path from the start node to node \(i\).

From node \(i\), traverse all outgoing edges \(i \to j\), and update \(dp[j][k] = \max(dp[j][k], dp[i][k] + (c == k))\), where \(c\) is the color of node \(j\).

The answer is the maximum value in the \(dp\) array.

If there is a cycle in the graph, it is impossible to visit all nodes, so return \(-1\).

The time complexity is \(O((n + m) \times |\Sigma|)\), and the space complexity is \(O(m + n \times |\Sigma|)\). Here, \(|\Sigma|\) is the size of the alphabet (26 in this case), and \(n\) and \(m\) are the number of nodes and edges, respectively.

Python3JavaC++GoTypeScript

class Solution:
    def largestPathValue(self, colors: str, edges: List[List[int]]) -> int:
        n = len(colors)
        indeg = [0] * n
        g = defaultdict(list)
        for a, b in edges:
            g[a].append(b)
            indeg[b] += 1
        q = deque()
        dp = [[0] * 26 for _ in range(n)]
        for i, v in enumerate(indeg):
            if v == 0:
                q.append(i)
                c = ord(colors[i]) - ord('a')
                dp[i][c] += 1
        cnt = 0
        ans = 1
        while q:
            i = q.popleft()
            cnt += 1
            for j in g[i]:
                indeg[j] -= 1
                if indeg[j] == 0:
                    q.append(j)
                c = ord(colors[j]) - ord('a')
                for k in range(26):
                    dp[j][k] = max(dp[j][k], dp[i][k] + (c == k))
                    ans = max(ans, dp[j][k])
        return -1 if cnt < n else ans

class Solution {
    public int largestPathValue(String colors, int[][] edges) {
        int n = colors.length();
        List<Integer>[] g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        int[] indeg = new int[n];
        for (int[] e : edges) {
            int a = e[0], b = e[1];
            g[a].add(b);
            ++indeg[b];
        }
        Deque<Integer> q = new ArrayDeque<>();
        int[][] dp = new int[n][26];
        for (int i = 0; i < n; ++i) {
            if (indeg[i] == 0) {
                q.offer(i);
                int c = colors.charAt(i) - 'a';
                ++dp[i][c];
            }
        }
        int cnt = 0;
        int ans = 1;
        while (!q.isEmpty()) {
            int i = q.pollFirst();
            ++cnt;
            for (int j : g[i]) {
                if (--indeg[j] == 0) {
                    q.offer(j);
                }
                int c = colors.charAt(j) - 'a';
                for (int k = 0; k < 26; ++k) {
                    dp[j][k] = Math.max(dp[j][k], dp[i][k] + (c == k ? 1 : 0));
                    ans = Math.max(ans, dp[j][k]);
                }
            }
        }
        return cnt == n ? ans : -1;
    }
}

class Solution {
public:
    int largestPathValue(string colors, vector<vector<int>>& edges) {
        int n = colors.size();
        vector<vector<int>> g(n);
        vector<int> indeg(n);
        for (auto& e : edges) {
            int a = e[0], b = e[1];
            g[a].push_back(b);
            ++indeg[b];
        }
        queue<int> q;
        vector<vector<int>> dp(n, vector<int>(26));
        for (int i = 0; i < n; ++i) {
            if (indeg[i] == 0) {
                q.push(i);
                int c = colors[i] - 'a';
                dp[i][c]++;
            }
        }
        int cnt = 0;
        int ans = 1;
        while (!q.empty()) {
            int i = q.front();
            q.pop();
            ++cnt;
            for (int j : g[i]) {
                if (--indeg[j] == 0) q.push(j);
                int c = colors[j] - 'a';
                for (int k = 0; k < 26; ++k) {
                    dp[j][k] = max(dp[j][k], dp[i][k] + (c == k));
                    ans = max(ans, dp[j][k]);
                }
            }
        }
        return cnt == n ? ans : -1;
    }
};

func largestPathValue(colors string, edges [][]int) int {
    n := len(colors)
    g := make([][]int, n)
    indeg := make([]int, n)
    for _, e := range edges {
        a, b := e[0], e[1]
        g[a] = append(g[a], b)
        indeg[b]++
    }
    q := []int{}
    dp := make([][]int, n)
    for i := range dp {
        dp[i] = make([]int, 26)
    }
    for i, v := range indeg {
        if v == 0 {
            q = append(q, i)
            c := colors[i] - 'a'
            dp[i][c]++
        }
    }
    cnt := 0
    ans := 1
    for len(q) > 0 {
        i := q[0]
        q = q[1:]
        cnt++
        for _, j := range g[i] {
            indeg[j]--
            if indeg[j] == 0 {
                q = append(q, j)
            }
            c := int(colors[j] - 'a')
            for k := 0; k < 26; k++ {
                t := 0
                if c == k {
                    t = 1
                }
                dp[j][k] = max(dp[j][k], dp[i][k]+t)
                ans = max(ans, dp[j][k])
            }
        }
    }
    if cnt == n {
        return ans
    }
    return -1
}

function largestPathValue(colors: string, edges: number[][]): number {
    const n = colors.length;
    const indeg = Array(n).fill(0);
    const g: Map<number, number[]> = new Map();
    for (const [a, b] of edges) {
        if (!g.has(a)) g.set(a, []);
        g.get(a)!.push(b);
        indeg[b]++;
    }
    const q: number[] = [];
    const dp: number[][] = Array.from({ length: n }, () => Array(26).fill(0));
    for (let i = 0; i < n; i++) {
        if (indeg[i] === 0) {
            q.push(i);
            const c = colors.charCodeAt(i) - 97;
            dp[i][c]++;
        }
    }
    let cnt = 0;
    let ans = 1;
    while (q.length) {
        const i = q.pop()!;
        cnt++;
        if (g.has(i)) {
            for (const j of g.get(i)!) {
                indeg[j]--;
                if (indeg[j] === 0) q.push(j);
                const c = colors.charCodeAt(j) - 97;
                for (let k = 0; k < 26; k++) {
                    dp[j][k] = Math.max(dp[j][k], dp[i][k] + (c === k ? 1 : 0));
                    ans = Math.max(ans, dp[j][k]);
                }
            }
        }
    }
    return cnt < n ? -1 : ans;
}

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