LeetCode 1857: Largest Color Value in a Directed Graph Solution

Master LeetCode problem 1857 (Largest Color Value in a Directed Graph), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1857. Largest Color Value in a Directed Graph

Hard

Hash Table Dynamic Programming Graph Topological Sort Memoization Counting

Problem Explanation

Explanation:

We can use dynamic programming to solve this problem.
We will iterate through the nodes in a topological order to avoid cycles.
For each node, we will calculate the color frequency for each color by considering the incoming edges.
We will update the maximum color value for each node based on the maximum color frequency of its incoming edges.
Finally, we return the maximum color value among all nodes.

Time Complexity:

The time complexity of this approach is O(n + m), where n is the number of nodes and m is the number of edges.

Space Complexity:

The space complexity is O(n) for storing color frequencies and color values.

Solution Code

class Solution {
    public int largestPathValue(String colors, int[][] edges) {
        int n = colors.length();
        List<Integer>[] graph = new ArrayList[n];
        int[][] colorFreq = new int[n][26];
        int[] inDegree = new int[n];
        int[] colorValue = new int[n];
        Arrays.fill(colorValue, -1);
        
        for (int i = 0; i < n; i++) {
            graph[i] = new ArrayList<>();
        }
        
        for (int[] edge : edges) {
            int from = edge[0];
            int to = edge[1];
            graph[from].add(to);
            inDegree[to]++;
        }

        Queue<Integer> queue = new LinkedList<>();
        for (int i = 0; i < n; i++) {
            if (inDegree[i] == 0) {
                queue.offer(i);
                colorValue[i] = 0;
            }
        }
        
        int maxColorValue = 0;
        int visited = 0;
        while (!queue.isEmpty()) {
            int node = queue.poll();
            visited++;
            
            colorFreq[node][colors.charAt(node) - 'a']++;
            int maxFreq = 0;
            for (int i = 0; i < 26; i++) {
                maxFreq = Math.max(maxFreq, colorFreq[node][i]);
            }
            colorValue[node] = maxFreq;
            maxColorValue = Math.max(maxColorValue, colorValue[node]);
            
            for (int neighbor : graph[node]) {
                for (int i = 0; i < 26; i++) {
                    colorFreq[neighbor][i] = Math.max(colorFreq[neighbor][i], colorFreq[node][i]);
                }
                inDegree[neighbor]--;
                if (inDegree[neighbor] == 0) {
                    queue.offer(neighbor);
                }
            }
        }
        
        return visited == n ? maxColorValue : -1;
    }
}

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