Sync LeetCode submission Runtime - 153 ms (15.12%), Memory - 123.8 MB (34.41%)

Author: jimit105

1986-largest-color-value-in-a-directed-graph/README.md

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+<p>There is a <strong>directed graph</strong> of <code>n</code> colored nodes and <code>m</code> edges. The nodes are numbered from <code>0</code> to <code>n - 1</code>.</p>
+
+<p>You are given a string <code>colors</code> where <code>colors[i]</code> is a lowercase English letter representing the <strong>color</strong> of the <code>i<sup>th</sup></code> node in this graph (<strong>0-indexed</strong>). You are also given a 2D array <code>edges</code> where <code>edges[j] = [a<sub>j</sub>, b<sub>j</sub>]</code> indicates that there is a <strong>directed edge</strong> from node <code>a<sub>j</sub></code> to node <code>b<sub>j</sub></code>.</p>
+
+<p>A valid <strong>path</strong> in the graph is a sequence of nodes <code>x<sub>1</sub> -&gt; x<sub>2</sub> -&gt; x<sub>3</sub> -&gt; ... -&gt; x<sub>k</sub></code> such that there is a directed edge from <code>x<sub>i</sub></code> to <code>x<sub>i+1</sub></code> for every <code>1 &lt;= i &lt; k</code>. The <strong>color value</strong> of the path is the number of nodes that are colored the <strong>most frequently</strong> occurring color along that path.</p>
+
+<p>Return <em>the <strong>largest color value</strong> of any valid path in the given graph, or </em><code>-1</code><em> if the graph contains a cycle</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<p><img alt="" src="https://assets.leetcode.com/uploads/2021/04/21/leet1.png" style="width: 400px; height: 182px;" /></p>
+
+<pre>
+<strong>Input:</strong> colors = &quot;abaca&quot;, edges = [[0,1],[0,2],[2,3],[3,4]]
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> The path 0 -&gt; 2 -&gt; 3 -&gt; 4 contains 3 nodes that are colored <code>&quot;a&quot; (red in the above image)</code>.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<p><img alt="" src="https://assets.leetcode.com/uploads/2021/04/21/leet2.png" style="width: 85px; height: 85px;" /></p>
+
+<pre>
+<strong>Input:</strong> colors = &quot;a&quot;, edges = [[0,0]]
+<strong>Output:</strong> -1
+<strong>Explanation:</strong> There is a cycle from 0 to 0.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == colors.length</code></li>
+	<li><code>m == edges.length</code></li>
+	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;= m &lt;= 10<sup>5</sup></code></li>
+	<li><code>colors</code> consists of lowercase English letters.</li>
+	<li><code>0 &lt;= a<sub>j</sub>, b<sub>j</sub>&nbsp;&lt; n</code></li>
+</ul>

1986-largest-color-value-in-a-directed-graph/solution.java

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+class Solution {
+    private int dfs(int node, String colors, Map<Integer, List<Integer>> adj, int[][] count,
+            boolean[] visit, boolean[] inStack) {
+        // If the node is already in the stack, we have a cycle.
+        if (inStack[node]) {
+            return Integer.MAX_VALUE;
+        }
+        if (visit[node]) {
+            return count[node][colors.charAt(node) - 'a'];
+        }
+        // Mark the current node as visited and part of current recursion stack.
+        visit[node] = true;
+        inStack[node] = true;
+
+        if (adj.containsKey(node)) {
+            for (int neighbor : adj.get(node)) {
+                if (dfs(neighbor, colors, adj, count, visit, inStack) == Integer.MAX_VALUE) {
+                    return Integer.MAX_VALUE;
+                }
+                for (int i = 0; i < 26; i++) {
+                    count[node][i] = Math.max(count[node][i], count[neighbor][i]);
+                }
+            }
+        }
+
+        // After all the incoming edges to the node are processed,
+        // we count the color on the node itself.
+        count[node][colors.charAt(node) - 'a']++;
+        // Remove the node from the stack.
+        inStack[node] = false;
+        return count[node][colors.charAt(node) - 'a'];
+    }
+
+    public int largestPathValue(String colors, int[][] edges) {
+        int n = colors.length();
+        Map<Integer, List<Integer>> adj = new HashMap<>();
+
+        for (int[] edge : edges) {
+            adj.computeIfAbsent(edge[0], k->new ArrayList<Integer>()).add(edge[1]);
+        }
+
+        int[][] count = new int[n][26];
+        boolean[] visit = new boolean[n];
+        boolean[] inStack = new boolean[n];
+        int answer = 0;
+        for (int i = 0; i < n; i++) {
+            answer = Math.max(answer, dfs(i, colors, adj, count, visit, inStack));
+        }
+
+        return answer == Integer.MAX_VALUE ? -1 : answer;
+    }
+}