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May 26, 2025
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66 changes: 66 additions & 0 deletions 1857. Largest Color Value in a Directed Graph
Original file line number Diff line number Diff line change
@@ -0,0 +1,66 @@
class Solution {
public:
int maxi = -1;
int dp[100001][26];
bool dfs(int node, vector<int> adj[], vector<int>& vis, string& colors) {
vis[node] = 1;
dp[node][colors[node] - 'a']++;
for (auto it : adj[node]) {
if (!vis[it]) {
if (dfs(it, adj, vis, colors))
return true;
}

for (int i = 0; i < 26; i++)
if (colors[node] - 'a' == i)
dp[node][i] = max(dp[node][i], dp[it][i] + 1);
else
dp[node][i] = max(dp[node][i], dp[it][i]);
}
for(int i=0;i<26;i++)maxi=max(maxi,dp[node][i]);
return false;
}
int largestPathValue(string colors, vector<vector<int>>& edges) {
int n = colors.size();
memset(dp, 0, sizeof(dp));
vector<int> adj[n];
queue<int> q, p;
vector<int> indegree(n, 0);
// creating adjacency matrix
for (auto it : edges) {
adj[it[0]].push_back(it[1]);
indegree[it[1]]++;
}
// topological sort using kahn's algo
for (int i = 0; i < n; i++)
if (indegree[i] == 0) {
q.push(i);
p.push(i);
}
vector<int> z;
while (!p.empty()) {
int node = p.front();
p.pop();
z.push_back(node);
for (auto it : adj[node]) {
indegree[it]--;
if (indegree[it] == 0)
p.push(it);
}
}
//if cycle is present return -1
if (z.size() != n)
return -1;
vector<int> vis(n, false), pathvis(n, false);
// making dfs calls only for nodes which have zero indegree
// other nodes will be included automatically in
// the dp table by dfc calls of these nodes
while (!q.empty()) {
int node = q.front();
q.pop();
if (dfs(node, adj, vis, colors))
return -1;
}
return maxi;
}
};
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