02. Max DAG Edges

✅ GFG solution to the Max DAG Edges problem: find maximum number of edges that can be added to a Directed Acyclic Graph (DAG) without forming cycles. 🚀

The problem can be found at the following link: 🔗 Question Link

🧩 Problem Description

Given a directed acyclic graph (DAG) with V vertices numbered from 0 to V-1 and E edges, represented as a 2D array edges[][], where each entry edges[i] = [u, v] denotes a directed edge from vertex u to vertex v, find the maximum number of additional edges that can be added to the graph without forming any cycles.

Note: The resulting graph must remain a DAG, meaning that adding any further edge would not create a cycle.

📘 Examples

Example 1

Input: V = 3, E = 2, edges[][] = [[0, 1], [1, 2]]
Output: 1
Explanation: The given DAG allows one more edge, 0 -> 2, which keeps the structure acyclic. 
Adding anything else would create a cycle.

Example 2

Input: V = 4, E = 4, edges[][] = [[0, 1], [0, 2], [1, 2], [2, 3]]
Output: 2
Explanation: Two additional edges (0 -> 3, 1 -> 3) can be added without forming cycles.

🔒 Constraints

• $1 \le V \le 10^3$

• $0 \le E \le \frac{V \times (V-1)}{2}$

• $0 \le \text{edges}[i][0], \text{edges}[i][1] < V$

✅ My Approach

The solution uses a mathematical formula based on graph theory properties to determine the maximum possible edges in a DAG:

Complete Graph Formula

1. Understanding Maximum Edges:

   • In a complete directed graph with V vertices, the maximum number of edges possible is when every vertex connects to every other vertex.

   • For a directed graph, this is V × (V - 1) / 2 edges (choosing any 2 vertices from V vertices).

2. Current Edge Count:

   • The graph already has E edges (given as edges.size()).

3. Available Slots:

   • Maximum additional edges = Total possible edges - Current edges

   • Formula: V × (V - 1) / 2 - E

4. Why This Works:

   • A DAG can have at most as many edges as a complete directed graph without creating cycles.

   • The directed acyclic nature allows for a topological ordering where all edges point "forward."

   • Subtracting existing edges from the maximum capacity gives us the remaining capacity.

📝 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(1), as we perform only a constant number of arithmetic operations (multiplication, division, and subtraction) regardless of the number of vertices or edges.

• Expected Auxiliary Space Complexity: O(1), as we only use a constant amount of additional space for storing the result, without any extra data structures.

🧑‍💻 Code (C++)

class Solution {
public:
    int maxEdgesToAdd(int V, vector<vector<int>>& edges) {
        return V * (V - 1) / 2 - edges.size();
    }
};

⚡ View Alternative Approaches with Code and Analysis

📊 2️⃣ Formula Expansion Approach

💡 Algorithm Steps:

1. Calculate maximum possible edges using combination formula C(V,2).

2. Compute total capacity as V vertices choosing 2 for each edge.

3. Subtract current edge count from maximum capacity.

4. Return the available slots for new edges.

class Solution {
public:
    int maxEdgesToAdd(int V, vector<vector<int>>& edges) {
        int maxCapacity = (V * V - V) >> 1;
        return maxCapacity - edges.size();
    }
};

📝 Complexity Analysis:

• Time: ⏱️ O(1) - Constant time arithmetic operations

• Auxiliary Space: 💾 O(1) - No extra space required

✅ Why This Approach?

• Bit shift operation for division by 2

• Mathematically equivalent but uses different formula form

• Demonstrates optimization techniques

📊 3️⃣ Step-by-Step Calculation

💡 Algorithm Steps:

1. Store vertex count and existing edge count separately.

2. Calculate complete graph edge capacity step by step.

3. Find difference between capacity and usage.

4. Return remaining edge capacity.

class Solution {
public:
    int maxEdgesToAdd(int V, vector<vector<int>>& edges) {
        int used = edges.size();
        int total = V * (V - 1) / 2;
        int available = total - used;
        return available;
    }
};

📝 Complexity Analysis:

• Time: ⏱️ O(1) - Fixed number of operations

• Auxiliary Space: 💾 O(1) - Only three integer variables

✅ Why This Approach?

• More readable with explicit variable names

• Easy to debug and trace values

• Good for understanding the logic flow

📊 4️⃣ Direct Computation

💡 Algorithm Steps:

1. Compute triangular number for V-1 to get max edges.

2. Use the formula directly without intermediate storage.

3. Inline the edge size calculation.

4. Return single expression result.

class Solution {
public:
    int maxEdgesToAdd(int V, vector<vector<int>>& edges) {
        return (V - 1) * V / 2 - static_cast<int>(edges.size());
    }
};

📝 Complexity Analysis:

• Time: ⏱️ O(1) - Single expression evaluation

• Auxiliary Space: 💾 O(1) - No variables needed

✅ Why This Approach?

• Most compact implementation possible

• Rearranged formula to prevent potential overflow

• Type casting for safety

🆚 🔍 Comparison of Approaches

🚀 Approach	⏱️ Time Complexity	💾 Space Complexity	✅ Pros	⚠️ Cons
🎯 Direct Formula	🟢 O(1)	🟢 O(1)	🚀 Simplest and cleanest	📖 Less explicit
🔧 Bit Shift	🟢 O(1)	🟢 O(1)	⚡ Micro-optimization	🔍 Less readable
📝 Step-by-Step	🟢 O(1)	🟢 O(1)	📖 Most readable	💾 Extra variables
🎪 Rearranged	🟢 O(1)	🟢 O(1)	🛡️ Overflow prevention	🤔 Non-standard formula

🏆 Best Choice Recommendation

🎯 Scenario	🎖️ Recommended Approach	🔥 Performance Rating
🏅 Production code	🥇 Direct Formula	★★★★★
📖 Learning/Teaching	🥈 Step-by-Step	★★★★★
🔧 Performance critical	🥉 Bit Shift	★★★★☆
🎯 Large vertex count	🏅 Rearranged	★★★★☆

☕ Code (Java)

class Solution {
    public int maxEdgesToAdd(int V, int[][] edges) {
        return V * (V - 1) / 2 - edges.length;
    }
}

🐍 Code (Python)

class Solution:
    def maxEdgesToAdd(self, V, edges):
        return V * (V - 1) // 2 - len(edges)

🧠 Contribution and Support

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: 📬 Any Questions?. Let's make this learning journey more collaborative!

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