LeetCode 2959: Number of Possible Sets of Closing Branches Solution

Master LeetCode problem 2959 (Number of Possible Sets of Closing Branches), 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.

2959. Number of Possible Sets of Closing Branches

Hard

Bit Manipulation Graph Heap (Priority Queue)Enumeration Shortest Path

Problem Explanation

Explanation

To solve this problem, we can use a bitmask dynamic programming approach. We will iterate through all possible subsets of branches and for each subset, we will check if all branches in that subset are reachable within the given maxDistance. We will use a bitmask to represent the subset of branches.

Initialize a 2D array dp where dp[mask] represents the number of possible sets of closing branches for the subset of branches represented by the bitmask mask.
For each subset represented by a bitmask mask, iterate through all branches i that are not closed. Check if branch i can be added to the subset represented by mask without violating the maxDistance constraint.
Update the dp array accordingly.
Finally, return the result stored in dp[(1 << n) - 1], which represents all branches being active.

Time complexity: O(2^n * n^2), where n is the number of branches
Space complexity: O(2^n)

Solution Code

class Solution {
    public int numberOfSets(int n, int maxDistance, int[][] roads) {
        int MOD = 1000000007;
        int[][] dp = new int[1 << n][n];
        dp[1][0] = 1;
        
        for (int mask = 1; mask < (1 << n); mask++) {
            for (int i = 0; i < n; i++) {
                if ((mask & (1 << i)) == 0) continue;
                
                for (int j = i + 1; j < n; j++) {
                    if ((mask & (1 << j)) == 0) continue;
                    
                    if (roads[i][j] <= maxDistance) {
                        dp[mask | (1 << j)][j] = (dp[mask | (1 << j)][j] + dp[mask][i]) % MOD;
                    }
                }
            }
        }
        
        int result = 0;
        for (int i = 0; i < n; i++) {
            result = (result + dp[(1 << n) - 1][i]) % MOD;
        }
        
        return result;
    }
}

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