LeetCode 3384: Team Dominance by Pass Success Solution

Master LeetCode problem 3384 (Team Dominance by Pass Success), 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.

3384. Team Dominance by Pass Success

Hard

Database

Problem Explanation

Explanation:

The problem asks us to find the team dominance by pass success for each team in a competition. Team dominance by pass success is defined as the number of other teams that a team has a higher pass success rate than.

To solve this problem, we can iterate through each team and calculate their pass success rate by comparing it with all other teams. For each team, we compare its pass success rate with all other teams and increment the count if the pass success rate of the current team is greater than the pass success rate of the other team.

Algorithm:

Create a result array to store the team dominance by pass success for each team.
Iterate through each team in the input array.
For each team, iterate through all other teams in the input array.
Calculate the pass success rate for the current team and the other team.
If the pass success rate of the current team is greater than the pass success rate of the other team, increment the count for team dominance by pass success.
Store the count in the result array for the current team.
Return the result array.

Time Complexity:

The time complexity of this algorithm is O(n^2), where n is the number of teams in the input array.

Space Complexity:

The space complexity of this algorithm is O(n) to store the result array.

Solution Code

class Solution {
    public int[] teamDominance(int[][] passSuccess) {
        int n = passSuccess.length;
        int[] result = new int[n];
        
        for (int i = 0; i < n; i++) {
            int count = 0;
            for (int j = 0; j < n; j++) {
                if (i != j) {
                    double team1SuccessRate = ((double) passSuccess[i][0] / passSuccess[i][1]);
                    double team2SuccessRate = ((double) passSuccess[j][0] / passSuccess[j][1]);
                    
                    if (team1SuccessRate > team2SuccessRate) {
                        count++;
                    }
                }
            }
            result[i] = count;
        }
        
        return result;
    }
}

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