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LeetCode 1395: Count Number of Teams Solution

Master LeetCode problem 1395 (Count Number of Teams), a medium 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.

Problem Explanation

Explanation

To solve this problem, we can iterate through all possible triplets of soldiers and check if they form a valid team according to the given conditions. We can keep track of the count of valid teams formed. We can do this by maintaining two arrays smaller and larger for each soldier, representing the count of soldiers with smaller and larger ratings respectively. For each soldier, we can count the number of valid teams that can be formed using that soldier as the middle element in the triplet.

  1. Initialize a variable count to keep track of the total number of valid teams.
  2. Initialize two arrays smaller and larger of size n, all initialized to 0.
  3. For each soldier at index i:
    • Iterate through all soldiers with index j < i:
      • If rating[j] < rating[i], increment smaller[i] by 1.
      • If rating[j] > rating[i], increment larger[i] by 1.
    • Iterate through all soldiers with index k > i:
      • If rating[k] < rating[i], increment the count by smaller[i] * larger[k].
      • If rating[k] > rating[i], increment the count by larger[i] * smaller[k].
  4. Return the count.

Solution Code

class Solution {
    public int numTeams(int[] rating) {
        int n = rating.length;
        int count = 0;
        int[] smaller = new int[n];
        int[] larger = new int[n];

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (rating[j] < rating[i]) {
                    smaller[i]++;
                } else if (rating[j] > rating[i]) {
                    larger[i]++;
                }
            }

            for (int k = i + 1; k < n; k++) {
                if (rating[k] < rating[i]) {
                    count += smaller[i] * larger[k];
                } else if (rating[k] > rating[i]) {
                    count += larger[i] * smaller[k];
                }
            }
        }

        return count;
    }
}

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