Leetcode 1395: Count Number of Teams
You are given a list of soldiers, each with a unique rating. You need to form teams of 3 soldiers from this list. A valid team is one where the soldiers’ ratings are either strictly increasing or strictly decreasing as we move from left to right in the team. The team must satisfy the condition that the indices (i, j, k) follow 0 <= i < j < k < n.
Problem
Approach
Steps
Complexity
Input: The input consists of a list of integers representing the soldiers' ratings.
Example: rating = [3, 6, 1, 4, 5]
Constraints:
• n == rating.length
• 3 <= n <= 1000
• 1 <= rating[i] <= 105
• All the integers in rating are unique.
Output: The output is the total number of valid teams of size 3 that can be formed based on the ratings.
Example: Output: 4
Constraints:
• The output will be an integer.
Goal: To count the number of valid teams that can be formed.
Steps:
• For each combination of three indices (i, j, k), check if the ratings are in increasing or decreasing order.
• Sum up all valid teams.
Goal: The problem must be solved efficiently given the constraints of n being as large as 1000.
Steps:
• n is at least 3 and at most 1000.
• The ratings are unique integers.
Assumptions:
• The ratings are given in an array, and no two soldiers have the same rating.
• You can form multiple teams with the same soldiers in different orders.
• Input: rating = [3, 6, 1, 4, 5]
• Explanation: The valid teams are: (3, 4, 5), (6, 1, 4), (6, 1, 5), and (6, 4, 1).
Approach: The problem can be solved by brute force, where we check each possible triplet of indices (i, j, k) and verify if they form a valid team.
Observations:
• We need to check all combinations of 3 soldiers from the list.
• We can optimize by using dynamic programming or a memoization approach to reduce redundant checks.
Steps:
• 1. Initialize a counter for the valid teams.
• 2. For each combination of indices (i, j, k), check if the ratings form an increasing or decreasing sequence.
• 3. Increment the counter for each valid combination.
• 4. Return the total count of valid teams.
Empty Inputs:
• Not applicable as n is always at least 3.
Large Inputs:
• Ensure that the approach scales for large inputs, up to n = 1000.
Special Values:
• Consider inputs where the ratings are in strictly increasing or strictly decreasing order.
Constraints:
• Avoid brute force solutions that involve checking all possible triplets for larger inputs.
int memo[1001][1001][4];
vector<int> rate;
int dp(int idx, int prv, int cnt) {
if(cnt == 3) return 1;
if(idx == rate.size()) return 0;
if(memo[idx][prv + 1][cnt] != -1) return memo[idx][prv + 1][cnt];
int ans = 0;
if(prv == -1 || rate[idx] > rate[prv]) {
ans += dp(idx + 1, idx, cnt + 1);
}
ans += dp(idx + 1, prv, cnt);
return memo[idx][prv + 1][cnt] = ans;
}
int numTeams(vector<int>& rate) {
this->rate = rate;
memset(memo, -1, sizeof(memo));
int res1 = dp(0, -1, 0);
reverse(this->rate.begin(),this->rate.end());
memset(memo, -1, sizeof(memo));
int res2 = dp(0, -1, 0);
return res1 + res2;
}
1 : Variable Initialization
int memo[1001][1001][4];
We create a memoization table 'memo' to store intermediate results of subproblems.
2 : Variable Initialization
vector<int> rate;
We declare a vector 'rate' to store the ratings of the players.
3 : Base Case Check
int dp(int idx, int prv, int cnt) {
The function takes three arguments: the current index 'idx', the previous index 'prv', and the count of players in the current team 'cnt'.
4 : Base Case Check
if(cnt == 3) return 1;
If we have selected three players, we return 1, as a valid team has been formed.
5 : Base Case Check
if(idx == rate.size()) return 0;
If we have processed all the ratings and haven’t formed a team, we return 0.
6 : Memoization Check
if(memo[idx][prv + 1][cnt] != -1) return memo[idx][prv + 1][cnt];
If the result for the current state is already computed, we return the stored result to avoid redundant calculations.
7 : Recursive Case
int ans = 0;
We initialize a variable 'ans' to store the number of valid teams.
8 : Recursive Case
if(prv == -1 || rate[idx] > rate[prv]) {
If the current rating is greater than the previous player's rating, we proceed to form a team.
9 : Recursive Case
ans += dp(idx + 1, idx, cnt + 1);
We call the 'dp' function recursively to include the current player and increase the team count.
10 : Recursive Case
ans += dp(idx + 1, prv, cnt);
We also recursively check if the current player can be skipped (no change in team).
11 : Memoization Storage
return memo[idx][prv + 1][cnt] = ans;
Return the computed result and store it in the memoization table for future reference.
12 : Main Function
int numTeams(vector<int>& rate) {
The 'numTeams' function calculates the total number of valid teams.
13 : Variable Setup
this->rate = rate;
We store the input ratings in the class variable 'rate'.
14 : Memoization Setup
memset(memo, -1, sizeof(memo));
We initialize the memoization table with -1 to signify that no subproblem has been solved yet.
15 : Recursive Call
int res1 = dp(0, -1, 0);
We call the 'dp' function to calculate the result for the first order of ratings.
16 : Reverse Setup
reverse(this->rate.begin(),this->rate.end());
We reverse the 'rate' vector to handle the reversed case.
17 : Memoization Setup
memset(memo, -1, sizeof(memo));
Reinitialize the memoization table before calling the 'dp' function again.
18 : Recursive Call
int res2 = dp(0, -1, 0);
We calculate the result for the reversed ratings.
19 : Final Calculation
return res1 + res2;
Return the total number of valid teams from both the original and reversed rating order.
Best Case: O(n^3)
Average Case: O(n^3)
Worst Case: O(n^3)
Description: Since we are checking each combination of 3 soldiers, the time complexity is cubic in the worst case.
Best Case: O(1)
Worst Case: O(1)
Description: We use a constant amount of space beyond the input list, so the space complexity is O(1).
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