Today I have done Count Number of Teams on leetcode and leetcode's December LeetCoding Challenge with cpp.

Count Number of Teams

Description

There are n soldiers standing in a line. Each soldier is assigned a unique rating value.

You have to form a team of 3 soldiers amongst them under the following rules:

• Choose 3 soldiers with index (i, j, k) with rating (rating[i], rating[j], rating[k]).
• A team is valid if: (rating[i] < rating[j] < rating[k]) or (rating[i] > rating[j] > rating[k]) where (0 <= i < j < k < n).

Return the number of teams you can form given the conditions. (soldiers can be part of multiple teams).

Example 1:

Input: rating = [2,5,3,4,1]
Output: 3
Explanation: We can form three teams given the conditions. (2,3,4), (5,4,1), (5,3,1). 

Example 2:

Input: rating = [2,1,3]
Output: 0
Explanation: We can't form any team given the conditions.

Example 3:

Input: rating = [1,2,3,4]
Output: 4

Constraints:

• n == rating.length
• 1 <= n <= 200
• 1 <= rating[i] <= 10^5

Solution

brute force

class Solution {
public:
    int numTeams(vector<int>& rating) {
        int ans = 0;
        int len = rating.size();
        for(int i = 0; i < len-2; ++i) {
            for(int j = i+1; j < len-1; ++j) {
                for(int k = j+1; k < len; ++k) {
                    if(rating[i] > rating[j] && rating[j] > rating[k]) ans += 1;
                    if(rating[i] < rating[j] && rating[j] < rating[k]) ans += 1;
                }
            }
        }
        return ans;
    }
};

December LeetCoding Challenge 17

Description

4Sum II

Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.

To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.

Example:

Input:
A = [ 1, 2]
B = [-2,-1]
C = [-1, 2]
D = [ 0, 2]

Output:
2

Explanation:
The two tuples are:
1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

Solution

two-way enumeration

class Solution {
public:
    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        map<int, int> AB, CD;
        for(auto i : A) {
            for(auto j : B) {
                AB[i+j] += 1;
            }
        }
        for(auto i : C) {
            for(auto j : D) {
                CD[i+j] += 1;
            }
        }
        int answer = 0;
        for(auto [num, cnt] : AB) {
            answer += cnt * CD[-num];
        }
        return answer;
    }
};