num_good_pairs.py

def number_of_good_pairs(nums):
    # keep track of my good pairs
    # iterate over the nums using a nested loop, using i and j as the indices
    # check if the nums at i are thes same as the nums at j
    # if that is the case then increment the good pairs
    # once the loops are finished return the good pairs to the caller
    
    # first pass solution (brute force) O(n^2) Quadratic runtime
    # good_pairs = 0
    # for i in range(len(nums)):
    #     for j in range(i + 1, len(nums)):
    #         if nums[i] == nums[j]:
    #             good_pairs += 1
    
    # return good_pairs
    # return ((len(nums) - 1) * len(nums)) / 2

    # TODO: 2nd pass solution
    good_pairs = 0
    freq = [0] * 101
        
    for i in range(len(nums)):
        freq[nums[i]] += 1
    
    for i in range(101):
        if freq[i] > 1:
            good_pairs += ((freq[i] - 1) * freq[i]) // 2
            
    return good_pairs


print(number_of_good_pairs([1,2,3,1,1,3])) # => 4
print(number_of_good_pairs([1, 1, 1, 1])) # => 6
print(number_of_good_pairs([1, 2, 1, 1, 2, 1])) # => 7