Good Pairs

Unit 2 Session 2 (Click for link to problem statements)

U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Why does the i < j thing matter?
  • It prevents you from counting the same pair twice!

P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: First, find out how many of each element are in the list. Then calculate how many possible pairs can exist for each value, and add these up.

1) First, create a frequency map for the list¹
2) Create a variable to hold pairs, initialized to 0
3) For each value in our frequency map:
  a) Calculate the possible pairs for that value
  b) Add the possible pairs to our pairs variable
4) Return the total number of pairs

¹ see Frequency Count for psuedocode

⚠️ Common Mistakes

• You will need to use a Combination Formula for this problem. If you're not brushed up on statistics, try looking up "n choose 2 formula".

I-mplement

def num_identical_pairs(nums):
    # Frequency map to count occurrences of each number
    frequency_map = {}
    for num in nums:
        if num in frequency_map:
            frequency_map[num] += 1
        else:
            frequency_map[num] = 1
            
    # Count the number of good pairs
    good_pairs = 0
    for count in frequency_map.values():
        good_pairs += count * (count - 1) // 2  # Combination of 2 from n
        
    return good_pairs