Grokking the Coding Interview: Patterns for Coding Questions
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Solution: Number of Good Pairs

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A

Anonymous

· 2 years ago

I realized there was a way to solve this in a single line (excluding the import) by using the triangular number formula. There's no real advantage here, except that we don't need to count the pairs separately:

Instead, we count the occurrences n for each number and calculate the number of pairs from 0…n-1

from collections import Counter


class Solution:
  def numGoodPairs(self, nums):
    return sum((v * (v - 1) // 2) for _, v in Counter(nums).items())
A

Anonymous

· 3 years ago

class Solution {
  numGoodPairs(nums) {
    let pairCount = 0;
    for(let i = 0; i < nums.length; i++) {
      for(let j = i + 1; j < nums.length; j++) {
        if (nums[i] == nums[j]) pairCount++;
      }
    }
    return pairCount;
  }
}
A

Anonymous

· 3 years ago

class Solution:
  def numGoodPairs(self, nums):
    pairCount = 0
    # TODO: Write your code here
    left, right = 0,0
    while left < len(nums) - 1:
      right += 1
      if nums[left] == nums[right] and left < right:
        pairCount += 1

      if right == len(nums) - 1:
        left += 1
        right = left

    return pairCount
Show 1 reply
A

Anonymous

· 3 years ago

and the examples don't really help. this could be explained a lot better.

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