Grokking the Coding Interview: Patterns for Coding Questions
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Number of Good Pairs (easy)

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A

Anonymous

· 2 years ago

def numGoodPairs(self, nums):
    pair_count = 0
    seen_nums = {}
    for num in nums:
      if num not in seen_nums:
        seen_nums[num] = 1
      else:
        pair_count += seen_nums[num]
        seen_nums[num] += 1
    return pair_count
A

Anonymous

· 3 years ago

if we use this approach time complexity is high?

public int numGoodPairs(int[] nums) {
    int pairCount = 0;
    // TODO: Write your code here
    for(int i=0; i<nums.length; i++){
      for (int j=i+1; j<nums.length; j++){
        if (nums[i]== nums[j]){
          pairCount++;
        }
      }
    }
    return pairCount;
  }
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A

Anonymous

· 3 years ago

class Solution {
  public int numGoodPairs(int[] nums) {
    int pairCount = 0;
    Map<Integer, Integer> frequency = new HashMap<>();

    for (Integer num : nums) {
      frequency.put(num, frequency.getOrDefault(num, 0) + 1);
    }

    for (Map.Entry<Integer, Integer> entry : frequency.entrySet()) {
      int value = entry.getValue();
      pairCount += value * (value - 1) / 2;
    }
    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

HI

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