Complete the code to sort the input list before processing.
nums = [3, 0, -2, -1, 1, 2] nums.[1]()
We use the sort() method to sort the list in place before finding triplets.
Complete the code to skip duplicate elements for the first pointer in the triplet search.
for i in range(len(nums)): if i > 0 and nums[i] == nums[1]: continue
We compare the current element with the previous one to skip duplicates.
Fix the error in the while loop condition to correctly move the left pointer.
while left < right and nums[left] == nums[1]: left += 1
We compare the current left element with the previous left element to skip duplicates.
Fill both blanks to correctly move the right pointer and skip duplicates after finding a valid triplet.
while left < right and nums[right] == nums[1]: right = [2]
We compare the current right element with the previous right element and move the right pointer leftwards to skip duplicates.
Fill all three blanks to complete the dictionary comprehension that counts the frequency of each number in the list.
count = [1](num: nums.count(num) for num in [2]) unique_nums = sorted(count.[3]())
list instead of dict for the comprehension.values() instead of keys().We use dict to create a dictionary, iterate over nums, and get the keys from the dictionary to sort unique numbers.