Скачать или смотреть LeetCode Python Solutions: 217 Contains Duplicate

ZeroStress LeetCode Python Problem : 217 Contains Duplicate #python #leetcode

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The leetcode problem we are solving in this video is Contains Duplicate. According to the problem description, we are given an integer array nums. Our solution should return true if any value in NUMS appears at least twice in the array, and return false if every element is distinct.

The intuition behind solving the "Contains Duplicate" problem is to check if any element in the given array appears at least twice. One approach to solving this problem is to use a hash set to keep track of the unique elements seen so far. As we iterate through the array, we can check if the current element is already in the hash set. If it is, then we have found a duplicate and can return True. If we iterate through the entire array without finding a duplicate, then we can return False. The time complexity of this solution is O(n), where n is the length of the input array, and the space complexity is also O(n) since we need to store all the unique elements in the hash set.

The specific implementation involves defining a function called containsDuplicate that takes in an array nums as its input. We initialize an empty hash set h. Then we loop through each element i in the nums array. For each i, we check if it's already in the h hash set. If it is, we return True. If it's not, we add it to the h hash set. If we reach the end of the loop without finding any duplicates, we return False.

The time complexity of this approach is O(n) because we iterate through the entire input array once. The space complexity is also O(n) because we use a hash set to store the unique elements seen so far in the input array.

Let’s take a look at the code:

The code starts by defining a class called "Solution" with a method called "containsDuplicate" that takes an argument "nums", which is a list of integers.

hset = set()
This line initializes an empty set called "hset" to store the unique elements of the list.

for idx in nums:
This line sets up a loop that iterates over each element of the input list "nums".

if idx in hset:
return True
else:
hset.add(idx)

This block of code checks if the current element "idx" is already in the set "hset". If it is, that means the element has already been encountered in the list and the function should return True . If the element is not already in the set, it is added to the set using the "add" method.

Overall, the logic of the code is to iterate over the list and check if each element has already been encountered by storing it in a set. If a duplicate is found, the function returns True. If no duplicates are found after iterating over the entire list, the function returns False.