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74 lines
2.2 KiB
Markdown
74 lines
2.2 KiB
Markdown
https://leetcode.com/problems/contains-duplicate/description/
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#leetcode
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#interview
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Given an integer array `nums`, return `true` if any value appears **at least twice** in the array, and return `false` if every element is distinct.
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**Example 1:**
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**Input:** nums = [1,2,3,1]
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**Output:** true
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**Explanation:**
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The element 1 occurs at the indices 0 and 3.
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**Example 2:**
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**Input:** nums = [1,2,3,4]
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**Output:** false
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**Explanation:**
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All elements are distinct.
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**Example 3:**
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**Input:** nums = [1,1,1,3,3,4,3,2,4,2]
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**Output:** true
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**Constraints:**
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- `1 <= nums.length <= 105`
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- `-109 <= nums[i] <= 109`
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Here's how you can solve the problem using Python. You can leverage the properties of sets to check for duplicates efficiently:
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```python
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class Solution:
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def containsDuplicate(self, nums: List[int]) -> bool:
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# Create an empty set to store unique elements
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seen = set()
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# Iterate through each number in the input list
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for num in nums:
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# If the number is already in the set, it means it's a duplicate
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if num in seen:
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return True
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# Otherwise, add the number to the set
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seen.add(num)
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# If no duplicates were found, return False
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return False
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```
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### Explanation:
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1. **Initialize a set**: `seen = set()`. Sets in Python do not allow duplicate values, which makes them useful for checking whether an element has already been seen.
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2. **Iterate through `nums`**: For each number in the input list `nums`, check if it already exists in the `seen` set.
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3. **Check for duplicates**:
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- If `num` is already in `seen`, it means we've encountered a duplicate, so return `True`.
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- If not, add `num` to the `seen` set.
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4. If we iterate through the entire list without finding a duplicate, return `False`.
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### Time Complexity:
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- The time complexity is O(n), where n is the number of elements in the list. Checking and inserting into a set both have an average time complexity of O(1).
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### Space Complexity:
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- The space complexity is O(n) because, in the worst case, we might need to store all elements of the input list in the set.
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This solution efficiently checks for duplicates using sets and is well-suited for the given constraints. |