Tackling the "Contains Duplicate" Problem: Solution and Complexity Analysis.

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:

Input: nums = [1,2,3,1]
Output: true
        

Example 2:

Input: nums = [1,2,3,4]
Output: false
        

Example 3:

Input: nums = [1,1,1,3,3,4,3,2,4,2]
Output: true        

The "Contains Duplicate" problem is a common challenge in coding interviews and competitive programming, often featured on platforms like LeetCode. The task is straightforward: given an array of integers, determine if the array contains any duplicates. This problem can be approached in several ways, each with its own complexity characteristics. Let's explore a typical solution and understand its time and space complexity.

Solution Steps

1. Use a Hash Set: A common and efficient way to solve this problem is by using a hash set (in some languages, this might be a hash table or a dictionary). As we iterate through the array, we check if an element is already in the set.
2. Iterate Through the Array: We pass through each element of the array. For each element, we perform a check against the hash set.
3. Check and Update the Hash Set:If the current element is already in the hash set, it means a duplicate has been found, and we return true.If the current element is not in the hash set, we add it to the set and continue.
4. Return the Result: If the iteration completes without finding any duplicates, we return false.

Time Complexity

1. Single Iteration: The solution involves a single pass through the array, which gives us a time complexity of O(n), where n is the number of elements in the array.
2. Hash Set Operations: Checking for the existence of an element in a hash set and inserting an element into the hash set both have an average time complexity of O(1).

Combining these, the overall time complexity of the "Contains Duplicate" solution is O(n).

Space Complexity

The space complexity of the solution depends on the hash set used to store the unique elements of the array.

1. Hash Set Storage: In the worst-case scenario, where there are no duplicates, the hash set will store every element of the array once. Therefore, the worst-case space complexity is O(n), where n is the number of elements in the array.

Conclusion

The "Contains Duplicate" problem's hash set-based solution is both efficient and straightforward. With a time complexity of O(n) and a space complexity of O(n), it effectively balances computational speed with memory usage. This approach is widely used due to its simplicity and effectiveness, making it an essential method for programmers to know in the context of coding interviews and algorithmic problem-solving...