Problem Description

In this problem, we are given an array nums which contains n distinct positive integers, meaning all the elements in the array are unique and start from index 0. Additionally, we are provided with a set of m operations, each representing a change we need to apply to the array. Each operation is a pair of integers where the first integer is a number that currently exists in the array and the second integer is the number we want to replace it with. The conditions are such that the first integer definitely exists in the array while the second is guaranteed not to be in the array before the operation is carried out. Our task is to perform all these operations on the array and return the final state of the array after all the replacements have been made.

Key Points:

• Each element in the array is unique.
• Each operation consists of replacing an existing element with a new one that is not already in the array.

The challenge is to apply these operations efficiently, keeping in mind that the naive approach of searching for each element to replace would lead to a less optimal solution.

Intuition

The intuition behind the solution is to optimize the search and replace process. If we try to look for the element to replace linearly every time, it would be inefficient, especially for a large array. Therefore, we use a hash map, in Python it's a dictionary, to keep track of the indices of the current elements in the array.

Here’s how we arrive at the solution step by step:

1. Create a hash map (dictionary) to map each value in the array to its index for constant time access. This allows us to quickly find where the replacement should occur without searching the entire array.

2. Iterate over the list of operations. For each operation:

   • Find the index of the current value we need to replace (since it exists in the nums array as per the problem's guarantee).
   • Replace the value with the new one in the array.
   • Update the hash map by changing the mapping of the old value to the new value, essentially changing the key while keeping the value (which is the index) the same.

This way, we ensure that all the replacements can be done in linear time relative to the number of operations, leading to an efficient solution.

Solution Approach

The implementation of the solution follows a straightforward approach leveraging a dictionary data structure for fast lookups and updates to the nums array based on the operations provided. Here's how the code brings the intuition to life:

1. Initialize a dictionary (hash map) named d. Use a dictionary comprehension to map each value v of the array nums to its corresponding index i. This is done in the form of {v: i for i, v in enumerate(nums)}, which will allow constant-time access to the indices of elements we need to update.

2. Iterate over all the given operations using a for loop. In each iteration, you are given a pair (a tuple in Python) containing two integers: [a, b] where a is the value you need to replace in nums, and b is the value you're going to replace it with.

3. Inside the loop, access the index of value a directly from the dictionary d – this gives us the exact position in the nums array where replacement is to take place, thus avoiding a full array scan. The operation nums[d[a]] accesses the element we want to replace and sets it to b.

4. Now, update the dictionary for the new value b to have the same index as a had before. The operation d[b] = d[a] ensures that going forward, you can now find the position of b just as easily as you could find a.

5. After the loop has processed all operations, we return the modified nums array, which now reflects all the replacements that have been made.

By using a hash map, the solution ensures that lookup and update operations are done in O(1) time, making the overall time complexity of the solution O(m), where m is the number of operations, as each operation involves a constant amount of work.

This approach is efficient and avoids the need for repeated array traversal, which would be necessary if we tried to find elements by value rather than by keeping track of their indices.

Example Walkthrough

Let's consider an example to illustrate the solution approach described above. Suppose we have the following input:

The nums array is [5, 3, 9, 7], and the list of operations is [(5, 10), (3, 15), (9, 20)]. Let's walk through each step of the approach.

1. We initialize a dictionary named d that will map each number in nums to its index: {5: 0, 3: 1, 9: 2, 7: 3}.

2. We begin iterating over the list of operations:

   • The first operation is (5, 10). We look up the index of 5 in d, which is 0. We replace the element at index 0 in the nums array with 10. The nums array now looks like [10, 3, 9, 7]. We then update d to reflect the change: {10: 0, 3: 1, 9: 2, 7: 3}.

   • The second operation is (3, 15). We look up the index of 3 in d, which is 1. We replace the element at index 1 in the nums array with 15. The nums array now looks like [10, 15, 9, 7]. We then update d to reflect the change: {10: 0, 15: 1, 9: 2, 7: 3}.

   • The third operation is (9, 20). We look up the index of 9 in d, which is 2. We replace the element at index 2 in the nums array with 20. The nums array now looks like [10, 15, 20, 7]. We then update d to reflect the change: {10: 0, 15: 1, 20: 2, 7: 3}.

3. After completing all the operations, we end with the final state of the nums array, which is [10, 15, 20, 7].

Following this approach, we have efficiently performed all replacements without having to search through the nums array multiple times, showcasing the benefits of maintaining a hash map to track indices of array elements.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code can be analyzed by looking at the two major steps in the function: the creation of the dictionary d that maps current values to their indices, and the iteration over the operations list to apply the value changes.

1. Constructing the dictionary d takes O(n) time, where n is the length of the nums array. This is because it involves iterating over each element once.
2. Iterating over the operations list takes O(m) time, where m is the number of operations because each operation involves a constant amount of work: updating a value in the list and updating a single entry in the dictionary d.

The total time complexity is the sum of these two parts: O(n + m).

Space Complexity

For space complexity, we consider the extra space used by the algorithm, not including the input and output.

1. The extra space comes from the dictionary d that maps values to indices, which contains at most n key-value pairs, where n is the length of nums. Thus, the space complexity for the dictionary is O(n).
2. No additional space that grows with the input size is used during the iteration over operations.

Hence, the total space complexity is O(n).

Learn more about how to find time and space complexity quickly using problem constraints.