2899. Last Visited Integers

Problem Description

The problem provides us with an array of strings where each element is either a string that represents a positive integer or the string "prev". We’re required to iterate through this array and, whenever we encounter the string "prev", we must find the last visited integer according to specific rules.

For a sequence of one or more "prev" strings, the number of consecutive "prev" strings (including the current "prev") is counted as k. We have to look at the integers seen so far in reverse order. The last visited integer is the (k-1)th integer in this reversed order. If k exceeds the total number of integers we’ve seen, then the last visited integer is to be considered -1.

Our goal is to return an array of integers, which includes the last visited integer for each "prev" string found in the input array.

For example, given an input array of ["1", "2", "prev", "prev", "3", "prev"], our output should be [2, 1, -1].

Intuition

The intuition behind the solution is to simulate the process described, keeping track of two essential pieces of information: the integers we have seen (in order) and the current sequence length of "prev" strings (k).

To achieve this, we can keep an array called nums to store all the integers we have encountered so far in the order they were visited. We also keep a count k, which is reset to 0 every time we encounter an integer and is incremented when we encounter a "prev". Whenever we bump into a "prev", we look k elements back from the end of our stored integers—if there are enough elements—and add that to the result array. If there are not enough elements (k is greater than the number of integers encountered), we add -1 to the result.

Essentially, we are creating a stack of numbers encountered, and for every "prev", we are performing a lookup that acts like indexing from the back of the stack by k positions, simulating the process of tracking the last visited integers.

Solution Approach

The solution follows a simple simulation approach that revolves around keeping track of the visited integers and the number of consecutive occurrences of "prev". This approach can be outlined as follows:

1. We initialize an array nums to store integers that have been seen. This array acts as a stack where we can easily add new integers and look up past integers.
2. A variable k is used to keep count of consecutive "prev" strings. It is reset to 0 whenever a non-"prev" string (i.e., an integer) is encountered.
3. We iterate through the words array, processing one string at a time.
4. If the current word is not "prev" (it is a positive integer), we reset k to 0 and append the integer form of the word to nums.
5. If the current word is "prev", we increment k by 1 to account for the new "prev" string in the sequence.
6. We then attempt to retrieve the last visited integer by looking k places from the end of nums using the index len(nums) - k. If k is within the range of nums, we append the desired integer to ans; otherwise, we append -1, indicating that there are not enough visited integers to satisfy the "prev" condition.
7. After processing all words in the array, we return the ans array, which contains the last visited integer for each "prev" encountered.

The solution uses straightforward array manipulation and conditions to achieve the objective, employing basic data structures (lists in Python), and index manipulation. The primary pattern this solution leverages is iteration with condition checks, ensuring the last visited integer is correctly identified and handled according to the problem's rules.

Here's a breakdown of the code corresponding to the steps above:

1class Solution:
2    def lastVisitedIntegers(self, words: List[str]) -> List[int]:
3        nums = []  # Step 1: initialize the stack of seen integers
4        ans = []   # Initialize the array for storing last visited integers
5        k = 0      # Initialize the count of consecutive "prev"
6        for w in words:  # Step 3: iterate through each word
7            if w == "prev":  # Step 4 and 5: handle "prev"
8                k += 1  # Increment count for consecutive "prev"
9                i = len(nums) - k  # Calculate index for the last visited integer
10                ans.append(-1 if i < 0 else nums[i])  # Append the last visited integer or -1
11            else:  # Step 6: reset k and add new integer to nums
12                k = 0
13                nums.append(int(w))  # Convert string to integer and add to nums
14        return ans   # Step 7: return the result

Example Walkthrough

Let's go through an example to illustrate the solution approach using a small input array: ["10", "prev", "20", "prev", "prev"].

1. Initialize nums as an empty array to represent the stack of seen integers, ans as an empty array for the last visited integers, and k as 0 for the count of consecutive "prev" strings.

2. Iterate through each element in words:

   • For the first element "10", since it is not "prev", reset k to 0 (it's already 0). Convert "10" to an integer and append it to nums. Now nums = [10].
   • The second element is "prev". Increment k to 1 (k was 0). Calculate the index: len(nums) - k which is 0. There is an available integer, so append nums[0] (which is 10) to ans. Now ans = [10].
   • The third element "20" is not "prev", so reset k to 0 and append 20 to nums. Now nums = [10, 20].
   • The fourth element is again "prev". Increment k to 1. Calculate index len(nums) - k which is 1. Append nums[1] (which is 20) to ans. Now ans = [10, 20].
   • The last element is "prev" again. Increment k to 2 (since we do not reset k as the element is "prev"). Calculate index len(nums) - k which is 0. Append nums[0] (which is 10) to ans. Now ans = [10, 20, 10].

3. Finally, we end up with ans = [10, 20, 10], which is the output, with each entry representing the last visited integer for each "prev".

By processing each element one by one and keeping track of the seen integers (nums) and the consecutive count of "prev" (k), we can efficiently simulate the process and determine the last visited integer or -1 when required.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code is O(n), where n is the length of the words array. We iterate over each word in the array once. In the iteration, the operations we perform, such as checking the value of w and updating the nums list or the index k, all have constant time complexity (O(1)). Even when we access nums[i] the time complexity is O(1) because list indexing is a constant time operation in Python. Hence, the main contributing factor to the time complexity is the single loop through the words array, resulting in O(n) time complexity.

Space Complexity

The space complexity of the code is also O(n). We use additional data structure nums to store the integers as they appear when they are not "prev". Thus, in the worst-case scenario where there is no "prev", nums will have the same number of integers as there are elements in the words list. The ans list at most will contain n "-1" integers when all elements in words are "prev". The sum of the size of nums and ans lists dictates the space complexity. Therefore, the space complexity is O(n) due to the storage requirements that scale linearly with the input size.

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