Problem Description

In this problem, we are given a list of strings called words that is arranged in a circular fashion. What this means is that if you were to look at the array of words, the word that comes after the last one is the first word of the list, creating a loop-like structure.

We are also given a target string that we need to find in the words list, starting the search from a given startIndex. The goal is to figure out the shortest distance from this starting index to the index where the target string is located.

The search can proceed in either direction - moving to the next word in the list or to the previous one. When searching, each step taken counts as 1, regardless of the direction.

To summarize:

• If the target string is present, we need to find and return the minimum number of steps required to reach it from the startIndex.
• If the target string is not in the list, we should return -1.

Keep in mind that the next element of words[i] is words[(i + 1) % n] and the previous element is words[(i - 1 + n) % n], where n is the total number of elements in words, effectively making it a circular array.

Intuition

To solve this problem, the intuition is to iterate through the entire circular list to locate the target string. As we encounter the given target, we calculate how far it is from our startIndex.

We consider two scenarios to find the shortest path:

1. Moving clockwise (forward) to reach the target.
2. Moving counter-clockwise (backward) to reach the target.

Let's consider an example where we have a list of 6 words and our startIndex is at position 1. If our target is at position 4, we can reach it in 3 steps by moving forward or in 3 steps by moving backward (since the array is circular). We always want the shortest number of steps.

The python code defines a function called closetTarget that calculates the minimum number of steps required to reach the target word from the startIndex. The distance is calculated in both directions for each occurrence of the target and the minimum is stored. After checking all the words, the function returns the smallest distance found. If the target word is absent, the function will return -1.

The key to this approach is realizing that the minimum distance to reach the target from the current index is either directly (forward or backward) or going the entire circle minus the direct distance, which effectively means going in the opposite direction.

Solution Approach

The implementation leverages a simple linear search along with modular arithmetic to handle the circular nature of the array.

The algorithm proceeds as follows:

1. We initialize a variable ans to n, which is the length of the words list. This initial value acts as a placeholder for the scenario where the target is not found.

2. We then loop over the words list using enumerate to get both the index i and the word w at each iteration.

3. For every word w that matches the target, we calculate the distance t from startIndex in two ways:

   • Direct distance: t = abs(i - startIndex)
   • Circular distance: n - t, which represents the distance if we were to go around the array in the opposite direction.

4. We update ans with the minimum of its current value, the direct distance t, and the circular distance n - t. This ensures that after scanning all words, ans holds the shortest distance required to reach target from startIndex.

5. If we finish the loop and ans remains equal to n, it means that the target word was not found in the words list. In this case, we return -1.

6. Otherwise, we return the value stored in ans, which is the shortest distance found.

An important detail to note is the use of modular arithmetic to deal with the circular indexing, but since we are using Python's abs function to calculate the direct distance, we do not explicitly use the modulus operator % in our distance calculations. Instead, the modulus operator would be necessary if we were manually wrapping around the indices.

This algorithm will work efficiently as it only requires a single pass over the list, giving it a time complexity of O(n), where n is the number of words in the list. We don't use any additional data structures, giving us a space complexity of O(1).

Example Walkthrough

Consider the following example to illustrate the solution approach:

• Let's assume words = ["apple", "banana", "cherry", "date", "elderberry", "fig"] is our list of words arranged in a circular fashion.
• The target string that we are looking for is "date".
• Our startIndex is 2, which means we start our search from the word "cherry".

Following the steps outlined in the solution approach:

1. We set ans = 6 as there are 6 words in our list. This is our initial best-case scenario value, which we will update if we find the word.

2. We enumerate through our list to get each word and its index:

   • Index 0, Word "apple": Not our target.
   • Index 1, Word "banana": Not our target.
   • Index 2, Word "cherry": This is our starting index.
   • Index 3, Word "date": This is our target!

3. Since we have found our target at index 3, we calculate the direct distance from our startIndex (2) to our target index (3):

   • Direct distance: t = abs(3 - 2) = 1

   We also calculate the circular distance by subtracting this direct distance from the total number of words (n):

   • Circular distance: n - t = 6 - 1 = 5

4. We update ans with the minimum of its current value (6), the direct distance (1), and the circular distance (5). This update gives us ans = min(6, 1, 5) = 1.

5. Since we successfully found the word, we don't need to return -1, and we can skip to step 6.

6. We return the value stored in ans, which is 1, indicating that the target word "date" is 1 step away from the starting index 2 (the word "cherry") in the list.

Thus, the function closetTarget will return 1 for this example, because the shortest path to the target "date" from startIndex 2 is by moving forward one step in the circular array.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the provided code is O(n), where n is the number of words in the input list words. This is because there is a single loop that iterates over each word in the list exactly once to check if it matches the target.

Space Complexity

The space complexity of the provided code is O(1) regardless of the size of the input. This is because the code only uses a few extra variables (n, ans, i, w, t) that do not depend on the size of the input list words.

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