Shift Distance Between Two Strings

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StringsArraysDynamic Programming

You are given two strings s and t of the same length, and two integer arrays nextCost and previousCost.

In one operation, you can pick any index i of s, and perform either one of the following actions:

Shift s[i] to the next letter in the alphabet. If s[i] == 'z', you should replace it with 'a'. This operation costs nextCost[j] where j is the index of s[i] in the alphabet.
Shift s[i] to the previous letter in the alphabet. If s[i] == 'a', you should replace it with 'z'. This operation costs previousCost[j] where j is the index of s[i] in the alphabet.

The shift distance is the minimum total cost of operations required to transform s into t.

Return the shift distance from s to t.

Example 1:

Input: s = "abab", t = "baba", nextCost = [100,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], previousCost = [1,100,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]

Output: 2

Explanation:

We choose index i = 0 and shift s[0] 25 times to the previous character for a total cost of 1.
We choose index i = 1 and shift s[1] 25 times to the next character for a total cost of 0.
We choose index i = 2 and shift s[2] 25 times to the previous character for a total cost of 1.
We choose index i = 3 and shift s[3] 25 times to the next character for a total cost of 0.

Example 2:

Input: s = "leet", t = "code", nextCost = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1], previousCost = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]

Output: 31

Explanation:

We choose index i = 0 and shift s[0] 9 times to the previous character for a total cost of 9.
We choose index i = 1 and shift s[1] 10 times to the next character for a total cost of 10.
We choose index i = 2 and shift s[2] 1 time to the previous character for a total cost of 1.
We choose index i = 3 and shift s[3] 11 times to the next character for a total cost of 11.

Constraints:

1 <= s.length == t.length <= 10⁵
s and t consist only of lowercase English letters.
nextCost.length == previousCost.length == 26
0 <= nextCost[i], previousCost[i] <= 10⁹

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Are the input strings case-sensitive, or should I treat them as case-insensitive?
If `string2` cannot be obtained by shifting `string1`, what should the function return (e.g., -1, null, or throw an exception)?
Can the input strings be empty or null? If so, how should I handle those cases?
By 'shifting', do you mean a circular shift, where the characters moved off one end reappear at the other, and is the shift only to the right?
Are there any constraints on the characters that can appear in the strings (e.g., only ASCII characters, Unicode characters, etc.)?

Brute Force Solution

Approach

The brute force strategy aims to find the minimum shift distance between two strings by exploring all possible shifts. This involves incrementally shifting one string and comparing it with the other string to see how well they align. The goal is to try every possible alignment and choose the best one.

Here's how the algorithm would work step-by-step:

Imagine the first string stays in place, and we're going to slide the second string alongside it.
Start by shifting the second string zero positions.
Compare the characters of both strings one by one to calculate the distance between them (like counting how many character positions are different).
Then shift the second string one position to the right.
Again, compare the characters of both strings to calculate the distance.
Keep shifting the second string further and further, calculating the distance each time.
Do the same thing by shifting the second string to the left. Keep track of the distance after each shift.
Once we've shifted the second string through all possible positions relative to the first string, we pick the smallest distance we found. That's our answer.

Code Implementation

def shift_distance_between_two_strings(first_string, second_string):
    length_of_first_string = len(first_string)
    length_of_second_string = len(second_string)
    minimum_distance = float('inf')

    # Iterate through all possible shifts
    for shift in range(-length_of_second_string + 1, length_of_first_string):
        current_distance = 0

        # Compare characters based on the current shift value
        for index in range(max(0, -shift), min(length_of_first_string, length_of_second_string - shift)):
            if first_string[index] != second_string[index + shift]:
                current_distance += 1

        # Update minimum distance if necessary
        minimum_distance = min(minimum_distance, current_distance)

    return minimum_distance

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all possible shifts of the second string relative to the first, which takes O(n) time where n is the length of the strings. For each shift, it compares the characters of the two strings, which also takes O(n) time. Since these operations are nested, the overall time complexity is O(n * n). Thus, the total operations approximate to n², which simplifies to O(n²).

Space Complexity

O(1) – The brute force strategy described doesn't create any auxiliary data structures that scale with the input size. It only involves shifting one string relative to the other and comparing characters, likely using a fixed number of variables for indexing and distance calculation. No temporary lists, hash maps, or significant recursion are mentioned or implied. Therefore, the space used is constant, regardless of the string lengths, which we can consider N.

Optimal Solution

Approach

The goal is to find the fewest shifts needed to transform one string into another by repeatedly moving the first character to the end. Instead of trying every possible shift, we can cleverly check if the second string is a rotation of the first.

Here's how the algorithm would work step-by-step:

First, check if the strings have the same length. If they don't, it's impossible for one to be a shifted version of the other.
Next, take the first string and duplicate it so it appears twice in a row. For example, if the string is 'abc', double it to get 'abcabc'.
Now, see if the second string is contained within this doubled string. If it is, that means the second string is a shifted version of the first.
If the second string is found inside the doubled string, count how many characters you need to move from the beginning of the doubled string to the beginning of the second string. This tells you the number of shifts needed to transform the first string to the second.
If the second string is not found inside the doubled string, it means it's not a rotated version of the first, so no number of shifts will work.

Code Implementation

def shift_distance(first_string, second_string):
    if len(first_string) != len(second_string):
        return -1

    # Concatenate first_string with itself to check for rotations.
    concatenated_string = first_string + first_string

    if second_string in concatenated_string:
        # Find the index to determine shift distance.
        shift_value = concatenated_string.find(second_string)

        return shift_value
    else:
        # Return -1 when second_string is not a rotation of first_string.
        return -1

Big(O) Analysis

Time Complexity

O(n) – The algorithm first checks if the strings have the same length, which takes O(1) time. Then, it concatenates the first string with itself, resulting in a string of length 2n, which takes O(n) time. The most significant operation is checking if the second string (length n) is a substring of the doubled string (length 2n). String searching algorithms like the built-in 'in' or 'find' functions in most languages are typically implemented with a time complexity of O(n) in this scenario, where n is the length of the longer string, in this case the doubled string since the search string is smaller. Therefore, the overall time complexity is dominated by the string concatenation and substring search, both being O(n).

Space Complexity

O(N) – The dominant space usage comes from creating the doubled string, which has a length of 2N, where N is the length of the input string. While there might be some constant space used for variables, the doubled string's size scales linearly with the input string's length. Therefore, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Both strings are null or empty	Return 0 as the shift distance between empty strings is conventionally 0, or throw IllegalArgumentException, depending on requirements.
One string is null or empty while the other is not	Return -1 or throw an exception indicating an invalid input state.
Strings have different lengths	Return -1 as it is impossible to shift one string to become another of differing lengths.
Strings are identical	Return 0 because no shift is required.
Strings contain non-alphanumeric characters or different character sets	Define the allowed character set and handle invalid characters by either rejecting the input or pre-processing by stripping them.
Strings are very long (potential performance bottleneck)	Ensure algorithm uses efficient search methods (e.g., KMP algorithm for substring matching) rather than naive shifting.
The shift distance wraps around multiple times (larger than string length)	Ensure that the shift amount is calculated using the modulo operator (%) to handle rotations exceeding the string length.
No possible shift exists that transforms string A into string B	Return -1 or a special value indicating no valid shift was found.

Both strings are null or empty

How to Handle:

Return 0 as the shift distance between empty strings is conventionally 0, or throw IllegalArgumentException, depending on requirements.

One string is null or empty while the other is not

How to Handle:

Return -1 or throw an exception indicating an invalid input state.

Strings have different lengths

How to Handle:

Return -1 as it is impossible to shift one string to become another of differing lengths.

Strings are identical

How to Handle:

Return 0 because no shift is required.

Strings contain non-alphanumeric characters or different character sets

How to Handle:

Define the allowed character set and handle invalid characters by either rejecting the input or pre-processing by stripping them.

Strings are very long (potential performance bottleneck)

How to Handle:

Ensure algorithm uses efficient search methods (e.g., KMP algorithm for substring matching) rather than naive shifting.

The shift distance wraps around multiple times (larger than string length)

How to Handle:

Ensure that the shift amount is calculated using the modulo operator (%) to handle rotations exceeding the string length.

No possible shift exists that transforms string A into string B

How to Handle:

Return -1 or a special value indicating no valid shift was found.