Problem Description

You are given two strings s and t of the same length, along with two integer arrays nextCost and previousCost. Each character in the strings s and t can be shifted either to the next letter or to the previous letter in the alphabet. The cost to shift a character s[i] to the next letter is determined by nextCost[j], where j is the position of s[i] in the alphabet, and the cost to shift to the previous letter is determined by previousCost[j]. If a character 'z' is shifted next, it wraps around to 'a', and similarly, 'a' shifting previous wraps around to 'z'. Your goal is to transform the string s into the string t with the minimum total cost of such operations. This minimum total cost is referred to as the "shift distance".

Intuition

To solve this problem, we need to determine the most cost-effective way to transform each character in s to the corresponding character in t. Since we can either shift to the next or previous letter in the alphabet, we can calculate the cost for each possible shift and choose the lesser cost.

1. Pre-calculate Costs: First, we compute the cumulative costs of shifting through the alphabet both forwards (using nextCost) and backwards (using previousCost). This allows us to quickly determine the cost of shifting from any character s[i] to any t[i].

2. Calculate Cost Efficiently: For each pair of characters s[i] and t[i], find:

   • The cost c1 to shift s[i] forward to reach t[i]. We need to wrap around if t[i] precedes s[i] in the alphabet.
   • The cost c2 to shift s[i] backward to reach t[i]. Similarly, wrap around if s[i] precedes t[i].

3. Compare and Sum Costs: For each character transformation, compare c1 and c2 and add the smaller value to the total cost. This ensures that we are always choosing the cheapest way to transform s into t.

By implementing these steps, the algorithm efficiently calculates the minimum shift distance for transforming s into t.

Learn more about Prefix Sum patterns.

Solution Approach

The solution employs an efficient approach using pre-computed cumulative costs for shifting characters both forwards and backwards through the alphabet. Below is a step-by-step breakdown of the solution:

1. Initialize Arrays for Cumulative Costs:

   • We use two arrays s1 and s2 to store the cumulative costs for shifting forward and backward, respectively. Each array has a size of 2 * m + 1 where m is the number of letters in the alphabet (26). This size accounts for wrapping around the alphabet.

2. Calculate Forward Cumulative Costs:

   • Iterate through the alphabet twice (m << 1) and populate s1 with cumulative costs using the nextCost array. The entry s1[i + 1] represents the cost to shift from 'a' to the character at position i.

3. Calculate Backward Cumulative Costs:

   • Similarly, populate s2 using previousCost. The entry s2[i + 1] represents the cost to move backwards from 'a' to character at position i, considering the wrap around effect.

4. Compute the Shift Distance:

   • For each character pair (s[i], t[i]), compute:
     • x and y as the alphabet indices of s[i] and t[i] respectively.
     • c1 as the cost to move from x to y forward, calculated as s1[y + m if y < x else y] - s1[x].
     • c2 as the cost to move from x to y backward, calculated as s2[x + m if x < y else x] - s2[y].
   • Add the minimum of c1 and c2 to the total cost.

5. Return the Total Shift Distance:

   • The final accumulated cost in ans represents the minimum total cost to transform s into t.

This method efficiently determines the shift distance by leveraging cumulative cost arrays to quickly ascertain the cost of any necessary transformations in constant time.

Example Walkthrough

Let's consider an example where s = "abc", t = "bcd", with nextCost and previousCost as follows:

• nextCost = [1, 1, 1, ..., 1] (i.e., a constant 1 cost to move forward)
• previousCost = [1, 1, 1, ..., 1] (i.e., a constant 1 cost to move backward)

The goal is to transform s into t with minimum cost:

1. Pre-calculate Costs:

   • We will use arrays s1 and s2 to hold cumulative costs for forward and backward movements.

2. Calculate Forward Cumulative Costs (s1):

   • For s1, iterate over twice the alphabet length (52 characters):

     s1[1] = s1[0] + nextCost[0] = 1
     s1[2] = s1[1] + nextCost[1] = 2
     s1[3] = s1[2] + nextCost[2] = 3
     ...

   • s1[i] captures cumulative cost to move from 'a' to any character.

3. Calculate Backward Cumulative Costs (s2):

   • Similarly, populate s2 for backward movements:

     s2[1] = s2[0] + previousCost[0] = 1
     s2[2] = s2[1] + previousCost[1] = 2
     s2[3] = s2[2] + previousCost[2] = 3
     ...

   • s2[i] captures cumulative cost to move back from 'a' to any character.

4. Compute the Shift Distance:

   • For each pair (s[i], t[i]), calculate costs:
     • For 'a' to 'b':
       • x = 0, y = 1
       • Forward cost c1: s1[1] - s1[0] = 1
       • Backward cost c2: s2[0+m] - s2[1] = 1 + 26 - 1 = 26
       • Choose minimum: add c1 = 1 to total cost.
     • For 'b' to 'c':
       • x = 1, y = 2
       • Forward cost c1: s1[2] - s1[1] = 1
       • Backward cost c2: s2[1+m] - s2[2] = 26 + 1 - 2 = 25
       • Choose minimum: add c1 = 1 to total cost.
     • For 'c' to 'd':
       • x = 2, y = 3
       • Forward cost c1: s1[3] - s1[2] = 1
       • Backward cost c2: s2[2+m] - s2[3] = 26 + 2 - 3 = 25
       • Choose minimum: add c1 = 1 to total cost.

5. Return the Total Shift Distance:

   • The total cost to transform s = "abc" to t = "bcd" is 1 + 1 + 1 = 3.

Thus, the minimum cost to transform s into t with the given costs is 3.

Solution Implementation

Time and Space Complexity

The time complexity of the shiftDistance function is O(n + m), where n is the length of the input strings s and t, and m is the fixed constant of 26 representing the number of letters in the English alphabet. The iteration over string characters contributes to the O(n) component, while the precomputation loops over costs contribute to O(m).

The space complexity is O(m), due to the auxiliary arrays s1 and s2 that store cumulative costs and have a size proportional to 2m + 1. Here, m is 26, which is a fixed constant, so in practical terms, this complexity simplifies to O(1).

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