LeetCode Problem Workspace

Shift Distance Between Two Strings

Find the minimum cost of transforming one string into another, considering operations between characters.

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Medium · Array plus String

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Answer-first summary

Find the minimum cost of transforming one string into another, considering operations between characters.

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The problem requires transforming string s into t with a minimum cost. Each transformation depends on two given cost arrays, nextCost and previousCost, and the operations available are restricted to shifting characters at specific indices. By considering these costs efficiently, we can solve the problem using dynamic programming or other optimized methods.

Problem Statement

You are given two strings, s and t, of equal length. You are also given two arrays: nextCost and previousCost, both with 26 elements each. The arrays represent the costs associated with shifting characters from one to another. The task is to find the minimum cost of transforming string s into string t by performing a series of shift operations on individual characters of s.

In each operation, you can pick any index i from string s and shift the character at that position to match the corresponding character in string t. The cost to shift one character to another is determined by the nextCost or previousCost arrays, depending on the direction of the shift. The goal is to minimize the total cost of transforming s into t.

Examples

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

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

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

Solution Approach

Dynamic Programming Approach

A dynamic programming approach is well-suited for this problem. We can create a table where each entry represents the minimum cost of transforming one part of string s into string t. By considering each shift operation, we can compute the minimum cost at each step using the provided cost arrays. This approach ensures that we explore the most cost-effective transformations.

Cost Calculation and Transition

For every unordered pair of characters (a, b), the cost of transforming one into the other is determined by the minimum of nextCost[a] + previousCost[b]. This allows us to evaluate the cost-effectiveness of each shift and decide the optimal transition between characters at each index. Efficiently calculating these transitions is key to solving the problem.

Optimizing for Space and Time Complexity

Considering the problem's constraints, the time and space complexity will depend on how we implement the solution. By optimizing the storage of intermediate results and carefully managing transitions, we can achieve a solution that handles large inputs effectively without running into performance issues.

Complexity Analysis

Metric Value
Time Depends on the final approach
Space Depends on the final approach

The time complexity depends on the approach used for dynamic programming and calculating costs for each transformation. The space complexity will also vary depending on how the results are stored during computation. A direct approach can lead to a time complexity of O(n^2), but with optimization, it can be reduced to linear time, O(n).

What Interviewers Usually Probe

  • Candidate understands the need for efficient transformations between characters.
  • Candidate can identify the importance of cost minimization in this string problem.
  • Candidate is familiar with the optimal use of dynamic programming to solve cost-related problems.

Common Pitfalls or Variants

Common pitfalls

  • Misunderstanding the directionality of the shift costs can lead to incorrect calculations.
  • Failing to optimize for space or time complexity when dealing with large inputs.
  • Not considering edge cases where no transformations are needed, which can affect the solution's correctness.

Follow-up variants

  • Consider the variant where the nextCost and previousCost arrays are not provided. How would you approach the problem?
  • What if the strings s and t have different lengths? How would the problem change?
  • How would the problem change if we were allowed to perform shifts on multiple characters at once?

FAQ

What is the primary pattern in the 'Shift Distance Between Two Strings' problem?

The problem primarily follows the 'Array plus String' pattern, focusing on transforming one string into another with specific operations and costs.

How can dynamic programming help in solving the problem?

Dynamic programming allows us to calculate the minimum cost of transforming string s into t step by step, minimizing recalculations and handling large inputs efficiently.

What are the time and space complexity considerations for this problem?

The time and space complexity depend on the approach taken for dynamic programming and how the cost calculations are optimized. A direct approach could result in O(n^2) complexity, but optimizations can reduce it.

How can the given cost arrays impact the solution?

The nextCost and previousCost arrays play a crucial role in determining the cost of each transformation. Efficiently calculating these costs leads to a minimal total transformation cost.

Can this problem be solved without using dynamic programming?

While dynamic programming is an efficient approach, it may be possible to solve this problem with alternative methods, such as greedy algorithms, though they may not guarantee optimal solutions.

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Solution

Solution 1

#### Python3

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class Solution:
    def shiftDistance(
        self, s: str, t: str, nextCost: List[int], previousCost: List[int]
    ) -> int:
        m = 26
        s1 = [0] * (m << 1 | 1)
        s2 = [0] * (m << 1 | 1)
        for i in range(m << 1):
            s1[i + 1] = s1[i] + nextCost[i % m]
            s2[i + 1] = s2[i] + previousCost[(i + 1) % m]
        ans = 0
        for a, b in zip(s, t):
            x, y = ord(a) - ord("a"), ord(b) - ord("a")
            c1 = s1[y + m if y < x else y] - s1[x]
            c2 = s2[x + m if x < y else x] - s2[y]
            ans += min(c1, c2)
        return ans
Shift Distance Between Two Strings Solution: Array plus String | LeetCode #3361 Medium