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.

#3361

Medium

auto_awesome数组·string

LeetCode 题解工作台

两个字符串的切换距离

给你两个长度相同的字符串 s 和 t ，以及两个整数数组 nextCost 和 previousCost 。一次操作中，你可以选择 s 中的一个下标 i ，执行以下操作之一：将 s[i] 切换为字母表中的下一个字母，如果 s[i] == 'z' ，切换后得到 'a' 。操作的代价为 next…

数组字符串前缀和

题目描述

给你两个长度相同的字符串 s 和 t ，以及两个整数数组 nextCost 和 previousCost 。

一次操作中，你可以选择 s 中的一个下标 i ，执行以下操作之一：

将 s[i] 切换为字母表中的下一个字母，如果 s[i] == 'z' ，切换后得到 'a' 。操作的代价为 nextCost[j] ，其中 j 表示 s[i] 在字母表中的下标。
将 s[i] 切换为字母表中的上一个字母，如果 s[i] == 'a' ，切换后得到 'z' 。操作的代价为 previousCost[j] ，其中 j 是 s[i] 在字母表中的下标。

切换距离 指的是将字符串 s 变为字符串 t 的最少操作代价总和。

请你返回从 s 到 t 的 切换距离 。

示例 1：

输入：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]

输出：2

解释：

选择下标 i = 0 并将 s[0] 向前切换 25 次，总代价为 1 。
选择下标 i = 1 并将 s[1] 向后切换 25 次，总代价为 0 。
选择下标 i = 2 并将 s[2] 向前切换 25 次，总代价为 1 。
选择下标 i = 3 并将 s[3] 向后切换 25 次，总代价为 0 。

示例 2：

输入：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]

输出：31

解释：

选择下标 i = 0 并将 s[0] 向前切换 9 次，总代价为 9 。
选择下标 i = 1 并将 s[1] 向后切换 10 次，总代价为 10 。
选择下标 i = 2 并将 s[2] 向前切换 1 次，总代价为 1 。
选择下标 i = 3 并将 s[3] 向后切换 11 次，总代价为 11 。

提示：

1 <= s.length == t.length <= 10⁵
s 和 t 都只包含小写英文字母。
nextCost.length == previousCost.length == 26
0 <= nextCost[i], previousCost[i] <= 10⁹

lightbulb

解题思路

方法一

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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

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

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

warning

常见陷阱

外企场景

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

swap_horiz

进阶变体

外企场景

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

help

常见问题

外企场景

继续练习

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