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g3301_3400.s3361_shift_distance_between_two_strings.readme.md Maven / Gradle / Ivy

3361\. Shift Distance Between Two Strings

Medium

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