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Merge pull request #1051 from 0xff-dev/2825
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Add solution and test-cases for problem 2825
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6boris authored Dec 24, 2024
2 parents caa2fd8 + 01c3d75 commit 8b2d9dc
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# [2825.Make String a Subsequence Using Cyclic Increments][title]

## Description
You are given two **0-indexed** strings `str1` and `str2`.

In an operation, you select a **set** of indices in `str1`, and for each index `i` in the set, increment `str1[i]` to the next character **cyclically**. That is `'a'` becomes `'b'`, `'b'` becomes `'c'`, and so on, and `'z'` becomes `'a'`.

Return `true` if it is possible to make `str2` a subsequence of `str1` by performing the operation **at most once**, and `false` otherwise.

**Note**: A subsequence of a string is a new string that is formed from the original string by deleting some (possibly none) of the characters without disturbing the relative positions of the remaining characters.

**Example 1:**

```
Input: str1 = "abc", str2 = "ad"
Output: true
Explanation: Select index 2 in str1.
Increment str1[2] to become 'd'.
Hence, str1 becomes "abd" and str2 is now a subsequence. Therefore, true is returned.
```

**Example 2:**

```
Input: str1 = "zc", str2 = "ad"
Output: true
Explanation: Select indices 0 and 1 in str1.
Increment str1[0] to become 'a'.
Increment str1[1] to become 'd'.
Hence, str1 becomes "ad" and str2 is now a subsequence. Therefore, true is returned.
```

**Example 3:**

```
Input: str1 = "ab", str2 = "d"
Output: false
Explanation: In this example, it can be shown that it is impossible to make str2 a subsequence of str1 using the operation at most once.
Therefore, false is returned.
```

## 结语

如果你同我一样热爱数据结构、算法、LeetCode,可以关注我 GitHub 上的 LeetCode 题解:[awesome-golang-algorithm][me]

[title]: https://leetcode.com/problems/make-string-a-subsequence-using-cyclic-increments
[me]: https://github.com/kylesliu/awesome-golang-algorithm
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package Solution

func Solution(x bool) bool {
return x
func Solution(str1 string, str2 string) bool {
if len(str2) > len(str1) {
return false
}
b := 0
for i := 0; i < len(str1) && b < len(str2); i++ {
t1 := str1[i] - 'a'
t2 := str2[b] - 'a'
if t1 == t2 || t2 == (t1+1+26)%26 {
b++
}
}
return b == len(str2)
}
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Expand Up @@ -10,30 +10,30 @@ func TestSolution(t *testing.T) {
// 测试用例
cases := []struct {
name string
inputs bool
s1, s2 string
expect bool
}{
{"TestCase", true, true},
{"TestCase", true, true},
{"TestCase", false, false},
{"TestCase1", "abc", "ad", true},
{"TestCase2", "zc", "ad", true},
{"TestCase3", "ab", "d", false},
}

// 开始测试
for i, c := range cases {
t.Run(c.name+" "+strconv.Itoa(i), func(t *testing.T) {
got := Solution(c.inputs)
got := Solution(c.s1, c.s2)
if !reflect.DeepEqual(got, c.expect) {
t.Fatalf("expected: %v, but got: %v, with inputs: %v",
c.expect, got, c.inputs)
t.Fatalf("expected: %v, but got: %v, with inputs: %v %v",
c.expect, got, c.s1, c.s2)
}
})
}
}

// 压力测试
// 压力测试
func BenchmarkSolution(b *testing.B) {
}

// 使用案列
// 使用案列
func ExampleSolution() {
}

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