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Add solution and test-cases for problem 2516
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0xff-dev committed Nov 20, 2024
1 parent dbd4dda commit 0c04504
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# [2516.Take K of Each Character From Left and Right][title]

## Description
You are given a string `s` consisting of the characters `'a'`, `'b'`, and `'c'` and a non-negative integer `k`. Each minute, you may take either the **leftmost** character of `s`, or the **rightmost** character of `s`.

Return the **minimum** number of minutes needed for you to take **at least** `k` of each character, or return `-1` if it is not possible to take `k` of each character.

**Example 1:**

```
Input: s = "aabaaaacaabc", k = 2
Output: 8
Explanation:
Take three characters from the left of s. You now have two 'a' characters, and one 'b' character.
Take five characters from the right of s. You now have four 'a' characters, two 'b' characters, and two 'c' characters.
A total of 3 + 5 = 8 minutes is needed.
It can be proven that 8 is the minimum number of minutes needed.
```

**Example 2:**

```
Input: s = "a", k = 1
Output: -1
Explanation: It is not possible to take one 'b' or 'c' so return -1.
```

## 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/take-k-of-each-character-from-left-and-right
[me]: https://github.com/kylesliu/awesome-golang-algorithm
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package Solution

func Solution(x bool) bool {
return x
import "sort"

func Solution(s string, k int) int {
if k == 0 {
return 0
}
l := len(s)
left, right := make([][3]int, l), make([][3]int, l)
left[0][s[0]-'a'] = 1
for i := 1; i < l; i++ {
left[i] = left[i-1]
left[i][s[i]-'a']++
}
right[l-1][s[l-1]-'a'] = 1
for i := l - 2; i >= 0; i-- {
right[i] = right[i+1]
right[i][s[i]-'a']++
}
ans := -1
for i := l - 1; i > 0; i-- {
a, b, c := right[i][0], right[i][1], right[i][2]
if a >= k && b >= k && c >= k {
if ans == -1 || ans > l-i {
ans = l - i
}
continue
}
idx := sort.Search(i, func(ii int) bool {
return left[ii][0]+a >= k && left[ii][1]+b >= k && left[ii][2]+c >= k
})
if idx == i {
continue
}
diff := l - i + idx + 1
if ans == -1 || ans > diff {
ans = diff
}
}

for i := 0; i < l-1; i++ {
a, b, c := left[i][0], left[i][1], left[i][2]
if a >= k && b >= k && c >= k {
if ans == -1 || ans > i+1 {
ans = i + 1
}
continue
}
ll := l - i - 1
start := i + 1
idx := sort.Search(ll, func(ii int) bool {
return right[start][0]+a >= k && right[start][1]+b >= k && right[start][2]+c >= k
})
if idx == ll {
continue
}
diff := i + 1 + l - i
if ans == -1 || ans > diff {
ans = diff
}
}
return ans
}
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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
expect bool
s string
k int
expect int
}{
{"TestCase", true, true},
{"TestCase", true, true},
{"TestCase", false, false},
{"TestCase1", "aabaaaacaabc", 2, 8},
{"TestCase2", "a", 1, -1},
}

// 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.s, c.k)
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.s, c.k)
}
})
}
}

// 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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