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Commit 35a05e0

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Merge pull request #388 from 0xff-dev/1962
Add solution and test-cases for problem 1962
2 parents fea22f8 + 276e571 commit 35a05e0

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‎leetcode/1901-2000/1962.Remove-Stones-to-Minimize-the-Total/README.md‎

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Original file line numberDiff line numberDiff line change
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# [1962.Remove Stones to Minimize the Total][title]
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> [!WARNING|style:flat]
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> This question is temporarily unanswered if you have good ideas. Welcome to [Create Pull Request PR](https://github.com/kylesliu/awesome-golang-algorithm)
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## Description
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You are given a **0-indexed** integer array `piles`, where `piles[i]` represents the number of stones in the i<sup>th</sup> pile, and an integer `k`. You should apply the following operation **exactly** `k` times:
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- Choose any `piles[i]` and **remove** `floor(piles[i] / 2)` stones from it.
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**Notice** that you can apply the operation on the **same** pile more than once.
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Return the **minimum** possible total number of stones remaining after applying the `k` operations.
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`floor(x)` is the **greatest** integer that is **smaller** than or **equal** to `x` (i.e., rounds `x` down).
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**Example 1:**
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```
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Input: a = "11", b = "1"
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Output: "100"
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Input: piles = [5,4,9], k = 2
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Output: 12
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Explanation: Steps of a possible scenario are:
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- Apply the operation on pile 2. The resulting piles are [5,4,5].
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- Apply the operation on pile 0. The resulting piles are [3,4,5].
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The total number of stones in [3,4,5] is 12.
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```
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## 题意
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> ...
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## 题解
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**Example 2:**
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### 思路1
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> ...
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Remove Stones to Minimize the Total
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```go
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```
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Input: piles = [4,3,6,7], k = 3
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Output: 12
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Explanation: Steps of a possible scenario are:
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- Apply the operation on pile 2. The resulting piles are [4,3,3,7].
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- Apply the operation on pile 3. The resulting piles are [4,3,3,4].
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- Apply the operation on pile 0. The resulting piles are [2,3,3,4].
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The total number of stones in [2,3,3,4] is 12.
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```
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## 结语
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package Solution
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func Solution(x bool) bool {
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import (
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"container/heap"
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)
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type Piles []int
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func (p *Piles) Len() int {
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return len(*p)
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}
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func (p *Piles) Less(i, j int) bool {
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a := (*p)[i]
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b := (*p)[j]
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if a&1 == 1 {
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a++
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}
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if b&1 == 1 {
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b++
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}
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return (*p)[i]-a/2 > (*p)[j]-b/2
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}
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func (p *Piles) Swap(i, j int) {
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(*p)[i], (*p)[j] = (*p)[j], (*p)[i]
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}
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func (p *Piles) Push(x interface{}) {
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*p = append(*p, x.(int))
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}
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func (p *Piles) Pop() interface{} {
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old := *p
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l := len(old)
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x := old[l-1]
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*p = old[:l-1]
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return x
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}
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func Solution(piles []int, k int) int {
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sum := 0
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for _, p := range piles {
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sum += p
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}
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h := Piles(piles)
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heap.Init(&h)
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for ; k > 0 && h.Len() > 0; k-- {
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x := heap.Pop(&h).(int)
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y := x
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if x&1 == 1 {
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x--
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}
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sum -= x / 2
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heap.Push(&h, y-x/2)
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}
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return sum
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}

‎leetcode/1901-2000/1962.Remove-Stones-to-Minimize-the-Total/Solution_test.go‎

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@@ -10,30 +10,31 @@ func TestSolution(t *testing.T) {
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// 测试用例
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cases := []struct {
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name string
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inputs bool
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expect bool
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inputs []int
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k int
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expect int
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}{
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{"TestCase", true, true},
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{"TestCase", true, true},
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{"TestCase", false, false},
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{"TestCase1", []int{5, 4, 9}, 2, 12},
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{"TestCase2", []int{4, 3, 6, 7}, 3, 12},
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{"TestCase3", []int{10, 88, 74, 63, 28, 99, 131, 74, 97, 201, 383, 274}, 6, 865},
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}
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// 开始测试
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for i, c := range cases {
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t.Run(c.name+" "+strconv.Itoa(i), func(t *testing.T) {
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got := Solution(c.inputs)
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got := Solution(c.inputs, c.k)
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if !reflect.DeepEqual(got, c.expect) {
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t.Fatalf("expected: %v, but got: %v, with inputs: %v",
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c.expect, got, c.inputs)
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t.Fatalf("expected: %v, but got: %v, with inputs: %v %v",
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c.expect, got, c.inputs, c.k)
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}
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})
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}
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}
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//压力测试
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//压力测试
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func BenchmarkSolution(b *testing.B) {
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}
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//使用案列
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//使用案列
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func ExampleSolution() {
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}

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