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Commit c45c772

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Merge pull request #401 from 0xff-dev/1856
Add solution and test-cases for problem 1856
2 parents 75ce266 + dcd6d86 commit c45c772

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‎leetcode/1801-1900/1856.Maximum-Subarray-Min-Product/README.md

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# [1856.Maximum Subarray Min-Product][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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The **min-product** of an array is equal to the **minimum value** in the array **multiplied by** the array's **sum**.
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- For example, the array `[3,2,5]` (minimum value is `2`) has a min-product of `2 * (3+2+5) = 2 * 10 = 20`.
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Given an array of integers `nums`, return the **maximum min-product** of any **non-empty subarray** of `nums`. Since the answer may be large, return it **modulo** 10<sup>9</sup> + 7.
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Note that the min-product should be maximized **before** performing the modulo operation. Testcases are generated such that the maximum min-product **without** modulo will fit in a **64-bit signed integer**.
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A **subarray** is a **contiguous** part of an array.
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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: nums = [1,2,3,2]
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Output: 14
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Explanation: The maximum min-product is achieved with the subarray [2,3,2] (minimum value is 2).
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2 * (2+3+2) = 2 * 7 = 14.
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```
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## 题意
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> ...
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**Example 2:**
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## 题解
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### 思路1
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> ...
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Maximum Subarray Min-Product
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```go
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```
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Input: nums = [2,3,3,1,2]
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Output: 18
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Explanation: The maximum min-product is achieved with the subarray [3,3] (minimum value is 3).
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3 * (3+3) = 3 * 6 = 18.
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```
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**Example 3:**
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```
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Input: nums = [3,1,5,6,4,2]
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Output: 60
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Explanation: The maximum min-product is achieved with the subarray [5,6,4] (minimum value is 4).
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4 * (5+6+4) = 4 * 15 = 60.
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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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return x
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const mod1856 = 1000000007
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func Solution(nums []int) int {
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l := len(nums)
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sum := make([]int64, l)
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sum[0] = int64(nums[0])
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for i := 1; i < l; i++ {
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sum[i] = int64(nums[i]) + sum[i-1]
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}
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nextSmaller := make([]int, l)
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prevSmaleer := make([]int, l)
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stack := make([]int, l)
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for i := 0; i < l; i++ {
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nextSmaller[i] = l
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prevSmaleer[i] = -1
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}
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stackIdx := -1
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for i := 0; i < l; i++ {
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for stackIdx >= 0 && nums[stack[stackIdx]] > nums[i] {
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nextSmaller[stack[stackIdx]] = i
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stackIdx--
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}
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stackIdx++
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stack[stackIdx] = i
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}
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stackIdx = -1
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for i := l - 1; i >= 0; i-- {
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for stackIdx >= 0 && nums[stack[stackIdx]] >= nums[i] {
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prevSmaleer[stack[stackIdx]] = i
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stackIdx--
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}
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stackIdx++
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stack[stackIdx] = i
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}
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n := int64(0)
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for i := 0; i < l; i++ {
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right := nextSmaller[i] - 1 // lastidx
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left := prevSmaleer[i] //
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s := sum[right]
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if left != -1 {
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s -= sum[left]
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}
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tmp := int64(nums[i]) * s
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if n == -1 || n < tmp {
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n = tmp
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}
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}
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return int(n % mod1856)
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}

‎leetcode/1801-1900/1856.Maximum-Subarray-Min-Product/Solution_test.go

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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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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{2, 3, 3, 1, 2}, 18},
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{"TestCase2", []int{3, 1, 5, 6, 4, 2}, 60},
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{"TestCase3", []int{2, 5, 4, 2, 4, 5, 3, 1, 2}, 50},
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}
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// 开始测试
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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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