Add solution 1656、1657、1658、1659

Author: halfrost

leetcode/1658.Minimum-Operations-to-Reduce-X-to-Zero/1658. Minimum Operations to Reduce X to Zero.go

@@ -0,0 +1,40 @@
+package leetcode
+
+func minOperations(nums []int, x int) int {
+	total := 0
+	for _, n := range nums {
+		total += n
+	}
+	target := total - x
+	if target < 0 {
+		return -1
+	}
+	if target == 0 {
+		return len(nums)
+	}
+	left, right, sum, res := 0, 0, 0, -1
+	for right < len(nums) {
+		if sum < target {
+			sum += nums[right]
+			right++
+		}
+		for sum >= target {
+			if sum == target {
+				res = max(res, right-left)
+			}
+			sum -= nums[left]
+			left++
+		}
+	}
+	if res == -1 {
+		return -1
+	}
+	return len(nums) - res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}

leetcode/1658.Minimum-Operations-to-Reduce-X-to-Zero/1658. Minimum Operations to Reduce X to Zero_test.go

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+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1658 struct {
+	para1658
+	ans1658
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1658 struct {
+	nums []int
+	x    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1658 struct {
+	one int
+}
+
+func Test_Problem1658(t *testing.T) {
+
+	qs := []question1658{
+
+		{
+			para1658{[]int{1, 1, 4, 2, 3}, 5},
+			ans1658{2},
+		},
+
+		{
+			para1658{[]int{5, 6, 7, 8, 9}, 4},
+			ans1658{-1},
+		},
+
+		{
+			para1658{[]int{3, 2, 20, 1, 1, 3}, 10},
+			ans1658{5},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1658------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1658, q.para1658
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, minOperations(p.nums, p.x))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/1658.Minimum-Operations-to-Reduce-X-to-Zero/README.md

@@ -0,0 +1,96 @@
+# [1658. Minimum Operations to Reduce X to Zero](https://leetcode.com/problems/minimum-operations-to-reduce-x-to-zero/)
+
+
+## 题目
+
+You are given an integer array `nums` and an integer `x`. In one operation, you can either remove the leftmost or the rightmost element from the array `nums` and subtract its value from `x`. Note that this **modifies** the array for future operations.
+
+Return *the **minimum number** of operations to reduce* `x` *to **exactly*** `0` *if it's possible, otherwise, return* `1`.
+
+**Example 1:**
+
+```
+Input: nums = [1,1,4,2,3], x = 5
+Output: 2
+Explanation: The optimal solution is to remove the last two elements to reduce x to zero.
+
+```
+
+**Example 2:**
+
+```
+Input: nums = [5,6,7,8,9], x = 4
+Output: -1
+
+```
+
+**Example 3:**
+
+```
+Input: nums = [3,2,20,1,1,3], x = 10
+Output: 5
+Explanation: The optimal solution is to remove the last three elements and the first two elements (5 operations in total) to reduce x to zero.
+
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 105`
+- `1 <= nums[i] <= 104`
+- `1 <= x <= 109`
+
+## 题目大意
+
+给你一个整数数组 nums 和一个整数 x 。每一次操作时，你应当移除数组 nums 最左边或最右边的元素，然后从 x 中减去该元素的值。请注意，需要 修改 数组以供接下来的操作使用。如果可以将 x 恰好 减到 0 ，返回 最小操作数 ；否则，返回 -1 。
+
+## 解题思路
+
+- 给定一个数组 nums 和一个整数 x，要求从数组两端分别移除一些数，使得这些数加起来正好等于整数 x，要求输出最小操作数。
+- 要求输出最小操作数，即数组两头的数字个数最少，并且加起来和正好等于整数 x。由于在数组的两头，用 2 个指针分别操作不太方便。我当时解题的时候的思路是把它变成循环数组，这样两边的指针就在一个区间内了。利用滑动窗口找到一个最小的窗口，使得窗口内的累加和等于整数 k。这个方法可行，但是代码挺多的。
+- 有没有更优美的方法呢？有的。要想两头的长度最少，也就是中间这段的长度最大。这样就转换成直接在数组上使用滑动窗口求解，累加和等于一个固定值的连续最长的子数组。
+- 和这道题类似思路的题目，209，1040(循环数组)，325。强烈推荐这 3 题。
+
+## 代码
+
+```go
+package leetcode
+
+func minOperations(nums []int, x int) int {
+	total := 0
+	for _, n := range nums {
+		total += n
+	}
+	target := total - x
+	if target < 0 {
+		return -1
+	}
+	if target == 0 {
+		return len(nums)
+	}
+	left, right, sum, res := 0, 0, 0, -1
+	for right < len(nums) {
+		if sum < target {
+			sum += nums[right]
+			right++
+		}
+		for sum >= target {
+			if sum == target {
+				res = max(res, right-left)
+			}
+			sum -= nums[left]
+			left++
+		}
+	}
+	if res == -1 {
+		return -1
+	}
+	return len(nums) - res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```

leetcode/1659.Maximize-Grid-Happiness/1659. Maximize Grid Happiness.go

@@ -0,0 +1,107 @@
+package leetcode
+
+import (
+	"math"
+)
+
+func getMaxGridHappiness(m int, n int, introvertsCount int, extrovertsCount int) int {
+	// lineStatus 					将每一行中 3 种状态进行编码，空白 - 0，内向人 - 1，外向人 - 2，每行状态用三进制表示
+	// lineStatusList[729][6] 		每一行的三进制表示
+	// introvertsCountInner[729] 	每一个 lineStatus 包含的内向人数
+	// extrovertsCountInner[729] 	每一个 lineStatus 包含的外向人数
+	// scoreInner[729] 			 	每一个 lineStatus 包含的行内得分（只统计 lineStatus 本身的得分，不包括它与上一行的）
+	// scoreOuter[729][729] 	 	每一个 lineStatus 包含的行外得分
+	// dp[上一行的 lineStatus][当前处理到的行][剩余的内向人数][剩余的外向人数]
+	n3, lineStatus, introvertsCountInner, extrovertsCountInner, scoreInner, scoreOuter, lineStatusList, dp := math.Pow(3.0, float64(n)), 0, [729]int{}, [729]int{}, [729]int{}, [729][729]int{}, [729][6]int{}, [729][6][7][7]int{}
+	for i := 0; i < 729; i++ {
+		lineStatusList[i] = [6]int{}
+	}
+	for i := 0; i < 729; i++ {
+		dp[i] = [6][7][7]int{}
+		for j := 0; j < 6; j++ {
+			dp[i][j] = [7][7]int{}
+			for k := 0; k < 7; k++ {
+				dp[i][j][k] = [7]int{-1, -1, -1, -1, -1, -1, -1}
+			}
+		}
+	}
+	// 预处理
+	for lineStatus = 0; lineStatus < int(n3); lineStatus++ {
+		tmp := lineStatus
+		for i := 0; i < n; i++ {
+			lineStatusList[lineStatus][i] = tmp % 3
+			tmp /= 3
+		}
+		introvertsCountInner[lineStatus], extrovertsCountInner[lineStatus], scoreInner[lineStatus] = 0, 0, 0
+		for i := 0; i < n; i++ {
+			if lineStatusList[lineStatus][i] != 0 {
+				// 个人分数
+				if lineStatusList[lineStatus][i] == 1 {
+					introvertsCountInner[lineStatus]++
+					scoreInner[lineStatus] += 120
+				} else if lineStatusList[lineStatus][i] == 2 {
+					extrovertsCountInner[lineStatus]++
+					scoreInner[lineStatus] += 40
+				}
+				// 行内分数
+				if i-1 >= 0 {
+					scoreInner[lineStatus] += closeScore(lineStatusList[lineStatus][i], lineStatusList[lineStatus][i-1])
+				}
+			}
+		}
+	}
+	// 行外分数
+	for lineStatus0 := 0; lineStatus0 < int(n3); lineStatus0++ {
+		for lineStatus1 := 0; lineStatus1 < int(n3); lineStatus1++ {
+			scoreOuter[lineStatus0][lineStatus1] = 0
+			for i := 0; i < n; i++ {
+				scoreOuter[lineStatus0][lineStatus1] += closeScore(lineStatusList[lineStatus0][i], lineStatusList[lineStatus1][i])
+			}
+		}
+	}
+	return dfs(0, 0, introvertsCount, extrovertsCount, m, int(n3), &dp, &introvertsCountInner, &extrovertsCountInner, &scoreInner, &scoreOuter)
+}
+
+// 如果 x 和 y 相邻，需要加上的分数
+func closeScore(x, y int) int {
+	if x == 0 || y == 0 {
+		return 0
+	}
+	// 两个内向的人，每个人要 -30，一共 -60
+	if x == 1 && y == 1 {
+		return -60
+	}
+	if x == 2 && y == 2 {
+		return 40
+	}
+	return -10
+}
+
+// dfs(上一行的 lineStatus，当前处理到的行，剩余的内向人数，剩余的外向人数）
+func dfs(lineStatusLast, row, introvertsCount, extrovertsCount, m, n3 int, dp *[729][6][7][7]int, introvertsCountInner, extrovertsCountInner, scoreInner *[729]int, scoreOuter *[729][729]int) int {
+	// 边界条件：如果已经处理完，或者没有人了
+	if row == m || introvertsCount+extrovertsCount == 0 {
+		return 0
+	}
+	// 记忆化
+	if dp[lineStatusLast][row][introvertsCount][extrovertsCount] != -1 {
+		return dp[lineStatusLast][row][introvertsCount][extrovertsCount]
+	}
+	best := 0
+	for lineStatus := 0; lineStatus < n3; lineStatus++ {
+		if introvertsCountInner[lineStatus] > introvertsCount || extrovertsCountInner[lineStatus] > extrovertsCount {
+			continue
+		}
+		score := scoreInner[lineStatus] + scoreOuter[lineStatus][lineStatusLast]
+		best = max(best, score+dfs(lineStatus, row+1, introvertsCount-introvertsCountInner[lineStatus], extrovertsCount-extrovertsCountInner[lineStatus], m, n3, dp, introvertsCountInner, extrovertsCountInner, scoreInner, scoreOuter))
+	}
+	dp[lineStatusLast][row][introvertsCount][extrovertsCount] = best
+	return best
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}

leetcode/1659.Maximize-Grid-Happiness/1659. Maximize Grid Happiness_test.go

@@ -0,0 +1,55 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1659 struct {
+	para1659
+	ans1659
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1659 struct {
+	m               int
+	n               int
+	introvertsCount int
+	extrovertsCount int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1659 struct {
+	one int
+}
+
+func Test_Problem1659(t *testing.T) {
+
+	qs := []question1659{
+
+		{
+			para1659{2, 3, 1, 2},
+			ans1659{240},
+		},
+
+		{
+			para1659{3, 1, 2, 1},
+			ans1659{260},
+		},
+
+		{
+			para1659{2, 2, 4, 0},
+			ans1659{240},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1659------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1659, q.para1659
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, getMaxGridHappiness(p.m, p.n, p.introvertsCount, p.extrovertsCount))
+	}
+	fmt.Printf("\n\n\n")
+}