Add solution 1674、1690

leetcode/1674.Minimum-Moves-to-Make-Array-Complementary/1674. Minimum Moves to Make Array Complementary.go

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+package leetcode
+
+func minMoves(nums []int, limit int) int {
+	diff := make([]int, limit*2+2) // nums[i] <= limit, b+limit+1 is maximum limit+limit+1
+	for j := 0; j < len(nums)/2; j++ {
+		a, b := min(nums[j], nums[len(nums)-j-1]), max(nums[j], nums[len(nums)-j-1])
+		// using prefix sum: most interesting point, and is the key to reduce complexity
+		diff[2] += 2
+		diff[a+1]--
+		diff[a+b]--
+		diff[a+b+1]++
+		diff[b+limit+1]++
+	}
+	cur, res := 0, len(nums)
+	for i := 2; i <= 2*limit; i++ {
+		cur += diff[i]
+		res = min(res, cur)
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}

leetcode/1674.Minimum-Moves-to-Make-Array-Complementary/1674. Minimum Moves to Make Array Complementary_test.go

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+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1674 struct {
+	para1674
+	ans1674
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1674 struct {
+	nums  []int
+	limit int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1674 struct {
+	one int
+}
+
+func Test_Problem1674(t *testing.T) {
+
+	qs := []question1674{
+
+		{
+			para1674{[]int{1, 2, 4, 3}, 4},
+			ans1674{1},
+		},
+
+		{
+			para1674{[]int{1, 2, 2, 1}, 2},
+			ans1674{2},
+		},
+
+		{
+			para1674{[]int{1, 2, 1, 2}, 2},
+			ans1674{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1674------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1674, q.para1674
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minMoves(p.nums, p.limit))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/1674.Minimum-Moves-to-Make-Array-Complementary/README.md

@@ -0,0 +1,97 @@
+# [1674. Minimum Moves to Make Array Complementary](https://leetcode.com/problems/minimum-moves-to-make-array-complementary/)
+
+## 题目
+
+You are given an integer array `nums` of **even** length `n` and an integer `limit`. In one move, you can replace any integer from `nums` with another integer between `1` and `limit`, inclusive.
+
+The array `nums` is **complementary** if for all indices `i` (**0-indexed**), `nums[i] + nums[n - 1 - i]` equals the same number. For example, the array `[1,2,3,4]` is complementary because for all indices `i`, `nums[i] + nums[n - 1 - i] = 5`.
+
+Return the ***minimum** number of moves required to make* `nums` ***complementary***.
+
+**Example 1:**
+
+```
+Input: nums = [1,2,4,3], limit = 4
+Output: 1
+Explanation: In 1 move, you can change nums to [1,2,2,3] (underlined elements are changed).
+nums[0] + nums[3] = 1 + 3 = 4.
+nums[1] + nums[2] = 2 + 2 = 4.
+nums[2] + nums[1] = 2 + 2 = 4.
+nums[3] + nums[0] = 3 + 1 = 4.
+Therefore, nums[i] + nums[n-1-i] = 4 for every i, so nums is complementary.
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,2,2,1], limit = 2
+Output: 2
+Explanation: In 2 moves, you can change nums to [2,2,2,2]. You cannot change any number to 3 since 3 > limit.
+```
+
+**Example 3:**
+
+```
+Input: nums = [1,2,1,2], limit = 2
+Output: 0
+Explanation: nums is already complementary.
+```
+
+**Constraints:**
+
+- `n == nums.length`
+- `2 <= n <= 105`
+- `1 <= nums[i] <= limit <= 105`
+- `n` is even.
+
+## 题目大意
+
+给你一个长度为 偶数 n 的整数数组 nums 和一个整数 limit 。每一次操作，你可以将 nums 中的任何整数替换为 1 到 limit 之间的另一个整数。
+
+如果对于所有下标 i（下标从 0 开始），nums[i] + nums[n - 1 - i] 都等于同一个数，则数组 nums 是 互补的 。例如，数组 [1,2,3,4] 是互补的，因为对于所有下标 i ，nums[i] + nums[n - 1 - i] = 5 。
+
+返回使数组 互补 的 最少 操作次数。
+
+## 解题思路
+
+- 这一题考察的是差分数组。通过分析题意，可以得出，针对每一个 `sum` 的取值范围是 `[2, 2* limt]`，定义 `a = min(nums[i], nums[n - i - 1])`，`b = max(nums[i], nums[n - i - 1])`，在这个区间内，又可以细分成 5 个区间，`[2, a + 1)`，`[a + 1, a + b)`，`[a + b + 1, a + b + 1)`，`[a + b + 1, b + limit + 1)`，`[b + limit + 1, 2 * limit)`，在这 5 个区间内使得数组互补的最小操作次数分别是 `2(减少 a, 减少 b)`，`1(减少 b)`，`0(不用操作)`，`1(增大 a)`，`+2(增大 a, 增大 b)`，换个表达方式，按照扫描线从左往右扫描，在这 5 个区间内使得数组互补的最小操作次数叠加变化分别是 `+2(减少 a, 减少 b)`，`-1(减少 a)`，`-1(不用操作)`，`+1(增大 a)`，`+1(增大 a, 增大 b)`，利用这前后两个区间的关系，就可以构造一个差分数组。差分数组反应的是前后两者的关系。如果想求得 0 ~ n 的总关系，只需要求一次前缀和即可。
+- 这道题要求输出最少的操作次数，所以利用差分数组 + 前缀和，累加前缀和的同时维护最小值。从左往右扫描完一遍以后，输出最小值即可。
+
+## 代码
+
+```go
+package leetcode
+
+func minMoves(nums []int, limit int) int {
+	diff := make([]int, limit*2+2) // nums[i] <= limit, b+limit+1 is maximum limit+limit+1
+	for j := 0; j < len(nums)/2; j++ {
+		a, b := min(nums[j], nums[len(nums)-j-1]), max(nums[j], nums[len(nums)-j-1])
+		// using prefix sum: most interesting point, and is the key to reduce complexity
+		diff[2] += 2
+		diff[a+1]--
+		diff[a+b]--
+		diff[a+b+1]++
+		diff[b+limit+1]++
+	}
+	cur, res := 0, len(nums)
+	for i := 2; i <= 2*limit; i++ {
+		cur += diff[i]
+		res = min(res, cur)
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```

leetcode/1690.Stone-Game-VII/1690. Stone Game VII.go

@@ -0,0 +1,65 @@
+package leetcode
+
+// 解法一 优化空间版 DP
+func stoneGameVII(stones []int) int {
+	n := len(stones)
+	sum := make([]int, n)
+	dp := make([]int, n)
+	for i, d := range stones {
+		sum[i] = d
+	}
+	for i := 1; i < n; i++ {
+		for j := 0; j+i < n; j++ {
+			if (n-i)%2 == 1 {
+				d0 := dp[j] + sum[j]
+				d1 := dp[j+1] + sum[j+1]
+				if d0 > d1 {
+					dp[j] = d0
+				} else {
+					dp[j] = d1
+				}
+			} else {
+				d0 := dp[j] - sum[j]
+				d1 := dp[j+1] - sum[j+1]
+				if d0 < d1 {
+					dp[j] = d0
+				} else {
+					dp[j] = d1
+				}
+			}
+			sum[j] = sum[j] + stones[i+j]
+		}
+	}
+	return dp[0]
+}
+
+// 解法二 常规 DP
+func stoneGameVII1(stones []int) int {
+	prefixSum := make([]int, len(stones))
+	for i := 0; i < len(stones); i++ {
+		if i == 0 {
+			prefixSum[i] = stones[i]
+		} else {
+			prefixSum[i] = prefixSum[i-1] + stones[i]
+		}
+	}
+	dp := make([][]int, len(stones))
+	for i := range dp {
+		dp[i] = make([]int, len(stones))
+		dp[i][i] = 0
+	}
+	n := len(stones)
+	for l := 2; l <= n; l++ {
+		for i := 0; i+l <= n; i++ {
+			dp[i][i+l-1] = max(prefixSum[i+l-1]-prefixSum[i+1]+stones[i+1]-dp[i+1][i+l-1], prefixSum[i+l-2]-prefixSum[i]+stones[i]-dp[i][i+l-2])
+		}
+	}
+	return dp[0][n-1]
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}

leetcode/1690.Stone-Game-VII/1690. Stone Game VII_test.go

@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1690 struct {
+	para1690
+	ans1690
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1690 struct {
+	stones []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1690 struct {
+	one int
+}
+
+func Test_Problem1690(t *testing.T) {
+
+	qs := []question1690{
+
+		{
+			para1690{[]int{5, 3, 1, 4, 2}},
+			ans1690{6},
+		},
+
+		{
+			para1690{[]int{7, 90, 5, 1, 100, 10, 10, 2}},
+			ans1690{122},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1690------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1690, q.para1690
+		fmt.Printf("【input】:%v       【output】:%v\n", p, stoneGameVII(p.stones))
+	}
+	fmt.Printf("\n\n\n")
+}