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feat: add solutions to lc problem: No.1674
No.1674.Minimum Moves to Make Array Complementary
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‎solution/1600-1699/1674.Minimum Moves to Make Array Complementary/README.md

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@@ -58,22 +58,169 @@ nums[3] + nums[0] = 3 + 1 = 4.
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：差分数组**
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我们不妨设 $a$ 为 $nums[i]$ 和 $nums[n-i-1]$ 的较小值，设 $b$ 为 $nums[i]$ 和 $nums[n-i-1]$ 的较大值。
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假设经过替换后，两数之和为 $x$。由题意，我们知道 $x$ 最小值为 $2$，即两个数替换为 $1$；最大值为 $2 \times limit$，即两个数都替换为 $limit$。因此 $x$ 的取值范围是 $[2,... 2 \times limit]$。
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如何求出对于不同的 $x$，需要替换的最少次数呢？
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我们分析发现：
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- 如果 $x = a + b$，那么我们需要替换的次数为 $0$，即当前的数对已经满足互补的要求；
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- 否则如果 $1 + a \le x \le limit + b $，那么我们需要替换的次数为 $1$，即把其中一个数替换即可；
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- 否则如果 $2 \le x \le 2 \times limit$，那么我们需要替换的次数为 $2$，即把两个数都替换。
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因此，我们可以遍历每一对数，执行如下操作：
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1. 先将 $[2,... 2 \times limit]$ 范围需要的操作次数加 $2$。
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1. 再将 $[1 + a,... limit + b]$ 范围需要的操作次数减 $1$。
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1. 最后将 $[a + b,... a + b]$ 范围需要的操作次数减 $1$。
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可以发现，这实际上是在对一个连续区间内的元素进行加减操作，因此我们可以使用差分数组来实现。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 `nums` 的长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def minMoves(self, nums: List[int], limit: int) -> int:
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d = [0] * (limit * 2 + 2)
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n = len(nums)
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for i in range(n >> 1):
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a, b = min(nums[i], nums[n - i - 1]), max(nums[i], nums[n - i - 1])
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d[2] += 2
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d[limit * 2 + 1] -= 2
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d[a + 1] -= 1
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d[b + limit + 1] += 1
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d[a + b] -= 1
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d[a + b + 1] += 1
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ans, s = n, 0
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for v in d[2: limit * 2 + 1]:
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s += v
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if ans > s:
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ans = s
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int minMoves(int[] nums, int limit) {
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int n = nums.length;
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int[] d = new int[limit * 2 + 2];
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for (int i = 0; i < n >> 1; ++i) {
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int a = Math.min(nums[i], nums[n - i - 1]);
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int b = Math.max(nums[i], nums[n - i - 1]);
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d[2] += 2;
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d[limit * 2 + 1] -= 2;
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d[a + 1] -= 1;
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d[b + limit + 1] += 1;
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d[a + b] -= 1;
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d[a + b + 1] += 1;
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}
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int ans = n, s = 0;
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for (int i = 2; i <= limit * 2; ++i) {
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s += d[i];
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if (ans > s) {
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ans = s;
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int minMoves(vector<int>& nums, int limit) {
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int n = nums.size();
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vector<int> d(limit * 2 + 2);
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for (int i = 0; i < n >> 1; ++i) {
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int a = min(nums[i], nums[n - i - 1]);
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int b = max(nums[i], nums[n - i - 1]);
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d[2] += 2;
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d[limit * 2 + 1] -= 2;
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d[a + 1] -= 1;
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d[b + limit + 1] += 1;
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d[a + b] -= 1;
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d[a + b + 1] += 1;
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}
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int ans = n, s = 0;
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for (int i = 2; i <= limit * 2; ++i) {
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s += d[i];
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if (ans > s) {
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ans = s;
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func minMoves(nums []int, limit int) int {
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d := make([]int, limit*2+2)
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n := len(nums)
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for i := 0; i < n>>1; i++ {
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a, b := min(nums[i], nums[n-i-1]), max(nums[i], nums[n-i-1])
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d[2] += 2
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d[limit*2+1] -= 2
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d[a+1] -= 1
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d[b+limit+1] += 1
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d[a+b] -= 1
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d[a+b+1] += 1
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}
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ans, s := n, 0
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for _, v := range d[2 : limit*2+1] {
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s += v
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if ans > s {
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ans = s
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}
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/1600-1699/1674.Minimum Moves to Make Array Complementary/README_EN.md

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@@ -57,13 +57,136 @@ Therefore, nums[i] + nums[n-1-i] = 4 for every i, so nums is complementary.
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### **Python3**
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```python
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class Solution:
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def minMoves(self, nums: List[int], limit: int) -> int:
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d = [0] * (limit * 2 + 2)
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n = len(nums)
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for i in range(n >> 1):
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a, b = min(nums[i], nums[n - i - 1]), max(nums[i], nums[n - i - 1])
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d[2] += 2
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d[limit * 2 + 1] -= 2
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d[a + 1] -= 1
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d[b + limit + 1] += 1
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d[a + b] -= 1
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d[a + b + 1] += 1
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ans, s = n, 0
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for v in d[2: limit * 2 + 1]:
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s += v
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if ans > s:
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ans = s
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int minMoves(int[] nums, int limit) {
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int n = nums.length;
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int[] d = new int[limit * 2 + 2];
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for (int i = 0; i < n >> 1; ++i) {
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int a = Math.min(nums[i], nums[n - i - 1]);
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int b = Math.max(nums[i], nums[n - i - 1]);
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d[2] += 2;
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d[limit * 2 + 1] -= 2;
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d[a + 1] -= 1;
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d[b + limit + 1] += 1;
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d[a + b] -= 1;
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d[a + b + 1] += 1;
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}
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int ans = n, s = 0;
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for (int i = 2; i <= limit * 2; ++i) {
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s += d[i];
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if (ans > s) {
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ans = s;
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int minMoves(vector<int>& nums, int limit) {
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int n = nums.size();
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vector<int> d(limit * 2 + 2);
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for (int i = 0; i < n >> 1; ++i) {
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int a = min(nums[i], nums[n - i - 1]);
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int b = max(nums[i], nums[n - i - 1]);
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d[2] += 2;
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d[limit * 2 + 1] -= 2;
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d[a + 1] -= 1;
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d[b + limit + 1] += 1;
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d[a + b] -= 1;
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d[a + b + 1] += 1;
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}
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int ans = n, s = 0;
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for (int i = 2; i <= limit * 2; ++i) {
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s += d[i];
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if (ans > s) {
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ans = s;
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func minMoves(nums []int, limit int) int {
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d := make([]int, limit*2+2)
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n := len(nums)
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for i := 0; i < n>>1; i++ {
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a, b := min(nums[i], nums[n-i-1]), max(nums[i], nums[n-i-1])
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d[2] += 2
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d[limit*2+1] -= 2
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d[a+1] -= 1
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d[b+limit+1] += 1
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d[a+b] -= 1
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d[a+b+1] += 1
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}
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ans, s := n, 0
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for _, v := range d[2 : limit*2+1] {
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s += v
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if ans > s {
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ans = s
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}
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/1600-1699/1674.Minimum Moves to Make Array Complementary/Solution.cpp

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class Solution {
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public:
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int minMoves(vector<int>& nums, int limit) {
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int n = nums.size();
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vector<int> d(limit * 2 + 2);
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for (int i = 0; i < n >> 1; ++i) {
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int a = min(nums[i], nums[n - i - 1]);
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int b = max(nums[i], nums[n - i - 1]);
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d[2] += 2;
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d[limit * 2 + 1] -= 2;
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d[a + 1] -= 1;
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d[b + limit + 1] += 1;
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d[a + b] -= 1;
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d[a + b + 1] += 1;
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}
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int ans = n, s = 0;
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for (int i = 2; i <= limit * 2; ++i) {
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s += d[i];
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if (ans > s) {
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ans = s;
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}
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}
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return ans;
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}
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};

‎solution/1600-1699/1674.Minimum Moves to Make Array Complementary/Solution.go

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func minMoves(nums []int, limit int) int {
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d := make([]int, limit*2+2)
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n := len(nums)
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for i := 0; i < n>>1; i++ {
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a, b := min(nums[i], nums[n-i-1]), max(nums[i], nums[n-i-1])
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d[2] += 2
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d[limit*2+1] -= 2
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d[a+1] -= 1
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d[b+limit+1] += 1
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d[a+b] -= 1
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d[a+b+1] += 1
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}
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ans, s := n, 0
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for _, v := range d[2 : limit*2+1] {
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s += v
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if ans > s {
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ans = s
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}
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}

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